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Chapter = The Analytical Process
You should read this chapter.
Chapter 1= Chemical Measurements
Section 1-1: SI Units and Prefixes
You should know this material.
Section 1-2: Conversion Between Units
You should know this material.
Section 1-3: Chemical Concentrations
Solution = solute + solvent
Solute = minor species in solution.
Solvent = major species in solution.
Most common solvent in CHM 3120: water.
Concentration = amount of solute contained in a given volume or mass of
solvent.
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Molarity (M)
M is the number of moles of solute per liter of solution
M = moles of solute / liters of solution
moles of solute = weight of solute (g)/formula weight of solute (g)
or
moles of solute = mass of solute (g)/formula mass of solute (g)
or
moles of solute = mass of solute (g)/molecular mass of solute (g)
Example: How many grams of Boric Acid [B(OH)3, FM 61.83] should be used to make
2.00 L of 0.0500M solution?
moles of solute = 2.00 L x 0.0500M = 0.1 mol
mass of solute = moles of solute x formula mass
mass of solute = 0.1 mol x 61.83 g = 6.183 g
Electrolyte
An electrolyte dissociates into ions in aqueous solution.
A strong electrolyte dissociates almost completely.
Example:
MgCl2 Mg+ + MgCl+
After dissociation, 89% exists in the form of Mg2+ and 11% in the form of MgCl+
The molarity of a strong electrolyte is referred to as Formal Concentration (F) to
indicate that the substance is really converted into other species in solution.
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A weak electrolyte is partially split into
ions in solution.
Example: Acetic Acid, CH3CO2H
Note: If you dissolve 0.01000 mol of acetic
acid in 1.000L, you will have:
a 0.01000 F solution
or
a 0.00959M solution
Why?
because 4.1% is dissociated into
CH3CO2- (acetate ion) and 95.9%
remains as CH3CO2H.
Nonetheless,
we usually say that the solution is
0.01M and understand that some of the
acid is dissociated
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Molality (m)
m is the number of moles of solute per kilogram of solvent (not solution!)
The advantage of molality over molarity is that molality does not change with
temperature.
The molarity of a solution changes with temperature because the volume of solution
changes with temperature.
Percent Composition
It is usually expressed as weight percent (wt%)
wt% = [mass of solute / mass of total solution or mixture] x 100
Converting weight percent into molarity
Example: Find the molarity of HCl in a reagent labeled 37.0 wt% HCl, density =
1.188g/mL.
37.0 wt% = There are 37.0 g of HCl in 100 g of reagent
density = 1.188g/mL = allows us to calculate the volume of 100 g of reagent
density (g/mL) = mass (g) / volume (mL)
volume (mL) = mass (g) / density (g/mL)
volume = 100 g / 1.188 g/mL = 84.1 mL
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We know that:
M = moles of solute / volume of solution in liters
Moles of solute = mass of solute (g) / formula mass of solute (g)
Formula mass of HCl = 36.46 g/mol
Mass of solute = 37 g
Volume of solution = 84.1 mL = 84.18 x 10-3 L
Substituting these values above:
Moles of solute = 37 g / 36.46 g/mol = 1.01 mol
M = 1.01 mol / 84.1 x 10-3 L = 12.01 M
Parts per Million (ppm) and Parts per Billion (ppb)
ppm = [mass of substance / mass of sample] x 106
ppb = [mass of substance / mass of sample] x 109
Note:
masses must be expressed in the same units in the nominator and
denominator
The density of a dilute aqueous solution is close to unity, so:
1g of water ~ 1mL of water 1 ppm = 1g/mL (1 x 10-6 g/mL)
1 ppb = 1ng/mL (1 x 10-9 g/mL)
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Converting Parts per Billion into Molarity
Example: The concentration of C20H42 (FM 282.55) in winter rainwater is
0.2ppb. Assuming that the density of rainwater is close to unity, find the
molar concentration of C20H42.
FYI = C20H42 is an n-alkane (CnH2n+2); its solubility is very low in water
and, therefore, its concentration in water is at the trace level.
0.2 ppb = 0.2 g of C20H42 per 109 g of rain water
or
0.2 x 10-9 g of C20H42 per g of rain water
[0.2 x 10-9 g of C20H42 / 1g of rain water] x 1000/1000
10-9 x 103 g of C20H42 /1000g of rain water
0.2 x 10-6 g of C20H42 / 1000 mL of rain water
0.2 x 10-6 g of C20H42 / 1L of rain water
number of moles = 0.2 x 10-6 g of C20H42 / 282.55 g/mol
0.2 x 10-6 g of C20H42 / 282.55 g/mol x 1L = 7 x 10-10 M
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0.2 x
Section 1-4 Preparing Solutions
Preparing a solution from a solid reagent:
Weigh out the correct mass of pure reagent
with an analytical balance
With the appropriate amount of solvent,
dissolve all the solute in a container
Transfer solution into a volumetric flask of
appropriate volume
Rinse container with small portions of
solvent
Dilute with more solvent to the desired final
volume
Mix well by inverting the flask many times
Note: This procedure is different that the
procedure recommended in the textbook.
This procedure is better because it assures
that all the solid is completely dissolved in
the appropriate volume of solution
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Example:
What are the steps to prepare 250 mL of a 0.1000M NaCl aqueous
solution?
FYI: NaCl is a solid.
1. Calculate the mass of NaCl needed. To calculate the mass, you need to
consider the following:
a. Reagent purity: If it is ACS reagent, purity is 100%
b. FW = 36.5 g/mol
c. Final volume = 250 mL = 0.25 L
d. M = 0.1000 M
M = number of moles of solute / volume of solution in liters
number of moles = mass (g) / formula weight (g/mol)
mass (g) = M x V(L) x FW (g/mol); FW = formula weight
mass = 0.1000 mol/L x 0.25 L x 36.5 g/mol = 0.9125 g of NaCl
2. Follow procedure in page 12
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Preparing solutions by dilution:
Dilute solutions can be prepared from
concentrated solutions
The procedure is the following:
a. A known volume of the concentrated solution
is transferred to a volumetric flask
b. The volumetric flask is filled to the mark with
the appropriate solvent
The molarity of the diluted solution (Mdil)
depends on the number of moles of
concentrated solution transferred to the
volumetric flask and the volume of the volumetric
flask. It is very important to measure the volume
of concentrated solution accurately!
We know that:
M = number of moles of solute / V of solution
(L)
number of moles of solute = Mconc x Vconc
V of solution = V of volumetric flask = Vdil
Mdil = Mconc x Vconc / Vdil
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Mdil x Vdil = Mconc x Vconc
The same is true for normality:
Ndil x Vdil = Nconc x Vconc
Example:
How many mL of concentrated HCl should be diluted to 1.00L to make a
0.100M HCl solution?
FYI: The molarity of concentrated HCl is 12.1M
Mconc = 12.1M
Mdil = 0.100M
Vdil = 1.00L
Mdil x Vdil = Mconc x Vconc Vconc = Mdil x Vdil / Mconc = 0.100 x 1 /
12.1 =
= 8.26 x 10-3 L = 8.26 mL
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Preparing solutions from a liquid reagent:
The procedure is the following:
a. A pipette is used to measure and transfer a known volume of liquid
reagent into the volumetric flask.
Attention: do not introduce the pipette into the reagent vessel!
If your pipette is contaminated, which is possible to happen without
you knowing, you will contaminate the entire reagent!
You should transfer some of the reagent into a beaker and you should
pipette the appropriate volume from the beaker
b. Fill the volumetric flask to the mark with the appropriate solvent
Example of calculations involved:
The density of concentrated ammonium hydroxide, which contains 28.0 wt
%
NH3 is 0.899 g/mL. What volume of this reagent should be diluted to 500
mL to make 0.250 M NH3?
1. We need to know the number of moles in the concentrated reagent
2. 28.0 wt% 28 g of NH3 in 100 g of reagent
3. The molecular weigh of NH3 is 17.03 g/mol 28 g / 17.03 g/mol = 1.64
moles of NH3 in 100 g of reagent
4. Density = 0.899 g/mL volume = 100 g / 0.899 g/mL = 111.23 mL
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There are 1.64 moles of NH3 in 111. 13 mL of reagent
M = 1.64 moles / 0.11113 L = 14.76 M
We want to prepare 500 mL of a 0.250 M solution
Mconc = 14.76 M
Vconc = ?
Mdil = 0.250 M
Vdil = 500 mL = 0.5 L
We know that: Mconc x Vconc = Mdil x Vdil
Vconc = 0.250 mol/L x 0.5 L / 14.76 mol/L = 8.45 x 10-3 L = 8.45 mL
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Section 1-5 The Equilibrium Constant
Consider the following reversible reaction:
aA + bB <=> cC + dD
The reaction reaches equilibrium when [A] / t = constant, [B] / t =
constant, [C] / t = constant and [D] / t = constant
When equilibrium is reached:
velocity of forward reaction (v1) = velocity of reverse reaction (v2)
We know that:
v1 = k1 [A]a[B]b, where k1 is the rate constant in the forward reaction
v2 = k2 [C]c[D]d, where k1 is the rate constant in the reverse reaction
At equilibrium:
k1 [A]a[B]b = k2 [C]c[D]d k1 / k2 = [C]c[D]d / [A]a[B]b
The ratio between the two rate constants is the equilibrium constant:
Keq = [C]c[D]d / [A]a[B]b
where a, b, c and d are stoichiometric coefficients
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You should remember the following:
1. The concentrations of solutes should be expressed as moles per liter
2. The concentrations of gases should be expressed as partial pressures
(units = bars)
3. The concentrations of pure solids, pure liquids, and solvents are omitted
because they are unity
FYI:
In chemical reactions: s = solid, aq = aqueous, g = gas, and l = liquid
Examples:
a. 3Ag+ (aq) + PO43- (aq) <=> Ag3PO4 (s)
K = 1 / [Ag+]3 [PO43-]
b. C6H6 (l) + 15/2 O2 (g) <=> 3H2O (l) + 6CO2 (g)
K = (pCO2)6 / (pO2)15/2
The value of K gives an indication of which reaction is favored in the
equilibrium:
K > 1 the forward reaction is favored [products] > [reactants]
K < 1 the reverse reaction is favored [products] < [reactants]
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Manipulating equilibrium constants:
1.
Constant of a reverse reaction:
Consider a weak acid that dissociates in water according to the reaction:
HA + H2O <=> H3O+ + AOr
HA <=> H+ + AThe dissociation constant is written as:
Ka = [H+] [A-] / [HA]
If the reverse reaction is written as:
H+ + A- <=> HA
Its equilibrium constant (K) is the inverse of the equilibrium constant of
the forward reaction:
K = [HA] / [H+] [A-] = 1/Ka
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2. If reactions are added, the new K is the product of the original Ks:
Consider the weak acid H3PO4 (phosphoric acid). This is a polyprotic
weak acid. Its dissociation in aqueous solution occurs in steps, each step
represents the dissociation of one proton:
H3PO4 <=> H+ + H2PO4Ka1 = [H+] [H2PO4-] / [H3PO4]
H2PO4- <=> H+ + HPO42-
Ka2 = [H+] [HPO42-] / [H2PO4-]
HPO42- <=> H+ + PO43Ka3 = [H+] [PO43-] / [HPO42-]
The total dissociation constant for the overall dissociation of the acid is
given by the equilibrium:
H3PO4 <=> 3H+ + PO43Ka = [H+]3 [PO43-] / [H3PO4]
The Ka value can be obtained from Ka1, Ka2, and Ka3:
H3PO4 <=> H+ + H2PO4+ H2PO4- <=> H+ + HPO42+ HPO42- <=> H+ + PO43H3PO4 <=> 3H+ + PO4316
Ka = [H+]3 [PO43-] / [H3PO4] = { [H+] [H2PO4-] / [H3PO4]} x {[H+] [HPO42-] / [H2PO4-]} x {[H+] [PO43-] /
[HPO42-]}
Ka = Ka1 x Ka2 x Ka3
This is true for all overall reactions that result from sum of reactions:
K overall = K1 x K2 x K3 x
From the equilibria:
H2O <=> H+ + OH
Kw = [H+] [OH-] = 1.0 x 10-14
NH3 + H2O <=> NH4+ + OH- KNH3 = [NH4+] [OH-] / [NH3] = 1.85 x 10-5
Find the equilibrium constant for the reaction:
NH4+ <=> NH3 + H+
Note:
1. The first equilibrium represents the auto-dissociation of water. This
equilibrium can be also represented as:
2 H2O <=> H3O+ + OH-
Kw = [H3O+ ] [OH-] = 1.0 x 10-14
2. The concentration of pure water is omitted from both 17
equilibrium
constants because it remains nearly constant ([H2O] 55.6 M)
The key is to identify the side of each chemical species:
NH4+ <=> NH3 + H+
=> H+ is in the products side
=> NH3 is in the products side
=> NH4+ is in the reactants side
(1) H2O <=> H+ + OH-
=> H+ is in the products side
(2) NH3 + H2O <=> NH4+ + OH-
=> NH3 is in the reactants side
=> NH4+ is in the products side
We should keep reaction (1) as it is and we should invert reaction number 2:
H2O <=> H+ + OH-
Kw = [H+] [OH-] = 1.0 x 10-14
+
NH4+ + OH- <=> NH3 + H2O
NH4+ <=> H+ + NH3
K = [NH3] / [NH4+ ] [OH-] = 1/ KNH3 = 1 / 1.85 x 10-5
K = Kw x K = Kw x 1/KNH3 = 1.0 x 10-14 / 1.85 x 10-5
K = 5.6 x 10-10
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Le Chteliers Principle
If a system at equilibrium is disturbed, the direction in which the system proceeds
back to equilibrium is such that the disturbance is partly offset
Do you really think Le Chtelier said that?
Example:
H3AsO3 + I2 + H2O <=> H3AsO4 + 2I- + 2H+
1.
What happens if the pH of the solution is lowered?
lowering the pH = increasing [H+]
The equilibrium will minimize its effect by increasing the velocity of the reverse
reaction to consume the excess of H+. We say the equilibrium shifts to the
left
2. What happens if iodine is added to the solution?
Adding iodine increases the concentration of I2
The equilibrium will minimize its effect by increasing the velocity of the forward
reaction to consume the excess of I2. We say the equilibrium shifts to the
right
3. What happens if H3AsO4 is removed from the solution?
The equilibrium will minimize its effect by increasing the velocity of the forward
reaction to replace the amount of H3AsO4 removed. We say the equilibrium
shifts to the right
4. What happens if H3AsO3 is removed from the solution?
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The equilibrium will minimize its effect by increasing the velocity of the reverse
After the perturbation, the system re-stores equilibrium and the value of the
equilibrium constant does not change
At equilibrium:
K = [H3AsO4 ] [I-]2 [H+]2 / [H3AsO3 ] [I2]
Immediately after the perturbation:
[H3AsO4 ] [I-]2 [H+]2 / [H3AsO3 ] [I2] = Q K
When the system re-stores equilibrium:
Q=K
You should always remember that the concentrations of products and reactants
will not be the same. Their concentrations will change with the perturbation but
the equilibrium constant will be the same
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Chapter 2: Tools of the Trade
You should read this chapter. The material presented in chapter 2 reinforces
lab practice
Chapter 3: Math Toolkit
Section 3-1: Significant Figures
Significant Figures: Minimum number of digits required to express a value in
scientific notation without loss of accuracy
Examples:
Number
Scientific Notation
142.7
1.427 x 102
4
142.70
1.4270 x 102
Significant Figures
5
Note 1: by writing the zero after the seven in 142.70 you imply that you know
the value of the digit after the seven, which is not the case for the number 142.7
0.000006302
6.302 x 10-6
4
Note 2: the zeros to the left of the 6 are merely holding decimal places and
they are not significant figures
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Number
92500
Scientific Notation
Significant Figures
9.25 x 104
3
9.250 x 104
4
9.2500 x 104
5
Note 3: You should write one of the three numbers to indicate how many
figures are actually known
Note 4: Zeros are significant figures when they are:
1. in the middle of a number
2. at the end of a number on the right-hand side of a decimal point
Note 5: The last (farthest to the right) significant figure in a measured quantity
always has some associated uncertainty. The minimum uncertainty is 1 in the last
digit
Examples:
1. Scale of a Spectronic 20 spectrophotometer:
The needle is between 0.23 and 0.24
We can estimate one more number
The value might be read as 0.233, 0.234 or 0.235
In any case we choose, the last digit is uncertain
because it is an estimate
In any case we choose, the value will have three significant figures
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2. Scale on a buret graduated to 0.1mL:
You should read the level to the nearest 0.01mL
Level of meniscus is between 9.6 and 9.7.
Because the meniscus is closer to seven than to six,
I would guess that the last figure is above five,
so the volume should be larger than 9.65mL. The
volume has 3 figures of merit and the last digit has
a 0.01mL uncertainty
9.67 0.01mL
3. Scale on an analytical balance with sensitivity 0.1mg:
Assume the weight of a beaker is 20.3218g. The last digit is uncertain.
Depending on the balance, the last digit might have an uncertainty of 0.1mg or
0.2mg.
If the uncertainty is 0.1mg, the weight of the beaker will be: 20.3218 0.0001g.
If the uncertainty is 0.2mg, the weight of the beaker will be 20.3218 0.0002g.
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Section 3-2: Significant Figures in Arithmetic
How many digits to retain in the answer after you have performed arithmetic
operations with your data?
General Rule: Rounding should be done only on the final answer (not
intermediate results), to avoid accumulating round-off errors
Addition and Subtraction
Note 6: If the numbers to be added or subtracted have equal numbers of digits,
the answer goes to the same decimal place as any of the individual numbers
1.362 x 10-4
+
3.111 x 10-4
4.473 x 10-4
3rd decimal place: same decimal place as 1.362 x 10-4 or 3.111 x 10-4
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Note 7: If the numbers to be added or subtracted have equal numbers of digits,
the number of significant figures in the answer may exceed or be less than that
in the original data
5.345
7.26 x 1014
+ 6.728
- 6.69 x 1014
12.073
0.57 x 1014
3rd decimal place:
same decimal place
as 5.345 or 6.728
2nd decimal place:
same decimal place
as 7.26 x 1014 or 6.69 x 1014
Note 8: If the numbers being added or subtracted do not have the same number
of significant figures, we are limited by the least certain one
Example: Calculation of the molecular mass of KrF2
Element Atomic mass
Atomic mass of KrF2
Kr
83.80
83.90
F
18.9984032
+ 18.998 403 2
18.998 403 2
121.796 806 4
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not significant
The not significant figures are removed from the answer by rounding off
the final number:
127.796 806 4 127.80
Note 9: When rounding off, look at all the digits beyond the last place desired;
then approximate your final answer to the closest number possible
In the case of:
127.796 806 4 is between 127.79 and 127.80
These two numbers are the only two possibilities because we are limited to
two decimal places only
Because 127.796 806 4 is closer to 127.80 than to 127.79, the final
answer is 127.80
Note 10: When the number is exactly halfway, round to the nearest even digit
Assume that, in the case of 43.550 00, we can retain only one decimal
place (or three significant figures)
This number is exactly half way between 43.500 00 and 43.600 00
So, by convention, we round off to 43.6
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Note 11: In adding or subtracting numbers expressed in scientific notation, you
should express all numbers with the same exponent
1.632 x 105
1.632
x 105
+ 4.107 x 103 => 0.04107 x 105
0.984 x 106
9.84
x 105
11. 51307 x 105
2nd decimal place: same decimal place as 9.84 x 105
Rounding off to the closest number possible (11.51 or 11.52); the final answer
is 11.51 x 105
Multiplication and Division
Note 12: In multiplication and division, the final result is limited to the number of
digits contained in the number with the fewest significant figures
Examples:
4.3179 x 1012
X 3.6 x 10 -19
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15.54444 x 1012-19 => 15.54444 x 10-7 => 1.554444 x 10-6 => 1.6 x
34.60
2.46287
14.0486 => 14.05
Logarithms and Antilogarithms
Consider the number:
n = 10a , where a is a real number (positive or negative)
The base 10 logarithm of n is equal to:
log n = log 10a
log n = a x log 10
The base 10 logarithm of 10 is 1:
log n = a
Examples:
log 101 = 1
log 100 = log 102 = 2
log 0.001 = log 10-3 = -3
Note: The number n is also called the antilogarithm of a
A logarithm is composed of a characteristic and a mantissa
Characteristic: it is the integer part
Mantissa: it is the decimal part
Example:
log 339 = 2.530
log 3.39 x 10-5 = -4.470
Characteristic = 2
Characteristic = -4
Mantissa = 0.530
Mantissa = 0.470
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Logarithm becomes important when dealing with pH and pOH of solutions,
pKa and pKb of weak acids and bases and solubility products of salts
(pKsp)
By definition:
pH = -log [H+]
pOH = -log [OH-]
pKa = -log Ka
pKb = -lok Kb
pKsp = -log Ksp
=> p = -log
You should know the following:
1. log a x b = log a + log b
2. log a/b = log a log b
3. log ab = b x log a
Example:
H2O <=> H+ + OH-
Kw = [H+][OH-] = 1.0 x 10-14
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log Kw = log 1.0 x 10-14 = log 1.0 + log 10-14 = 0 + -14 x log 10 = -14
-log Kw = -(-14) = 14
pKw = 14
Similarly:
log [H+][OH-] = log [H+] + log [OH-]
- log [H+][OH-] = - log [H+] - log [OH-]
-log Kw = - log [H+] - log [OH-]
pKw = pH + pOH
Or
pH + pOH = 14
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Section 3-3: Types of Error
Experimental error: it is the uncertainty associated to each measurement
There are two types of experimental errors:
1. Systematic errors
2. Random errors
Systematic errors
Systematic errors affect the accuracy of your method:
Note:
> Accuracy is the difference between the experimental value and the
true
value
> Absolute error is a measure of accuracy: x-
Common sources of systematic errors:
> Non-calibrated instrumentation
> Spectral and/or chemical interference in the method of analysis
Common ways to determine and correct systematic errors:
> Use a reference standard material (RSM)
> Run the blank of your sample
> Compare the experimental result of the suspicious method of analysis to
the experimental result obtained with a method known to provide accurate results
(standard method of analysis)
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RSM:
> RSM are sold by the U.S. National Institute of Standards and Technology
(NIST)
> There are more than 1,000 RSM available that include metals,
chemicals, rubber, plastics, engineering materials, radioactive substances and
environmental and clinical standards
> Example:
Assume you suspect that a method of analysis for mercury (II) in soil is
giving inaccurate results.
You can buy a soil sample contaminated with mercury (II). The
concentration of Hg (II) is accurately known and it is provided to you by NIST.
The full composition of the soil sample is also known to account for potential
interference. The concentration of mercury (II) in the RSM assumes the role of
true value ()
You analyze the RSM with the suspicious method and compare the
experimental result (x) to the true value.
Then you calculate the absolute error: x
> Checking for non-calibrated instrumentation usually employs a pure
chemical of known concentration that gives a known instrumental response
Example: a pH meter is calibrated with a buffer solution of known pH value
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Typical pH values used for calibration: 2, 4, 7, 9, 11
Sample Blank or Blank:
> Blank = Concomitants Analyte
Assuming a sample with five chemical components: a, b, c, d and e: a,
b, c, d, e
If a is the species of interest:
=> a = analyte
=> b, c, d, and e = concomitants
b, c, d, e
=> Blank =
Note:
In many cases, it is almost
impossible to prepare an exact
blank
If the species of interest is d:
=> d = analyte
=> a, b, c and e = concomitants
a, b, c, e
=> Blank =
> To check for spectral and/or chemical interference, you should analyze the blank sample. If
33
you observe a non-zero result, then your method responds to more species than the analyte
How do you identify the presence of systematic errors?
> Identical repetitions of the same experimental procedure provide
experimental results always above or below the true value
Repetition Experimental Result
Absolute Error
1
x1
x1 > 0
2
x2
x2 > 0
3
x3
x3 > 0
x1
x1
Similarly:
x2
x3
x2
x3
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Random errors
> Random errors are also called indeterminate errors
> Random errors result from our natural limitation of making physical measurements
> Random errors affect the reproducibility of a measurement or the precision of measurements
> Assume that you perform identical repetitions of the same experimental procedure:
Repetition
Experimental Value
1
x1
2
x2
3
x3
4
x4
Because of the existence of random errors, the probability to obtain the same experimental
value in each repetition (x1 = x2 = x3 = x4) is very low
> If one calculates the average experimental value of the measurements:
xaverage = (x1 + x2 + x3 + x4) / 4
The probability of each individual value to be higher or lower than the average is the same. In
other words, random errors have an equal chance to be positive or negative
> Random errors can not be eliminated from an experimental result. As a consequence,
random errors are included in the experimental result as part of its uncertainty
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Absolute and Relative Uncertainty
Absolute Uncertainty expresses the margin of uncertainty associated with a
measurement
> Examples:
1.
The estimated uncertainty in reading a calibrated:
5 mL buret is 0.01 mL
10 mL buret is 0.02 mL
Each time you use either one of these burets to measure a volume, your
measurement carries a random error of 0.01 mL for the 5 mL buret or 0.02
mL for the 10 mL buret
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2.
Each time you use one of these volumetric flasks to prepare a solution, the
final volume of solution carries a random error with it. For a volumetric flask of:
10 mL the associated uncertainty is 0.02 mL
50 mL the associated uncertainty is 0.05 mL
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Relative Uncertainty compares absolute uncertainty with its associated
measurement
> Relative uncertainty = absolute uncertainty / magnitude of measurement
> Percent relative uncertainty = 100 x relative uncertainty
> Examples:
1. Assuming that a 10 mL class A buret is used to perform one reading of a
5 mL volume. The percent relative uncertainty of the measurement is:
From Table 2-1, the tolerance of a 10 mL buret is 0.02 mL
=> absolute uncertainty = 0.02 mL
=> magnitude of measurement = 5 mL
=> percent relative uncertainty = 100 x 0.02 mL / 5 mL = 0.4%
2. Assuming that the same buret is used to perform one reading of a 10 mL
volume. The percent relative uncertainty of the measurement is:
=> absolute uncertainty = 0.02 mL
=> magnitude of measurement = 10 mL
=> percent relative uncertainty = 100 x 0.02 mL / 10 mL = 0.2%
> Comparison of examples 1 and 2 shows that the uncertainty of a
measurement decreases with the increase of the magnitude of the
measurement. This is always the case!
38
Section 3-4:Propagation of Uncertainty
Addition and Subtraction
> Assume you want to perform the following arithmetic operation:
1.76 0.03
+
1.89 0.02
-
0.59 0.02
3.06 ?
The uncertainty associated with the answer (3.06) is given by:
[ (0.03)2 + (0.02)2 + (0.02)2]1/2 = 0.04
So, the correct answer for the operation above is: 3.06 0.04
Note: the uncertainty of the answer is larger than each individual
uncertainty. This is always the case!
> In general terms:
For addition or subtraction of experimental values (xi) with experimental
uncertainties (ei): x1 e1; x2 e2; x3 e3; x4 e4
Uncertainty in addition or subtraction: e = [e12 + e22 + e32 + e42]1/2
39
Examples:
1. The volume delivered by a buret is always the difference between the
final and the initial readings
> If the uncertainty in each reading of a buret is 0.02 mL, what is the
uncertainty in the volume delivered?
> Suppose the initial reading is 0.00 0.02 mL and the final reading is
17.83 0.02 mL. The volume delivered is the difference:
17.83 0.02 mL
-
0.00 0.02 mL
17.83 e
=> e = [(0.02)2 + (0.02)2]1/2 = 0.03
The correct answer is 17.83 0.03 mL
The uncertainty in the volume delivered is 0.03 mL
This is also the case for graduated pipettes!
40
2. The weight of a chemical measured with an analytical balance is always the
difference between two readings
> Weighing procedures:
a. To weigh a non-hygroscopic chemical:
a.1. Place a clean receiving vessel on the balance pan. The mass of
empty vessel is called the tare.
a.2. Add chemical to the vessel and read the new mass
a.3. The mass of the empty vessel should be subtracted from that of
the filled vessel
> Calculation of uncertainty:
Assuming the weight of the mass in:
a.1 was 19.8342 g
a.2 was 21.3253 g
Assuming that the uncertainty of each measurement was 0.1mg:
a.1 was 19.8342 0.0001 g
a.2 was 21.3253 0.0001 g
The uncertainty in the weigh of chemical will be:
21.3253 0.0002 g
19.8342 0.0002 g
41
1.4911[(0.0002)2+(0.0002)2]1/2 = 1.49110.00028 = 1.49110.0003
Note: If an electronic balance is used, the tare can be set to zero by
pressing a button. The calculation of uncertainty is not affected. When the tare
sets the balance to the value zero, the zero value still carries the uncertainty of
any measurement that is made with the balance
b. To weigh a hygroscopic chemical:
> Because hygroscopic reagents rapidly absorb moisture from the air, the
steps are the following:
b.1. Weigh a capped bottle containing dry reagent
b.2. Quickly poor some reagent from the weighing bottle into a
receiver
b.3. Cap the weighing bottle and weigh it again
b.4. The difference is the mass of reagent
> Calculation of uncertainty: Because the mass of reagent is obtained from
the difference of two masses, the uncertainty in the mass of chemical is
calculated as in a
42
Multiplication and Division
Consider the following operations:
1.76 ( 0.03) x 1.89 ( 0.02) = 5.64 ?
0.59 ( 0.02)
> To find the uncertainty of an answer involving multiplication and/or
division:
1. The 1st step is to convert all uncertainties into percent relative
uncertainties:
0.03 / 1.76 x 100 = 0.0170 x 100 = 1.7%
0.02 / 1.89 x 100 = 0.0106 x 100 = 0.011 x 100 = 1.1%
0.02 / 0.59 x 100 = 0.0338 x 100 = 0.034 x 100 = 3.4%
2. The 2nd step is to find the relative uncertainty of the answer:
e % = [(1.7)2 + (1.1)2 + (3.4)2]1/2 = 4.0%
3. The 3rd step is relative uncertainty into the absolute uncertainty:
> We know that :
percent relative uncertainty = uncertainty/magnitude of measurement
x 100
43
=> %e = e / 5.64 x 100
=> e = %e x 5.64 /100
=> e = 4.0% x 5.64 / 100
=> e = 0.040 x 5.64 / 100 = 0.23
4. Drop the insignificant digits:
5.6 0.2
answer reported in terms of absolute uncertainty
5.6 0.4%
answer reported in terms of relative uncertainty
44
Example:
You need to prepare 250mL of a 0.0100M solution of Na2CO3 using an analytical
balance with 0.2mg uncertainty. Assume the mass of analyte is measured with only
one reading. The uncertainty of the volumetric flask is 0.10mL.
Considering significant figures, answer the following:
a) What is the formula weight of Na2CO3?
Atomic weights:
Na: 22.989 770 x 2 =
C:
45.979 540
12.010 7
x1=
+ 12.010 7
O: 15.999 4
x3=
+ 47.998 2
105.988 440
Your answer should have only four decimal digits
Rounding-off 105.988440: this number is between 105.9884 and 105.9885
Because it is closer to 105.9884, this should be your answer
45
(V)
b) What is the mass of Na2CO3 you need to weigh?
Molarity (M) = number of moles of solute (n) / volume of solution in litters
=> n = M x V
=> n = 0.0100 x 0.250 = (1.00 x 10-2) x (2.50 x 10-1) = 2.50 x 10-3
moles of Na2CO3
Note: the number of digits is the same for both numbers and it should
be retained in the
answer
n = mass of solute (g) / FW of solute (g/mol)
=> m = n (mol) x FW (g/mol)
=> m = (2.50 x 10-3)x (105.9884) = (2.50 x 10-3)x (1.059884 x 102) =
2.64971 x10-1g
Note: 2.50 x 10-3 limits the number of digits in the answer to 3
Rounding off:
2.649 x10-1 is between 2.64 x 10-1 and 2.65 x 10-1. Because it is
closer to 2.65 x 10-1, your answer should be: 2.65 x 10-1 = 0.265 g
Note: you have to consider the uncertainty of the analytical balance
( 0.2mg). This analytical tool gives you the ability to measure
up to
four decimal digits (0.0002g). The mass you need to weigh is 0.2650 g
46
c) What are the percent relative uncertainties of the mass of solute and the volume
used to prepare your solution?
> percent relative uncertainty = uncertainty / magnitude of measurement x 100
mass:
uncertainty = 0.0002 = 2 x 10-4g
magnitude of measurement = 0.2650g = 2.650 x 10-1g
> percent relative uncertainty of mass = 2 x10-4 / 2.650 x 10-1 x 100 = 0.07547%
Note: 2 x10-4 limits the number of digits in the answer to one digit
Rounding off: 0.07547 is between 0.07 and 0.08. Because it is closer to 0.08, your
answer should be:
percent relative uncertainty of mass = 0.08 %
volume:
uncertainty: 0.10mL = 1.0 x 10-1mL = 1.0 x 10-1mL x 10-3L/mL = 1.0 x 10-4L
magnitude of measurement = 250mL = 2.50 x 102mL x 10-3L/mL = 2.50 x 10-1L
>percent relative uncertainty of volume = 1.0 x 10-4L / 2.50 x 10-1L x 100 = 4.00 x 102
Note: 1.0 x10-4 limits the number of digits in the answer to two digits
>percent relative uncertainty of volume = 4.0 x 10-2 = 0.040 %
47
d) Calculate the percent relative uncertainty of the molar concentration of your
solution
M = n / V = m (g) / FW (g/mol) x V (L)
M = 0.2650 g 0.08% / 105.9884 g/mol x 0.250 L 0.040%
e% = [(0.08)2 + (0.040)2]1/2 = [0.0064 + 0.0016]1/2 = [0.0080]1/2 =
0.08944
Note: 0.08 limits the number of decimal digits in the final answer to two
Rounding off:
0.0894 is between 0.08 and 0.09. Because it is closer to 0.09, your answer
should be: 0.09%
e) Calculate the absolute uncertainty in the final molarity
e% = e / M x 100 => e = e% x M / 100 = 0.09 x 0.0100 / 100 = 0.0 =
0.000009
M = 0.0100 0.000009 M in terms of absolute molarity
Rounding off:
0.000009 0.00001
=> M = 0.01000 0.00001M
Ask Yourself:
How do I calculate the error associate with a dilution?
48
Chapter 4: Statistics
Section 4-1:The Gaussian Distribution
Whenever analytical measurements
are repeated on the same sample, the
data obtained are scattered as shown
in the example of Table a1-1.
The arithmetic mean or the average
of the experimental results is defined
as the sum of the measured values
divided
by
the
number
of
measurements:
Mean = x av = in xi / n
where xi = x1, x2, x3, x4, are the
individual experimental measurements
and n is the number of measurements
In Table a1-1:
x1 = 0.488; x2 = 0.480; x3 = 0.486;
; x50 = 0.479
There
are
50
individual
measurements (n = 50)
49
The experimental results in Table a1-1
can be organized in another table
according to the number of times each
result repeats itself
Experimental results in Table a1-1 can
be
organized
into
equal-size,
contiguous data groups, or cells, as
shown in the 1st column of Table a1-2
The 2nd column of Table a1-2 shows
the number of experimental results in
Table a1-1 that fit within any given cell.
For
example:
results
within
absorbance range 0.469 and 0.471
appear in trial 14, 30 and 38
The 3rd column of Table a1-3 shows
the relative frequency of occurrence of
data in each cell. For example:
0.469 0.471 => 3/50 = 0.06
0.490 0.492 => 4/50 = 0.08
50
The relative frequency of occurrence can be
plotted as in Figure a1-1A to give a bar
graph called histogram
If the number of repetitions (or experimental
results were made much larger) and the
size of the cells were made much smaller, a
smooth curve such as that shown in Figure
a1-1B would be obtained
A smooth curve of this type is called a
Gaussian curve or a Normal Error curve
A Gaussian curve
characteristics:
has
the
following
1. The most frequently observed result is
the mean of the set of data
2. The results cluster
around this mean value
symmetrically
3. Small divergences from the central
mean value are found more frequently than are
large divergences
4. In the absence of systematic errors, the
mean of a large set of data approaches the true
value
= lim n in xi / n
51
The standard deviation is a
measure of the width of the
distribution
For an infinite
repetitions:
number
of
standard deviation =
Gaussian curves for an infinite
number of repetitions showing the
effect of standard deviations
= [ ni =1 (xi )2 / n ]1/2
For
a
limited
repetitions:
number
of
standard deviation = s
s = [ ni =1 (xi xav)2 / n -1 ]1/2
The
smaller
the
standard
deviation, the narrower the
distribution. This is true for both an
infinite and a limited number of
repetitions,
i.e.
and
s
respectively
52
Whatever the values of and are:
68% of the population lie within 1
of
95% of the population lie within 2
of
99.7% of the values lie within 3 of
The agreement is also true for a limited number of
repetitions:
68% of the experimental results lie within 1s of xav
95% of the experimental results lie within 2s of xav
99.7% of the experimental results lie within 3s of xav
53
What have I told you so far?
There is a difference between the meanings of true value and experimental
average:
> The true value is the mean (or most frequent value) of an infinite
number of repetitions (n 30)
> The experimental average is the mean (or most frequent value) of a
limited number of repetitions (n < 30)
The difference between the experimental average and the true value is
called the absolute error
> Absolute error = x
In the absence of systematic errors, the absolute error is improved by
increasing the number of repetitions
> x- zero when n
Only for an infinite number of repetitions (n n 30), the absolute
error is zero:
> x = 0 => x =
Parameter under measurement
xav
54
The standard deviation of a limited number of measurements represents the
spread of individual results around the experimental average
xav
s1
x1-4
s2
s1 < s2
In the absence of systematic errors, the standard deviation is a statistical
parameter that represents random errors
Random errors affect the reproducibility of measurements
Random errors standard deviation reproducibility
Other ways of expressing random errors:
> Variance = s2
> Relative Standard Deviation (RSD) = [s / xav] x 100
55
Confidence Intervals
In most cases, analysts perform a limited number of repetitions:
=> we obtain an experimental average (xav) and an standard deviation
(s)
=> the true value () remains unknown
The confidence interval is a range of experimental results within which there
is a specified probability of finding the true value:
> = xav ts / n1/2
- ts / n1/2
xav
+ ts / n1/2
t = Students t: it is a statistical parameter used to evaluate probability
56
n -1
57
Examples:
> Problem 4-4 b:
For a given set of measurements, will the 95% confidence interval be
larger or smaller than the 90% confidence interval? Why?
= xav ts / n1/2
For any given set of measurements:
xav, s and n will be constant
From the table, we see that t 95% > t 90% for all degrees of freedom
=> t 95% x s / n1/2 > t 90% x s / n1/2
=> the confidence interval is larger for a probability of 95% than for a
probability of 90%
=> consider xav = 45; s = 2.5 and n = 3
= 4 5 t 95% x 2.5 / 31/2 = 4 5 4.303 x 2.5 / 31/2 = 45 6.2
=> there is 95% of probability of finding the true value within 38.8
51.2
= 4 5 t 90% x 2.5 / 31/2 = 45 2.920 x 2.5 / 31/2 = 45 4.2
=> there is 90% of probability of finding the true value within 40.8
49.2
the larger the confidence interval <=> the probability of finding the true value
58
> Problem 4.5:
For the numbers 116.0, 97.90, 114.2, 106.8, and 108.3 find the mean, standard
deviation, and the 90% confidence interval for the mean
xi
xi - xav
(xi-xav)2
116.0
116.0 108.6 = 7.4
(7.4)2 = 54.7
97.9
97.9 108.6 = -10.7
(-10.7)2 = 114.5
114.2
114.2 108.6 = 5.6
(5.6)2 = 31.4
106.8
106.8 108.6 = -1.8
(-1.8)2 = 3.2
108.3
108.3 108.6 = -0.3
(-0.3)2 = 0.09
xi = 543.2
(xi-xav)2 =
203.9
xav = xi / n
=> xav = 116.0 + 97.9 + 114.2 + 106.8 + 108.3 / 5 = 108.6
s = [ ni =1 (xi xav)2 / n -1 ]1/2 = [203.9 / 5-1]1/2 = 7.1
90% confidence interval = 108.6 2.132 x 7.1 / 51/2 = 108.6 6.8
=> There is 90% of finding true value within 101.8 and 115.4
59
Chapter 6: Good Titrations
Titration
The titrant is added to the analyte until the reaction is complete
The titrant is the reagent solution with known concentration. It is
usually delivered from a buret
Common titrations are based on acid-base, oxidation-reduction,
complex formation, or precipitation reactions
Equivalence point
It is reached when the quantity of titrant added is the exact
amount necessary for stoichiometric reaction with the analyte
The equivalence point is the ideal result that we seek in a titration.
However, what we actually measure is the end point
End point
The end point is marked by a sudden change in a physical
property of the solution
Methods of determining the end point include:
a) Detecting a sudden change in voltage or current between
a pair of electrodes
b) Observing an indicator color change
c) Monitoring the pH
d) Monitoring the absorbance of light by species in the
reaction
60
Titration error
It is the difference between the equivalence point and the end point
The ability of the analyst to detect the end point has a direct impact on the
titration error
The better the analyst, the smaller the titration error
The accuracy of a titration method depends on knowing the exact
concentration of one of the reactants used. The solution which
concentration is accurately known is called the standard solution
Standard solutions
A standard solution is prepared by dissolving an accurately weighed
quantity of a highly pure material called a primary standard and diluting it to
an accurately known volume in a volumetric flask
A primary standard should fulfill the following requirements:
a. It should be 100% pure. The highest tolerance for impurities is
0.02%. In this case the impurity content should be accurately known
b. It should be stable to room temperature and drying temperatures.
The primary standard is always dried before weighing
c. It should be readily available and fairly inexpensive
d. It should have a high formula weight
e. If it is used in a titration, it should possess the properties required for
a titration
61
62
When an appropriate primary standard is not available to titrate the analyte,
a secondary standard is used. Before using a secondary standard in a
titration, the secondary standard should be:
a. Standardized by titration against a primary standard
You should also known than the concentration of a secondary standard is
less accurate than the concentration of a primary standard because of
titration errors
Volumetric calculations
Volumetric calculations depend on whether the titration is a direct titration or
a back titration
Direct titrations
The titrant is added to the analyte until the end point is observed
Volumetric calculations in direct titrations using molarity
a. For 1:1 reactions
b. For reactions that are not 1:1
63
1:1 reactions
> Examples:
Acid-Base reactions:
1HCl + 1NaOH NaCl + H2O
1HNO3 + 1KOH KNO3 + H2O
Precipitation reactions:
1NaCl + 1AgNO3 AgCl + NaNO3
General formula to calculate the percentage of analyte that reacts on a 1:1
mole basis with the titrant:
> 1mol of titrant 1 mol of analyte
n moles of titrant n moles of analyte
> At the equivalence point:
n moles of titrant = n moles of analyte
Manalyte x Vanalyte = Mtitrant x Vtitrant
64
Manalyte = Mtitrant x Vtitrant / Vanalyte
where :
M titrant = molarity of standard solution used for titration
V titrant = volume of titrant, usually measured with the burette
V analyte = aliquot of analyte solution, usually measured with a pipette
Example:
0.100M AgNO3 is used to obtain the molar concentration of a NaCl
solution. Assuming that 25mL of titrant are used in the titration,
calculate:
a)
The number of moles in the titrated aliquot
Reaction: NaCl + AgNO3 AgCl + NaNO3
At the equivalence point:
nanalyte = Manalyte x Vanalyte = Mtitrant x Vtitrant
nanalyte = 0.100 mol/L x 25 x 10-3 L = 25 x 10-4 moles in the
aliquot
b) Assuming that the aliquot is 10 mL of analyte solution, what is the
mass of NaCl in the aliquot?
nNaCl = mass of NaCl (g) / FW (g/mol)
mass of NaCl (g) = nNaCl x FW (g/mol)
mass of NaCl (g) = 2.5 x 10-3 moles x 58.44 g/mol = 0.146 g
65
c) Assuming that the weight of reagent used to prepare the aliquot
titrated in b was 1.5g, what is the percent of NaCl in the reagent?
% NaCl = [mass of NaCl (g) / mass of reagent (g)] x 100
% NaCl = 0.146 g / 1.5 g x 100 = 9.73 %
Reactions that are not 1:1
When the reaction is not 1:1, a conversion factor is used to equate the
moles of analyte and titrant
Example: A 0.2638g soda ash sample is analyzed by titrating the sodium
carbonate with the standard 0.1288M hydrochloride solution, requiring
38.27 mL. Calculate the percent sodium carbonate in the sample.
The reaction is:
1Na2CO3 + 2HCl H2CO3 + 2NaCl
1 mol of Na2CO3 2 moles of HCl
n NaCl x 1 = n Na2CO3 x 2
n Na2CO3
nNaCl
nNa2CO3 = x nNaCl
> Conversion factor is 1/2
66
The number of moles of sodium carbonate will be:
nNa2CO3 = x MHCl x VHCl = x 0.1288 (mol/L) x 38.27 x 10-3 (L)
= 2.4645 x10-3 moles
The mass of sodium carbonate in the sample will be:
massNa2CO3 (g) = nNa2CO3 (mol) x FWNa2CO3 (g/mol) = 2.4645
x10-3 x 105.99
= 0.2612 g
% Na2CO3 = 0.2612 g / 0.2638g x 100 = 99.02 % of Na2CO3
General formulas for titration reactions involving an analyte A that reacts on a
a:t mole basis with titrant:
aA + tT P
where A = analyte, T = titrant, P = product, a and t stoichiometric coefficients of
balanced equation
number of moles of analyte:
nanalyte = a/t [Mtitrant x Vtitrant]
mass of analyte:
manalyte = a/t [Mtitrant x Vtitrant] x FWanalyte
Percent of analyte:
% analyte = a/t [Mtitrant x Vtitrant] x FWanalyte x 100
mass of aliquot
67
Example:
Consider the following titration reaction:
5 HO-CO-CO-OH + 2MnO4- + 6H+ 10CO2 + 2Mn2+ + 8H2O
Analyte
Titrant
If an unknown quantity of H2C2O4 consumed 23.45mL of a 0.01M
MnO4- solution, calculate:
a) The number of moles of oxalic acid:
analyte = H2C2O4
titrant = 0.01M MnO4nanalyte = a/t [Mtitrant x Vtitrant]
a=5
t=2
=> n H2C2O4 = 5/2 [0.01 x 23.45 x 10-3] = 5/2 x 2.345x10-4 = 5.862 x
10-4 moles
b) The mass of oxalic acid in the titrated aliquot:
manalyte = a/t [Mtitrant x Vtitrant] x FWanalyte
m H2C2O4 = 5/2 [0.01 x 23.45 x 10-3] x FW H2C2O4
m H2C2O4 = 5.862 x 10-4 moles x 90.0345 g/mol
=> m H2C2O4 = 0.0528g
68
1.
c) Assuming that the total mass of titrated aliquot was 0.0832g, calculate
the percent of oxalic acid in the sample:
% analyte = a/t [Mtitrant x Vtitrant] x FWanalyte x 100
mass of sample
% H2C2O4 = 5.862 x 10-4 moles x 90.0345g/mol x 100
0.0832 g
% H2C2O4 = 0.0528g / 0.0832g x 100 = 63.46%
You need to remember the following:
The number of moles in a titrated aliquot could be different than the total
number of moles (nt) in the sample:
Suppose that:
A certain mass of sample (g) is used to prepare 100mL of analyte
solution and only 10mL of this solution are used for titration:
nanalyte 10mL
nt x 10mL = nanalyte x 100mL
nt
100mL
nt = nanalyte x 100mL/10mL
or
nt = nanalyte x Vsolution/Valiquot
69
2. The same is true if only a portion of the total mass of sample is used
for titration:
nt = nanalyte x mass of sample (g) / mass of aliquot (g)
Back Titrations
A known excess of a standard reagent is added to the analyte:
Analyte
+
Reagent 1
Product + Excess of reagent 1
Unknown quantity
Known quantity
Unknown quantity
Then a second standard reagent is used to titrate the excess of the first
reagent:
Excess of reagent 1
Unknown quantity
+
Reagent 2
Product <= Back Titration
Known quantity
Back titrations are useful when:
a) the end point of the back titration is clearer the end point of the direct
titration
b) when an excess of the first reagent is required for a complete
reaction with the analyte
70
Example of titrimetric analysis with back titration
A 1.876 g sample containing oxalic acid (H2C2O4) requires 38.84 mL of
0.1032M NaOH for titration:
H2C2O4 + 2NaOH Na2C2O4 + 2H2O
If 1.38 mL of 0.0992 M HCl is used in back titration, calculate the
percentage of H2C2O4 in the sample.
reagent 1 = NaOH; used in excess in direct titration
reagent 2 = HCl; used to titrate excess of NaOH
Back titration: NaOH + HCl NaCl + H2O
> From back titration, at equivalence point:
n HCl = n NaOH excess
M HCl x V HCl = n NaOH in excess
0.0992 x 1.38 x 10-3 = n NaOH in excess
n NaOH in excess = 1.36896 x 10-4 moles
> From the direct titration, at the equivalence point:
n H2C2O4 = n NaOH = (M NaOH x V NaOH - n NaOH in excess)
71
n H2C2O4 = (0.1032 x 38.84x10-3 - 1.36896 x 10-4 ) = (4.008288 x 10-3 1.36896 x 10-4)
n H2C2O4 = 1.9367 x10-3
n H2C2O4 = mass / FW => mass H2C2O4 = n x FW = 1.9367 x10-3 x 90.0349
mass H2C2O4 = 0.1743 g
> % H2C2O4 = 0.1743 g / 1.876 g x 100 = 9.289%
General Formula:
nanalyte = a/t (Mtitrant x Vtitrant ntitrant in excess)
where: a/t is stoichiometric ratio in direct titration
ntitrant in excess is obtained in back titration
72
=> Section 6-4: Solubility Product
If a colorless solution of Pb(NO3)2 is added to a
colorless solution of KI, three main observations
can be made:
1. Initially, no change is observed. Small
volumes of Pb(NO3)2 do not change the color of
the solution and only one phase (liquid) exists
2. After adding a certain volume of
Pb(NO3)2 the appearance of a yellow precipitate
(solid) is observed. The solution has now two
phases, solid and liquid
3. The more Pb(NO3)2 is added to the
solution, the more precipitate (ppt) is formed
We could ask ourselves the following questions:
1. Why is it formed a ppt?
Pb(NO3)2 + 2KI PbI2 (s) + 2KNO3
2. Why is it that the ppt is not formed right
away, i.e. in step 1 of the process?
Because minimum concentrations of lead
(II) and iodide ions are needed in solution to
form the solid
73
The ppt is only formed when the solution has reached the saturation point
Saturated solution is a solution that contains all the solid capable of being
dissolved
At the saturation point, there is an equilibrium between the solid and the
ions in solution:
PbI2 (s) <=> Pb2+ + 2I-
The equilibrium constant for that reaction is called the solubility product
(Ksp):
Ksp = [Pb2+][I-]2
Solubility product constants for most species are tabulated
See appendix A of your textbook (page 538)
Note:
Their values are reported for room temperature (25C) and aqueous
solvent
> Large changes in temperature will change the Ksp values
Typically, an increase in temperature increases the Ksp
> Changes in solvent will change Ksp values
74
Ksp values are useful to find out the solubility of salts
What is the concentration of Pb2+ in a solution saturated with PbI2?
The dissociation equilibrium is given by:
PbI2 (s) <=> Pb2+ + 2I-
Initial concentration
solid
0
0
Final concentration
solid
x
Ksp = [Pb2+][I-]2 = 7.9 x 10-9
2x
=> [Pb2+] = x moles/L; [I-] = 2x moles/L
Substituting in the Ksp expression:
=> (x) (2x)2 = Ksp => 4x3 = Ksp => x = (Ksp/4)1/3 = 1.2 x 10-3 moles/L
=> [Pb2+] = 1.2 x 10-3 M
=> [I-] = 2 x 1.2 x 10-3 moles/L = 2.4 x 10-3 M
Ksp values are very useful to find out:
A. Solubility of salts
B. Predict precipitation of salts
75
Solubility of salts
Example 1: Compare the solubility of CuSCN and AgSCN.
> From Appendix A, page 539:
CuSCN Ksp = 4.0 x 10-14
AgSCN Ksp = 1.1 x 10-12
> Both salts have the same stoichiometry:
CuSCN (s) <=> Cu+ + SCNAt equilibrium
solid - x
x
x
AgSCN (s) <=> Ag+ + SCNAt equilibrium
solid - x
x
x
> [Cu+] = [Ag+] = x moles/L and [SCN-] = x moles/L
> From the Ksp:
[Cu+] [SCN-] = Ksp CuSCN => (x) (x) = x2 = Ksp CuSCN
[Ag+] [SCN-] = Ksp AgSCN => (x) (x) = x2 = Ksp AgSCN
> x = [Cu+] = [SCN-] = (Ksp CuSCN )1/2 = 2 x 10-7 moles/L
> x = [Ag+] = [SCN-] = (Ksp AgSCN ) = 1.05 x 10-6 moles/L
=> AgSCN is more soluble than CuSCN
Note: in cases where the stoichiometry of the dissociation reaction is the
same, the comparison of solubility can be made directly from the Ksp values
76
Example 2: Compare the solubility of La2(C2O4)3 and Th(C2O4)2.
> From Appendix A, page 539:
La2(C2O4)3 Ksp = 1 x 10-25
Th(C2O4)2 Ksp = 4.2 x 10-22
> Reactions have different stoichiometries:
La2(C2O4)3 (s) <=> 2La3+ + 3C2O42At equilibrium solid - x
2x 3x
Th(C2O4)2 (s) <=> Th2+ + 2C2O42At equilibrium solid - x
x
2x
> [La3+] = 2x moles/L and [C2O42-] = 3x moles/L
[Th2+ ] = x moles/L and [C2O42-] = 2x moles/L
> From Ksp:
[La3+]2[C2O42-]3 = Ksp La2(C2O4)3
[Th2+][C2O42-]2 = Ksp Th(C2O4)2
> Substituting concentrations for La2(C2O4)3 in its Ksp:
(2x)2(3x)3 = Ksp La2(C2O4)3 => 4x2.27x3 = Ksp La2(C2O4)3 = 108x5
x = (Ksp La2(C2O4)3/108)1/5 = 3.9 x 10-6 moles/L is the solubility
> [La3+] = 2 x 3.9 x 10-6 = 7.8 x 10-6 M
[C2O42-] = 3 x 3.9 x 10-6 = 11.7 x 10-6 M = 1.17 x 10-5M
77
> Substituting concentrations for Th(C2O4)2 in its Ksp:
(x)(2x)2 = Ksp Th(C2O4)2 => 4x3 = Ksp Th(C2O4)2 => x = (Ksp
Th(C2O4)2/4)1/3
x = 4.7 x 10-8 moles/L is the solubility
[Th2+] = 4.7 x 10-8 M
[C2O42-] = 2 x 4.7 x 10-8 = 9.4 x 10-8 M
Although La2(C2O4)3 has the lowest Ksp value of the two salts, it is still the
most soluble of the two!
Note: in cases where the stoichiometry of the dissociation reaction is not the
same, the straightforward comparison of Ksp values might lead to
erroneous conclusions on solubility
Predicting precipitation of salts
Example 3: What is the concentration of Ca2+ and CO32- necessary to
prepare a saturated solution of calcite (CaCO3)?
> From Appendix:
CaCO3 <=> Ca2+ + CO32Ksp = [Ca2+ ][CO32-] = 4.5 x 10-9 = (x)(x) = x2 => x = (4.5 x 10-9)1/2
x = 6.7 x 10-5 moles/L
78
Example 4: Assuming that you have a 0.01M CaCl2 solution. What is the
concentration of Na2CO3 needed to start the precipitation of calcite?
> CaCl2 Ca 2+ + 2Cl- => [Ca2+ ] = 0.01M
> From Ksp:
CaCO3 (s) <=> Ca2+ + CO32Ksp = [Ca2+][CO32-] => [CO32-] = Ksp / [Ca2+] = 4.5 x 10-9 / 0.01
[CO32-] = 4.5 x 10-7M
> Na2CO3 2Na+ + CO32=> [Na2CO3] = [CO32-]; the concentration needed to start the precipitation of
calcite should be: [Na2CO3] > 4.5 x 10-7M
Example 5: Assuming you have a mixture of BaCl2 and SrCl2 in aqueous solution.
Both salts are at the 0.01M concentration. If Na2CO3 is added to the mixture, which
carbonate will precipitate first?
> BaCO3 <=> Ba2+ + CO32- Ksp = [Ba2+ ][CO32-] = 5.0 x 10-9
[CO32-] = 5.0 x 10-9/ [Ba2+ ] = 5.0 x 10-9/ 0.01 = 5.0 x 10-7M
> SrCO3 <=> Sr2+ + CO32- Ksp = [Sr2+ ][CO32-] = 9.3 x 10-10M
[CO32-] = 9.3 x 10-10 / [Sr2+ ] = 9.3 x 10-10 / 0.01 = 9.3 x 10-8M
> The concentration needed to start SrCO3 precipitation is lower than the one
needed
to start BaCO3 precipitation; so SrCO3 will start to precipitate first.
79
Example 6: Assuming that you have a mixture of AgCl and Hg2Cl2. Both
salts are present at the 0.1M concentration. Is it possible to separate Ag+
from Hg22+ using NaCN as the precipitating agent?
> AgCl Ag+ + Cl- => [Ag+ ] = 0.1M
Hg2Cl2 Hg22+ + 2Cl- => [Hg22+ ] = 0.1M
> From Ksp:
AgCN (s) <=> Ag+ + CN-
Ksp = [Ag+ ][CN-] = 2.2 x 10-16
Hg2(CN)2 <=> Hg22+ + 2CN-
Ksp = [Hg22+ ][CN-]2 = 5 x 10-40
> The concentration of CN- needed to start the precipitation of each salt is:
AgCN => [CN-] = 2.2 x 10-16 / [Ag+ ] = 2.2 x 10-16 / 0.1 = 2.2 x 1015M
Hg2(CN)2 => [CN-] = (5 x 10-40 / [Hg22+ ])1/2 = (5 x 10-40 / 0.1)1/2
[CN-] = 7.07 x 10-20M
> Hg2(CN)2 starts precipitating first. However, to have a successful
separation, AgCN should start precipitating only after all Hg2(CN)2 has
precipitated.
Note: an ion is considered fully precipitated if its concentration in solution is
equal or smaller than 10-6M
80
> When the [Hg2 2+] = 10-6M:
[CN-] = (5 x 10-40 /10-6)1/2 = 2.24 x 10-17M
> This CN- concentration is smaller than the one needed to start the
precipitation of AgCN, which is 2.2 x 10-15M
=> It is possible to separate Ag+ from Hg22+
81
Chapter 7: Gravimetric and Combustion Analysis
In gravimetric analysis, the mass of a product is used to calculate the quantity of
analyte in the original sample
=> Section 7-1: Examples of Gravimetric Analysis
A simple example of gravimetric analysis is the determination of Cl- by precipitation
with Ag+:
Ag+ + Cl- AgCl (s)
The mass of AgCl tells us how many moles of AgCl were produced
Based on the stoichiometry of the reaction: for every mol of AgCl produced in the
reaction, there must be 1 mole of Cl- in the unknown sample
A 10.00 mL solution containing Cl- was treated with excess AgNO3 to precipitate
0.4368 g of AgCl (FW = 143.321g/mol). What was the molarity of Cl- in the unknown?
> the number of moles of AgCl precipitated is given by:
nAgCl = mass / FW = 0.4368 g / 143.321 g/mol = 3.048 x 10-3 mol AgCl
> Because the reaction is a 1:1 reaction, the number of moles of Cl- in the original
sample is the same:
nAgCl = n Cl- = 3.048 x 10-3 moles => [Cl-] = 3.048 x 10-3 moles / 10 x 10-3
[Cl-] = 0.3048 M
82
83
=> Section 7-2: Precipitation
You should read this section. Descriptive information on experimental parameters
affecting precipitation
=> Section 7-3: Examples of Gravimetric Calculations
In the usual gravimetric procedure a precipitate is weighed, and from this value the
weight of analyte in the sample is calculated
So, the question is:
Mass of
gravimetric
precipitate
?
Quantity of analyte
in original sample
The percentage of analyte (A) in the sample is calculated by the equation:
%A =
weight of A
x 100
weight of sample
Or
%A =
weight of precipitate x gravimetric factor x 100
weight of sample
Where:
weight of precipitate x gravimetric factor = weight of A
84
Utilization of the Gravimetric Factor in Gravimetric Calculations
Example 1: Assume that 0.4852 g of sample of iron ore is dissolved in acid, and iron is oxidized
to the +3 state and then precipitated as hydrous oxide, Fe2O3.xH2O. The precipitate is filtered,
washed, and ignited to Fe2O3, which is found to weigh 0.2481 g. Calculate the percentage of
iron (Fe) in the sample.
> The chemical steps are the following:
Iron ore Fe2+ and Fe3+ 2Fe3+ Fe2O3.xH2O Fe2O3
> The percentage of iron in the sample will be obtained from:
%Fe = weight of Fe x 100
weight of sample
or
%Fe = weight of Fe2O3 x gravimetric factor x 100
weight of sample
where:
weight of Fe = weight of Fe2O3 x gravimetric factor
> We have the weight of sample and the weight of precipitate, so:
%Fe = 0.2481 g x gravimetric factor x 100
0.4852 g
> Gravimetric factor:
2 moles of Fe 1 mol of Fe2O3
2 AW of Fe 1 FW of Fe2O3
weight of Fe weight of Fe2O3
=> weight of Fe x 1 FW of Fe2O3 = weight of Fe2O3 x 2 AW of Fe
=> weight of Fe = weight of Fe2O3 x 2 AW of Fe
85
FW of Fe2O3
Gravimetric factor = 2 AW of Fe = 2 x 55.85 g/mol = 0.6995
> Substituting in the equation above:
%Fe = 0.2481 g x 0.6995 x 100 = 0.1735 g x 100 = 35.76
0.4852 g
0.4852 g
For the reaction:
aA
t ppt
a AW of A
t FW of ppt
weight of A weight of ppt
=> weight of A x t FW of ppt = a AW of A x weight of ppt
=> weight of A = weight of ppt x a AW of A
t FW of ppt
=> gravimetric factor = a AW of A
t FW of ppt
> if A is a compound, molecule or ion:
=> gravimetric factor = a FW of A
t FW of ppt
86
Example 2: From Table 7-1, obtain the gravimetric factor for Mg2+:
> Species analyzed: Mg2+
ppt: Mg2P2O7
> Write the equation: a A t ppt
Mg2+ Mg2P2O7
> Balance the species of interest (Mg2+) to obtain a and t:
2 Mg2+ 1Mg2P2O7
> Write the rule of three:
2 AW Mg2+
1 FW Mg2P2O7
weight of Mg2+
weight of ppt
=> weight of Mg2+ x 1 FW Mg2P2O7 = 2 AW Mg2+ x weight of ppt
weight of Mg2+ = weight of ppt x 2 AW Mg2+
1 FW Mg2P2O7
=> gravimetric factor = 2 AW Mg2+
1 FW Mg2P2O7
87
Example 3: Solid residue weighing 8.4448 g from an aluminum refining process was
dissolved in acid, treated with 8-hydroxyquinoline, and ignited to give Al2O3 weighing
0.85554 g. Find the weight percent of Al in the original mixture
> The chemical steps are the following:
Al Al3+ Al2O3
> %Al = weight of ppt x gravimetric factor x 100
weight of sample
weight of ppt (Al2O3) = 0.85554 g
weight of sample = 8.4448 g
> gravimetric factor:
1. Write the equation: a A t ppt
Al Al2O3
2. Balance the species of interest (Al) to obtain a and t:
2 Al 1 Al2O3
3. Write the rule of three:
2 AW Al
1 FW Al2O3
weight of Al weight of ppt
=> weight of Al x 1 FW Al2O3 = weight of ppt x 2 AW Al
=> weight of Al = weight of ppt x 2 AW Al
1 FW Al2O3
gravimetric factor = 2 AW Al
1 FW Al2O3
88
> Knowing that:
AW Al = 26.982
FW Al2O3 = 101.961
=> gravimetric factor = 2 x 26.982 = 0.529
1 x 101.961
> %Al = weight of ppt x gravimetric factor x 100
weight of sample
%Al = 0.85554 g x 0.529 x 100 = 5.359
8.4448 g
FYI:
The reaction between Al3+ and 8-hydroxyquinoline is as follows:
89
Many ions can be analyzed in the same manner, i.e. by gravimetric
precipitation with an organic agent
90
Calculation of the amount of precipitating agent required
Example 4: Calculate the number of milliliters of ammonia, density 0.99 g/mL, 2.3% by weight
NH3, which will be required to precipitate as Fe(OH)3 the iron in a 0.70g sample that contains
25% Fe2O3.
Note: The precipitation reaction is:
Fe3+ + 3NH3 + 3H2O Fe(OH)3 (s) + 3NH4> 1 mol of Fe 3+ 3 moles of NH3
> % Fe2O3 = weight of Fe2O3 x 100 = 25
weight of sample
> weight of sample = 0.70 g
=> weight of Fe2O3 = 25 x 0.70 g/ 100 = 0.175 g
> 2 moles of Fe3+ 1 mol of Fe2O3
2 AW of Fe3+
1 FW of Fe2O3
weight of Fe3+ weight of Fe2O3
=> weight of Fe3+ = weight of Fe2O3 x 2 AW of Fe3+
1 FW of Fe2O3
=> weight of Fe3+ = 0.175 x 2 (55.847) = 0.122 g
1 (159.69)
> 1 AW of Fe 3+ 3 FW of NH3
0.122 g of Fe 3+ weight of NH3
=> weight of NH3 = 0.122 g of Fe 3+ x 3 FW of NH3
1 AW of Fe 3+
=> weight of NH3 = 0.122 g x 3 (17.03) = 0.112 g
91
> 2.3% wt NH3:
2.3 g of NH3 100 g of reagent
0.112 g
weight of reagent = 0.112 g x 100 g / 2.3 g = 4.486 g
> mass (g) / volume (mL) = density (g/mL)
volume (mL) = mass (g) / density (g/mL) = 4.486 g / 0.99 g/mL
volume = 4.92 mL
Calculation of the optimum amount of sample size
Example 5:
What size sample containing 12.0% chlorine (Cl) should be taken for analysis if the
chemist (or forensic scientist) wishes to obtain a precipitate of AgCl which weighs 0.500
g?
% Cl = weight of ppt x gravimetric factor x 100
weight of sample
> %Cl = 12
> weight of ppt (AgCl) = 0.500 g
> gravimetric factor:
ppt reaction: Ag+ + Cl- AgCl
1 AW Cl 1 FW AgCl
weight of Cl- weight of AgCl
=> weight of Cl- = weight of AgCl x 1 AW Cl92
1 FW AgCl
=> gravimetric factor = 1 AW Cl1 FW AgCl
= 35.4527 = 0.2473
143.3395
> weight of sample = weight of ppt x gravimetric factor x 100
% Cl
=> weight of sample = 0.500 g x 0.2473 x 100 = 1.03 g
12
93
Calibration methods for instrumental analysis
Direct Comparison
Calibration Curve
Section 4-4, 4-5 and 5-2 of your textbook
Standard Additions
Section 5-3 of your textbook
Internal Standards
Section 5-4 of your textbook, but we will not cover it in this course
Direct Comparison
Experimental procedure
1. A standard solution of known analyte concentration (Cs) is prepared
2. A solution of the unknown is prepared (Cu)
3. The signal of the standard is measured with the instrument (Ss)
4. The signal of the unknown is measured with the instrument (Su)
94
Data Analysis
Cs Ss
Cu Su
=> Cu x Ss = Cs x Su
=> Cu = Cs x Su / Ss
A more accurate result is obtained if the signal of the standard and the
signal of the unknown are corrected for instrumental response
The instrumental response is obtained by measuring the signal of the blank
The experimental procedure is as follows:
1. A standard solution of known analyte concentration (Cs) is prepared
2. A solution of the unknown is prepared (Cu)
3. A blank solution is prepared (no analyte)
4. The signal of the standard is measured with the instrument (Ss)
5. The signal of the blank is measured with the instrument (Sb1)
6. The signal of the unknown is measured with the instrument (Su)
7. The signal of the blank is measured with the instrument (Sb2)
95
Data Analysis
1. The fist blank signal is subtracted from the signal of the standard
Ss Sb1 = Ss
2. The second blank signal is subtracted from the signal of the unknown
Su Sb2 = Su
3.
Cs Ss
Cu Su
=> Cu x Ss = Cs x Su
=> Cu = Cs x Su / Ss
If only one blank measurement is made, the same blank signal is used to
correct for the instrumental response in the signal of the standard and the
unknown:
1. Ss Sb = Ss
2. Su Sb = Su
3. Cu = Cs x Su / Ss
Among the three procedures, the procedure that carries two blank
measurements intercalated within the measurement of the standard and the
sample is preferred
96
Even better is to perform at least three measurements of each
solution, in such a way that each signal is an average of at least three
measurements
Sample
Average
Standard
Ss1; Ss2; Ss3
Ssav sdstandard
Blank
Sb1,1; Sb1,2; Sb1,3
Sb1av sdblank1
Unknown
Su1; Su2; Su3
Suav sdunknown
Blank
Measurement
Sb2,1; Sb2,2; Sb2,3
Sb2av sdblank2
Data Analysis:
1. (Ssav sdstandard) (Sb1av sdblank1) = Ssav sds
sds = [(sdstandard)2 + (sdblank1)2]1/2
2. (Suav sdunknown) (Sb2av sdblank2) = Suav sdu
sdu = [(sdunknown)2 + (sdblank2)2]1/2
97
3. Cu = Cs x (Suav sdu) / (Ssav sds)
In this case, we can define a confidence interval for the concentration:
= Cu ts / n1/2
where:
% s = [(%sds)2 + (%sdu)2]1/2
where:
% s = (s / Cu) x 100 => s = (%s x Cu) / 100
%sds = (sds / Ssav) x 100
% sdu = (sdu / Suav) x 100
n = total number of measurements
Example:
Assume that the following measurements were done by direct comparison:
Standard: 25.00, 23.00, 26.00 arbitrary units (a.u.)
Blank 1: 2.00, 1.00, 3.00 a.u.
Unknown: 16.00, 18.00, 17.00 a.u.
Blank 2: 2.00, 2.00, 4.00 a.u.
a. If the concentration of the standard is 10 mg/mL, what is the
concentration of the unknown and its confidence interval (95%)?
b. What would the error be if the blank measurements were not done?
98
a.
> Cu = Cs x (Suav sdu) / (Ssav sds)
Cu = 10 mg/mL x (Suav sdu) / (Ssav sds)
Ssav sdstandard = 24.67 1.53
Sb1av sdblank1 = 2.00 1.00
=> Ssav sds = (24.67 1.53) (2.00 1.00) = (24.67 2.00) [(1.53)2 + (1.00)2]1/2
Ssav sds = 22.67 1.83
Suav sdunknown = 17.00 1.00
Sb2av sdblank2 = 2.67 1.15
=> Suav sdu = (17.00 1.00) (2.67 1.15) = (17.00 2.67) [(1.0)2 + (1.15)2]1/2
Suav sdu = 14.33 1.53
> Cu = 10 mg/mL x 14.33 1.53 = 10 mg/mL x 14.33 10.68%
22.67 1.83
22.67 8.07%
Cu = 6.32mg/mL [(10.68%)2 + (8.07%)2]1/2
Cu = 6.32 mg/mL 13.38%
or
Cu = 6.32 0.84 mg/mL; t (n=12; 95%) = 2.228 (n-1=10)
=> = 6.32 2.228 x 0.84 / (12)1/2 = 6.32 0.54 mg/mL
99
b. If the blank measurements had not been done:
Cu = Cs x Su / Ss
Cs = 10 mg/mL
Suav sdunknown = 17.00 1.00
Ssav sdstandard = 24.67 1.53
> Because we are interest in the absolute error, we will ignore standard
deviations:
Cu = 10 mg/mL x 17.00/24.67 = 6.89 mg/mL
> Absolute error:
- x = 6.32 6.89 = - 0.57 mg/mL
> Relative error:
( - x / ) x 100 = (-0.57 mg/mL / 6.32 mg/mL) x 100 = -9.02 %
100
Calibration Curve Method
Experimental Procedure
a)
Several standards of known concentrations of analyte are introduced into
the instrument and the instrumental response is recorded
b)
The instrumental response is obtained with a blank
)
Note:
The calibration curve method will only give accurate results if:
1. There are no interferents in the unknown sample
Or
2. All the interferents in the unknown sample are known and their
concentrations are also known. In this case:
a. All the standard solutions should contain all the interferents
b. The concentrations of interferents in the standard solutions should
match their respective concentrations in the unknown sample
101
Data Treatment
Analyte
Concentration
I0 s0
I1,1; I1,2; I1,3
I1 s1
C2
I2,1; I2,2; I2,3
I2 s2
C3
I3,1; I3,2; I3,3
I3 s3
C4
I0,1; I0,2; I0,3
C1
Signal Average
Zero
Signal
I4,1; I4,2; I4,3
I4 s4
It is assumed that there is no random
error in concentration
The signal intensity is plotted as a
function of analyte concentration. If the
relationship between signal and analyte
concentration is linear, the perfect
calibration curve should look as the
figure
The unknown concentration can be
obtained by interpolation
102
=> Section 4-4: Finding the Best
Straight Line
The method of least squares finds the
best line in a set of experimental data
The equation of the straight line is
given by:
y = mx + b
where:
m = slope
b = y-intercept
The method of least squares
calculates the slope and the intercept
based on the following formulas:
m = n (xiyi) - xi yi
D
b = (xi2) yi - (xiyi) xi
D
where:
D = n (xi2) (xi)2
n = # of data points in the graph
103
When calculating m and b the best approach is to come up with a table
correlating the following parameters: xi, yi, xiyi and xi2
Example:
From the values in the table:
? = Later
> D = = n (xi2) (xi)2 = 4.62 142 = 52
> m = n (xiyi) - xi yi = 4.57 14.14 = 0.61538
D
52
> b = (xi2) yi - (xiyi) xi = 62.14 57.14 = 1.34615
D
52
=> y = 0.61538x + 1.34615
104
Uncertainties in the slope and the
intercept
The uncertainties in the slope and the
intercept are given by:
a. Standard deviation of slope:
sm = sy(n/D)1/2
b. Standard
intercept:
deviation
of
the
sb = sy((xi2)/D)1/2
where sy is given by:
sy = ((di2)/n-2)1/2
di is the vertical deviation. The vertical
deviation is the difference between
the real intensity (yi) and the intensity
predicted by the best straight line (y)
di = yi-y = yi (mxi+b) = yi mxi b
For the example at hand the di and
the di2 values are calculated in the
last two columns of table 4-5
105
From table 4-5:
(di2) = 0.076923
=> sy = ((di2)/n-2)1/2 = (0.076923 / 4-2)1/2 = 0.19612
=> sm = sy(n/D)1/2 = 0.19612 (4 / 52)1/2 = 0.054394
=> sb = sy((xi2)/D)1/2 = 0.19612 (62 / 52)1/2 = 0.21415
Combining these results with m and b:
> Slope: 0.61538 0.05439
> Intercept: 1.34615 0.21415
Rule for significant figures: the first decimal place of the standard deviation
is the last significant figure of the slope or intercept
=> m = 0.61538 0.05439 = 0.615 0.054 = 0.61 0.05
=> b = 1.34615 0.21415 = 1.35 0.21= 1.3 0.2
106
Section 4-5: Constructing a
Calibration Curve
Be careful: inspect your data
and use judgment before
mindlessly using a computer
to draw a calibration curve
or, in other words: plot
experimental data to find out
linear dynamic range
Example:
Spectrophotometric analysis
of a protein by the Lowry
method
107
The best straight line for the experimental points used in the plot is
obtained calculating the slope and the intercept for the equation y = mx + b
m = n (xiyi) - xi yi ; sm = sy(n/D)1/2
D
b = (xi2) yi - (xiyi) xi ; sb = sy((xi2)/D)1/2
D
where:
D = n (xi2) (xi)2
n = # of data points in the graph = 14 = includes the three blank
readings and excludes:
one of the data points (392) for sample with 15 g
three data points for sample with 25 g
sy = ((di2)/n-2)1/2
=> m = 0.01630
b = 0.0047
sy = 0.0059
sm = 0.00022
sb = 0.0026
108
Finding the protein in an unknown
Assuming that the reading of absorbance of a sample containing an
unknown amount of protein is 0.373.
a. How many micrograms of protein does it contain?
y = mx + b
> you should always keep in mind that:
y = instrumental reading
x = concentration
Isolating x in the equation above:
yb/m=x
The sample reading needs to be corrected for the blank response:
0.373 0.0993 average of three blank readings
=> y = 0.2737
Substituting this value with m and b in the equation above:
x = 0.2737 0.0047
0.01630
x = 16.50 g of protein
109
b. What uncertainty is associated with its concentration?
Uncertainty in x is calculated by the formula:
(sy/ |m|) [1 + x2n/D + (xi2)/D 2xxi/D]1/2
Applying the formula to the problem:
uncertainty in x = 0.38 g
Note:
> This uncertainty is equivalent to the standard deviation of the
concentration
=> x = 16.5 0.4 g of protein
> You should not confuse it with its confidence interval:
= x t s / n1/2
110
It is also correct to plot the calibration curve with the average of intensities
for each concentration and the signal of the blank:
In this case it is not necessary to correct
the intensities for the blank value
n = the number of points
used data to construct the best straight line
n = 8 points
Blank
signal
111
Section 5-2: Validation of an Analytical Procedure
Limit of Detection (LOD)
> The LOD is the smallest quantity of analyte that a method can measure
Example:
method 1: LOD = 20 ng.mL-1
method 2: LOD = 80 ng.mL-1
If a sample contains 40 ng.mL-1 of analyte, method 2 will not detect its
presence
In order to reach the LOD of method 2, the sample needs to be preconcentrated
In case of liquid samples, pre-concentration can be done by:
a. Solvent evaporation
b. Solid-liquid extraction
c. Liquid-liquid extraction
> All these steps add analysis time and might cause analyte loss
Typically:
The smaller LOD of a method, the better
112
LOD calculation according to your textbook:
The experimental procedure consists on the following steps:
1. Prepare a sample solution whose concentration is 1 to 5 times the
detection limit
Note: if you do not have previous knowledge of the detection limit, use
your blank as the basis, i.e. prepare a solution whose signal is 3 to 5x the
blank signal
2. Measure the signal from n replicate samples (n 7)
3. Calculate the standard deviation of n measurements (s)
4. Measure the signal from n blanks and find the mean value, yblank
Data analysis:
> The minimum detectable signal , ydl, is defined as:
ydl = y blank + 3s
> The LOD is calculated as:
LOD = 3s / m
where m is the slope of the calibration curve
See example in page 97 of your textbook
113
> Another definition of LOD widely employed is the definition adopted by
IUPAC (International Union of Pure and Applied Chemistry)
The LOD of a method is calculated with the following equation:
LOD = 3sb / m
where:
sb is the standard deviation of 7 or more (n 7) blank measurements
m = slope of calibration curve (y = mx + b)
Notes:
> The relative standard deviation at the LOD concentration level is of the
order of 30%
> All you can say at the LOD level of concentration is that the presence of
analyte has been detected or, in other words, that your instrument is
detecting a signal statistically different from the blank signal
114
The signal detection limit (ydl) is the signal equivalent to the LOD
> It is defined as:
ydl = y blank + 3sb
The limit of quantitation (LOQ) is a more reliable parameter. The LOQ is the
smallest concentration of analyte that can be measured with reasonable
accuracy and precision:
LOQ = 10sb / m (IUPAC) or LOQ = 10s / m (textbook)
Example:
Signals from seven replicate measurements with a concentration about
three times the detection limit were 5.0, 5.0, 5.2, 4.2, 4.6, 6.0 and 4.9 nA.
Reagent blanks gave values of 1.4, 2.2, 1.7, 0.9, 0.4, 1.5, and 0.7 nA. The
slope of the calibration curve was m = 0.229 nA/M.
Find the following:
a. The signal detection limit
b. The limit of detection
c. The limit of quantitation
According to IUPAC
=> blank average = 1.26 nA
standard deviation = 0.62 nA
a. ydl = yblank + 3sb = 1.26 nA + 3x 0.62 nA = 3.12 nA
b. LOD = 3sb / m = 3 x 0.62 nA / 0.229 nA/M = 8.2 M
115
c. LOQ = 10sb / m = 10 x 0.62 nA / 0.229 nA/ M = 27.07 M
According to your textbook:
=> standard deviation of analyte samples: s = 0.56 nA
a. ydl = yblank + 3s = 1.26 nA + 0.56 nA = 2.94 nA
b. LOD = 3s / m = 3 x 0.56 nA / 0.229 nA/ M = 7.3 M
c. LOQ = 10s / m = 10 x 0.56 nA / 0.229 nA/M = 24.45 M
Comparing the two LOD:
LOD IUPAC = 8.2 M
LOD book = 7.3 M
> The difference comes from s and sb.
> In most cases, sb > s because the blank signal is closer to the
instrumental noise than the analyte signal from a small concentration.
> Because the LOD IUPAC < LOD book, the probability to attribute a blank
signal to the presence of analyte is lower with the IUPAC definition.
116
Section 5-3: Standard Addition
The Standard Addition method is used when the sample matrix is too
complex or unknown to reproduce it in standard solutions
The idea is then to make a standard solutions out of the sample matrix
Experimental procedure:
1. Measure the signal from an aliquot of the unknown sample
Concentration of analyte in the unknown sample = [X]i
intensity of unknown sample = IX
Signal
where i = initial
2. Add a known volume of standard solution with known analyte
concentration ([S]i) to a known volume of sample. Measure the signal of the
sample after standard addition
Final concentration of analyte in the sample = [X]f = [X]i + [S]f
Signal intensity of unknown after standard addition = IX+S
Total volume of sample after standard addition = V = V0 + VS
where V0 is the initial volume of aliquot and VS is the volume of standard
solution added to the sample
117
Data analysis:
Assuming that there is a linear relationship between signal intensity and
analyte concentration:
[X]i IX
[X]f IX+S
=> [X]i x IX+S = [X]f x IX
=> [X]i x IX+S = ([X]i + [S]f) x IX
=> [X]i x IX+S = [X]I x IX + [S]f x IX
=> [X]i x IX+S - [X]I x IX = [S]f x IX
=> [X]i (IX+S IX) = [S]f x IX
=> [X]i = [S]f x IX / (IX+S IX)
Considering that:
[S]f = [S]i x
VS
V0 + VS
where VS / V0 + VS is the dilution factor of the standard solution
=> [X]i = [S]i x (
V0 + VS
VS
)x(
IX
IX+S IX
)
118
Example:
Ascorbic acid (vitamin C) in a 50.0 mL sample of orange juice was
analyzed by an electrochemical method that gave a detector current of 1.78
A. A standard addition of 0.400 mL of 0.279M ascorbic acid increased the
current o 3.35 A. Find the concentration of ascorbic acid in the orange juice
> V0 = 50.0 mL
IX = 1.78 A
VS = 0.400 mL
[S]i = 0.279M
IX+S = 3.35 A
> [X]i = [S]i x (
VS
)x(
IX
V0 + VS
=> [X]i = 0.279 M x (
IX+S IX
0.400 mL
50.0 + 0.4 mL
)
)x(
1.78 A
)
3.35 A - 1.78 A
=> [X]i = 0.279 M x (0.400 mL / 50.4 mL) x (1.78 A / 1.57 A)
=> [X]i = 2.51 x 10-3M = 2.51 mM
119
Multiple standard additions
The concentration obtained by multiple standard additions is more reliable
than the concentration obtained with the single standard addition because it
is based on more experimental points
Rearranging previous equation one obtains:
IS+X (V0 + VS) = IX + IX
V0
[X]I
Function to plot
on y-axis
[S]i (VS)
V0
Function to plot
on x- axis
When IS+X (V0 + VS) =
zero
V0
IX + IX
[X]i
[S]i (VS) = zero => IX
V0
[S]i (VS) = -IX => [S]i (VS)= - [X]i
120
[X]i
V0
Graphically:
Corrected Signal
Sample Signal
0
[S]I (VS / V0)
-[X]I
121
Often, all solutions in a standard addition experiment are made up to the
same total volume by addition of water or other solvent. In this case:
IS+X
= IX + IX [S]f
[X]I
where: [S]f = [S]I (VS / VS + V0 + Vsolvent)
If all solutions are made up to the same final volume, we plot signal (IX+S)
versus [S]f:
Signal
Sample Signal
0
[S]f
-[X]I
=>[X]I = [S]f
122
Example: Ten-milliliter aliquots of a natural water sample were pipetted into
50.00mL volumetric flasks. Exactly 0.00, 5.00, 10.00, 15.00, 1nd 20.00mL of
a standard solution containing 11.1ppm of Fe3+ were added to each aliquot,
followed by an excess of thiocyanate ion to give the red complex
Fe(SCN)2+. After dilution to volume, absorbance for the five solutions,
measured with a photometer equipped with a green filter, were found to be
0.240, 0.437, 0.621, 0.809, and 1.009 respectively. What is the
concentration of Fe3+ in the water sample?
> All solutions are made up to the same final volume, so the plot should be
as follows:
Signal
where:
[S]f = Signal
Sample [S]I (VS / VS + V0 + Vsolvent)
and
0
Signal = IX+S
[S]f
-[X]I
123
[S]f = [S]I (VS / VS + V0 + Vsolvent)
> [S]I = 11.1ppm
> VS + V0 + Vsolvent = 50mL
> VS varies with standard addition
Table:
VS (mL) [S]f (ppm)
Absorbance (a.u.)
0.00
5.00
(11.1ppm)(5mL/50mL) = 1.11
10.00
(11.1ppm)(10mL/50mL) = 2.22 0.621
15.00
(11.1ppm)(15mL/50mL) = 3.33 0.809
20.00
0
0.240
(11.1ppm)(20mL/50mL) = 4.44 1.009
0.437
Plot [S]f versus Absorbance:
> [S]f is the independent variable = x
> Absorbance is the dependent variable = y
124
125
Chapter 18: Let There Be Light
18-1: Properties of Light
Two models exist to describe the behavior of light and its interaction with
matter:
a) Wave
b) Particle
Both behaviors are complementary. In some cases is better to think of light
as a wave, in other as a particle
Light waves
Light waves consist of two field components, namely a magnetic and an
electric field
The magnetic and the electric fields are oscillating fields that propagate in
perpendicular planes to each other
126
Because the interaction of the magnetic field with matter is too weak in comparison to
the interaction of the electric field with matter, analytical spectroscopic techniques
neglect the contribution of the magnetic field
The electric field of a wave is characterized by the following parameters:
a) Wavelength (): It is the distance between two maxima or two minima
> The unit of depends on the spectral region. In the ultraviolet and visible regions,
the most used unit is nanometers (nm)
1 nm = 10-9m
b) Frequency () : It is the number of complete oscillations that the wave makes
each second
> The unit of frequency is reciprocal seconds (s-1) or hertz (Hz)
127
The frequency of a wave does not change with the medium of propagation but the
Two-dimensional representation
of the electric field of a
monochromatic beam of radiation
Effect of change of medium on a monochromatic beam of radiation
128
The velocity of propagation of a wave (v) is equal to the product of its
frequency and its wavelength
v=
In vacuum, the velocity of propagation is equal to the speed of light (c):
c = = 2.998 x 108 m/s 3 x 108 m/s = 3 x 1010 cm.s-1
Because the of a wave changes with the medium of propagation and its
frequency remains constant, its velocity of propagation changes too
Particle behavior of a wave
When a wave is thought to behave as a particle, the name given to such
particles is photons
The energy of a photon (wave particle) is directly proportional to the
frequency of its wave:
E = h
where h is Plancks constant (h = 6.626 x 10-34 Js)
The energy of a photon is inversely proportional to the wavelength and
directly proportional to the velocity of propagation of its wave
> From v = => = v / and from E = h => E = hv /
Another useful parameter: wavenumber: = 1 /; units = cm-1
129
When waves interact with matter, several phenomena occur. Absorption is
one of them
Depending on the wavelength of the absorbed electromagnetic radiation,
different molecular processes will occur:
130
Absorption of a photon with appropriate energy (or wavelength) by an atom
or a molecule promotes the atom or the molecule to an energy state of
higher energy
Energy
Energy state of higher energy
Energy state of lower energy
At room temperature (25C) and normal pressure (1 atm), most atoms and
molecules are in their lowest energy state. The lowest energy state is called
the ground state
An energy state with higher energy than the ground state is called an
excited state
131
In many cases, the absorption of a photon is followed by the emission of a
photon. The emission of radiation brings the atom or molecule to an energy
state of lower energy than the excited state. If the processes involve the
ground state, they can be represented as follows:
132
The absorption of light can be monitored with a spectrophotometer
The wavelength selector selects one wavelength of radiation. Radiation of
one wavelength only is called monochromatic radiation
> Monochromatic = one color = one wavelength
> Wavelength selector = monochromator or spectrometer
= band-pass filter
= interference (or Fabry-Perot) filter
When a fraction of the incident beam, with radiant power (P0), is absorbed
by the sample, the radiant power of the emerging bean (P) is lower than the
radiant power of the incident beam => P < P0
133
Transmittance is the fraction of radiant power that passes through the
sample
T = P / P0
If the sample does not absorb radiation:
P = P0 => T = 1
If the sample absorbs all the incident radiation:
P = 0 => T = 0
The transmittance lies between o and 1:
0T1
Percent transmittance (%T) is another way to express transmittance
%T = 100T
%T lies between 0% and 100%
A transmittance of 30% means:
> Only 30% of the light pass through the sample
> 70% of the light does not pass through the sample
134
The most useful quantity for
chemical analysis is absorbance
P/P0
%T
A
1
100
0
A = -log T = -log (P / P0)
0.1
10
1
The higher that transmittance, the
lower the absorbance
0.01
1
2
Absorbance is defined as:
A plot of transmittance intensity (y
coordinate) versus wavelength (x
coordinate)
is
called
a
transmittance spectrum
A plot of absorbance intensity (y
coordinate) versus wavelength (x
coordinate) is called an absorption
spectrum
Both types of spectra are obtained
with a monochromator as a
wavelength selector. Filters do not
allow one to record spectra!
135
Beers Law
Absorbance is proportional to the
concentration of light absorbing
molecules in the sample
The figures show that the absorbance
of KMnO4 is proportional to its
concentration over four orders of
magnitude
Absorbance is also proportional to the
path-length of substance through
which light travels
Beers law:
A=axbxC
where:
A = absorbance (dimensionless)
b = optical pathlength (cm)
C = concentration (g.L-1)
a = absorptivity (L.g-1.cm-1
If C is in mol.L-1
A=xbxC
where:
= molar absorptivity (L.mol-1.cm-1)
136
The maximum absorption wavelength (s) is (are) characteristic of the
absorbing species
The spectral profile is characteristic of the absorbing species
Spectrum of Ozone
Maximum absorption at 264 nm
Absorption spectrum of
sunscreen lotion
UV-A: 320-400nm
UV-B: 280-320nm
137
Assuming that a calibration curve is built subtracting the blank signal from
each absorbance value plotted in the graph and that the best straight line is
obtained through the least squares method:
Best straight line: y = mx + b
y = absorbance
x = concentration
b = blank absorbance = zero
Absorbance
Concentration
According to Beers law: A = x b x C
Comparing Beers law to the equation of the best straight line:
m=xb
Because b is known (length of the cuvette), one can use the slope of the
calibration curve to find out the molar absorptivity of an absorbing species:
=m/b
The same is true for absorptivity (a)
138
Note 1: You should understand that if the blank absorbance is not subtracted
from the individual absorbance values plotted in the graph, the relationship =
m / b is still valid:
A
Best straight line with blank included
Best straight line with blank subtracted
C
In subtracting the blank, all the absorbance values are reduced by the same
amount and the slope does not change
139
Note 2: You should understand that the absorptivity or molar absorptivity of
an absorbing species is a constant, but its value depends on the
wavelength of absorption:
at maximum absorption
wavelength = max
500
Because the absorption at the maximum wavelength of absorption is
stronger than at 500nm:
max > 500
140
If two calibration curves are built at the two wavelengths:
A
Best straight line at max
Best straight line at = 500nm
C
The value of the intercept will depend on the absorption of the blank;
most likely it will be different at the two wavelengths
Note 3: Knowing that the limit of detection is defined as LOD = 3sB / m, and
considering a similar standard deviation at both wavelengths:
LOD = 3sB / m = 3sB / x b
=> The best LOD is obtained at max
=> Increasing the optical path (length of cuvette) also improves LOD
141
Note 4: Other parameters that affect absorptivity:
> Temperature
> Solvent
Using Beers Law:
Ask Yourself 18-B:
a) What is the absorbance of a 2.33 x 10-4M solution of a compound with
a molar absorptivity of 1.05 x 103M-1 cm-1 in a 1.00-cm cell?
A=xbxC
A = 1.05 x 103M-1 cm-1 x 1.00 cm x 2.33 x 10-4M = 0.245
b) What is the percent transmittance of the solution in part a?
A = - log T => T = 10-A => T = 10 -0.245 = 0.569 = 56.9%
c) Find A and %T when the path-length is doubled to 2.00 cm
> A = x b x C => A should double too because A and b are directly
proportional and their relationship is linear
A = 1.05 x 103M-1 cm-1 x 2.00 cm x 2.33 x 10-4M = 0.490
142
> A = - log T => T = 10-A = 10 x b x C
=> The relationship between T and b is not linear. The same is true
between T and C and T and . Their relationship is exponential
=> This is the main reason T versus C plots are not used for quantitative
analysis
> a linear relationship can be obtained if:
T = 10 x b x C
log T = log (10 x b x C )
log T = x b x C (log 10)
-log T = x b x C
For a given (constant) and C
> The same is true for for a
constant b and C
b
> The same is true for C for a
constant and b
143
Going back to the question:
T = 10 x b x C = 10 -1.05 x 103M-1 cm-1 x 2.00 cm x 2.33 x 10-4M =
10 -0.490 = 0.324
%T = 32.4%
d) Find A and %T when the path-length is 1.00 cm but the concentration is
doubled
> A = x b x C; if A = x b x 2C = 2 x x b x C, the absorbance should
double too
A = 1.05 x 103M-1 cm-1 x 1.00 cm x 4.66 x 10-4M = 0.489
> T = 10-A = 10-0.489 = 0.324 => %T = 32.4%
e) What would it be the absorbance in part (a) for a different compound with
twice as great molar absorptivity (2.10 x 10-3M-1cm-1)? The concentration and
path-length are unchanged from part (a)
> A = x b x C; if A = 2 x b x C = 2 x x b x C, the absorbance should
double too
A = 2.10 x 103M-1 cm-1 x 1.00 cm x 4.66 x 10-4M = 0.489
> T = 10-A = 10-0.489 = 0.324 => %T = 32.4%
144
A Closer Look at Instrumentation (Adapted from Chapter 19)
Light Sources:
> UV: Deuterium lamp
> Visible: Tungsten lamp
Monochromators:
> Czerny-Turner Design is the most
popular design in current instrumentation
145
Detectors:
> Photomultiplier tube (PMT) is a
more sensitive detector than the phototube
> The spectral response of the PMT
depends on the photo-emissive response
of the photocathode
146
Types of spectrophotometers:
> Single Beam Spectrophotometer:
> Double Beam Spectrophotometer:
The main advantage over single beam spectrophotometers is that
the double beam spectrophotometer corrects for drift in the source
intensity and/or detector response
147
Example of double beam spectrophotometer: Varian Carry 3E
148
> Photodiode Array Spectrophotometer:
1. Main differences with respect to single-beam and double-beam
spectrophotometers:
a) Instead of a PMT, the detector is a linear photodiode array (LPDA)
b) Instead of a monochromator, it uses a spectrograph
c) The spectrograph is located after the sample compartment
2. Main advantage: it records spectra in real-time
149
Chapter 8: Introducing Acids and Bases
Water dissociates according to the following equilibrium:
2H2O <=> H3O+ + OH-
Kw = [H3O+][OH-] = 1 x 10-14
For the sake of simplicity, we can also write:
H2O <=> H+ + OHwhere H+ represents H3O+
In pure water:
[H+] = [OH-]
From Kw:
[H+] x [H+] = 1 x 10-14 => [H+]2 = 1 x 10-14 => [H+] = (1 x 10-14 )1/2
[H+] = [OH-] = 1 x 10-7M
In pure water, the concentration of hydronium ions is equal to the
concentration of hydroxyl ions, i.e. 1 x 10-7M
150
Arrhenius Theory
> Arrhenius, as a graduate student, introduced the first definition of acids
and bases (1894)
> He received the Nobel Prize in Chemistry for this theory
Acid:
> Any substance that ionizes (partially or completely) in water to give
hydrogen ions, which associate with the solvent to give hydronium ions:
HA + H2O <=> H3O+ + A
Base:
> Any substance that ionizes (partially or completely) in water to give
hydroxyl ions:
B + H2O <=> BH+ + OH
According to Arrhenius theory:
> An acid increases the concentration of H3O+ in water
Example: HCl + H2O H3O+ + Cl=> [H+] in an acidic solution is higher than 1 x 10-7M
> A base increases the concentration of OH- in water
Example: NaOH (aq) Na+(aq) + OH-(aq)
=> [OH-] in a basic solution is higher than 1 x 10-7M
151
Bronsted and Lowry Theory
Arrhenius theory is restricted to aqueous solutions
Many reactions in Chemistry are acid-base reactions that do not involve the
reaction of a proton (H+) with water to for the hydronium ion
Example: HCl + NH3 NH4Cl
Acid:
> An acid is a proton (H+) donor
HCl + H2O H3O+ + ClNote: According to Arrhenius theory HCl is also an acid
Base:
> A base is a proton (H+) acceptor
HCl + NH3
NH4Cl
acid
base
salt
Note:
1. NH4Cl is a salt. A salt can be thought as the product of an acid-base
reaction:
Acid + Base Salt Neutralization reaction
2. Most salts are strong electrolytes. Strong electrolytes dissociate
completely in aqueous solution:
152
NH4Cl (s) NH4+(aq) + Cl-(aq)
Conjugate Acids and Bases:
The products of a reaction between an acid and a base are also acids and
bases:
Note:
> Acetate ion is a base because it can accept a proton to make acetic acid
> Methylammoniun ion is an acid because it can donate a proton
Conjugate acid-base pairs are related to each other by the gain or loss of
one H+
> Acetic acid and acetate ion
> Methylamine and methylammonium ion
153
Relation between [H+], [OH-], and pH
The autoprotolysis of water and it dissociation constant provide a tool we
can always use to find the concentration of H+ and OH-:
2H2O <=> H3O+ + OH-
Kw = [H3O+][OH-] = 1 x 10-14
> If [H+] is known:
[OH-] = Kw / [H+]
> If [OH-] is known:
[H+] = Kw / [OH-]
From Kw, the following is clear:
> The product between [H+] and [OH-] is always constant
> If [H+] increases in solution, [OH-] has to decrease to keep the product
constant
> If [OH-] increases in solution, [H+] has to decrease to keep the product
constant
pH is approximately defined as:
pH = -log[H+]
pOH is defined as:
pOH = -log[OH-]
154
In a neutral solution:
[H+] = [OH-] = 1 x 10-7M and pH = 7
In an acidic solution:
[H+] > [OH-] => [H+] > 10-7M and pH < 7
In a basic solution:
[OH-] > [H+] and pH > 7
Changing the pH by one unit changes [H+] by a factor of 10
Real-world examples:
155
8-B. Ask Yourself:
A solution of 0.050M Mg2+ is treated with NaOH until Mg(OH)2 is
precipitates. (a) At what concentration of OH- does this occur? (b) At what pH
does this occur?
> Reactions:
NaOH (aq) Na+ + OHMg 2+ + 2OH- Mg(OH)2 (s)
> The precipitate will only be formed when the concentrations of Mg 2+
and OH- in solution reach the solubility product of Mg(OH)2:
1. [Mg2+] = 0.050M
2. The concentration of OH- needed to reach the solubility product can be
obtained from:
Mg(OH)2 (s) <=> Mg2+ + 2OHKsp = [Mg2+][OH-]2 = 7.1x10-12
=> [OH-] = (7.1x10-12 / [Mg2+])1/2
[OH-] = (7.1x10-12 / 0.05)1/2
[OH-] = 1.19x10-5M
> Knowing the concentration of OH-, we use Kw to find out the
concentration of H+:
[H+] = Kw / [OH-] = 1x10-14/1.19x10-5= 8.39x10-10M
pH = -log8.39x10-10 = 9.08
156
Strengths of Acids and Bases
Strong Acids and Bases:
Dissociate completely in aqueous solution
HCl (aq) H+ + ClKOH (aq) K+ + OH-
> Virtually, no undissociated HCl or KOH exist in
aqueous solution
Common strong acids and bases:
>You should memorize acids and bases in Table 8-1
The concentration of H+ due to the presence of a strong
acid in aqueous solution is obtained directly from the
acid concentration:
HBr (aq) H+ + BrIf [HBr] = 0.01M => [H+]HBr = 0.01M because:
a. HBr is a strong acid, so no undissociated acid
remains in solution
b. The stoichiometry of the reaction is 1:1
The same is true for the concentration of OH- in
aqueous solution due to the presence of a strong base:
KOH (aq) K+ + OHIf [KOH] = 0.02M => [OH-]KOH = 0.02M
157
For any given acid in aqueous solution, the total concentration of H+ is
given by:
[H+]Total = [H+]H2O + [H+]Acid
For a concentrated strong acid:
[H+]Total [H+]Acid
> Because [H+]H2O is negligible in comparison to [H+]Acid
Example:
Calculate the pH of a 0.01M HCl solution:
HCl H+ + ClH2O <=> H+ + OH=> [H+]Total = [H+]H2O + [H+]HCl
[H+]HCl = 0.01M
[H+]H2O = ?
The [H+]H2O is smaller than 10-7M because the presence of strong acid
inhibits the auto-dissociation of water:
[H+]Total = (<1x10-7M) + 0.01M = 0.01M
or
[H+]H2O = [OH-]H2O => [OH-]H2O = Kw / [H+]Total 1x10-14/0.01=
1x10-12M
158
=> [H+]Total = 1x10-12M + 0.01M = 0.01M
Similar reasoning can be applied for any given base in aqueous solution
The total concentration of OH- is given by:
[OH-]Total = [OH-]H2O + [OH-]Base
For a concentrated strong base:
[OH-]Base >> [OH-]H2O
=> [OH-]Total [OH-]Base
However, for a diluted strong base, the concentration of OH- due to the
auto-dissociation of water should not be disregarded
Example: What is the pH and pOH of a 10-7M NaOH solution?
This is an approximation:
NaOH (aq) Na+ + OHIt is assumed that the presence
of diluted NaOH does not
H2O <=> H+ + OHaffect considerably
the dissociation equilibrium
=> [OH-]Total = [OH-]H2O + [OH-]NaOH
of water
[OH-]NaOH = 10-7M
[OH-]H2O = ?
From Kw: [H+]H2O = Kw / [OH-]Total 1.0x10-14 / 2.0 x 10-7 = 0.5 x 107M = 5 x 10-8M
=> [OH-]Total = 10-7M + 5 x 10-8M = 1.5 x 10-7M => pOH = -log1.5x10-7
=> [H+]Total = Kw/[OH-]Total = 1x10-14/1.5x10-7= => pH = -log
159
The same is true for diluted strong acids
What is the minimum concentration of strong acid or strong base that one can still
disregard water as a source of H+ or OH- ions?
Strong Acid (HA):
[HA]
[OH-]H2O = Kw / [HA]
[H+]H2O = [OH-]H2O
[H+]T= [H+]H2O + [H+]HA
10-5M
10-9M
10-9M
10-5M + 10-9M 10-5M
10-6M
10-8M
10-8M
10-6M + 10-8M = 1.01x10-6M
=> For acid concentrations lower than 10-5M, we should consider H2O as a
source of protons
The same is true for strong bases
160
Weak Acids and Bases
All weak acids react with water by donating a proton to H2O:
HA + H2O <=> H3O+ + A- Dissociation of weak acid
Or
HA <=> H+ + A
Dissociation of weak acid
The dissociation of a weak acid is an equilibrium with an equilibrium
constant called acid dissociation constant (Ka):
Ka = [H+][A-] / [HA]
Acetic acid is a typical examples of weak acid:
161
Most carboxylic acids are weak acids, and most carboxylate anions are
weak bases:
All weak bases react with water by abstracting (grabbing) a proton from
H2O:
B + H2O <=> BH+ + OH- Base hydrolysis
The equilibrium constant is called the base hydrolysis constant (Kb):
Kb = [BH+][OH-] / [B]
Methylamine is a typical weak base:
162
Methylamine is a representative amine
Amines are nitrogen containing compounds:
Amines
Ammonium ions:
Primary: RNH2
Secondary: R2NH
Tertiary: R3N
RNH3+
R2NH2+
R3NH+
Their base behavior results from a pair of free electrons on the nitrogen
atom
Amines are weak bases and ammonium ions are weak acids
The parent of all mines is ammonia. Ammonia reacts with water according
to the reaction:
NH3 + H2O <=> NH4+ + OH- Kb = [NH4+][OH-] / [NH3] = 1.75 x 10-5
Ammonium is a weak acid:
NH4+ + H2O <=> NH3 + H3O+
Ka = [NH3][H3O+] / [NH4+]
Ammonia and ammonium ion are a conjugated pair, i.e. they differ by one
proton: NH3 / NH4+
163
The same is true for all the methylamines. Their conjugate pairs are:
RNH2 / RNH3+
R2NH / R2NH2+
R3N / R3NH+
Similar to ammonia, methylamines react with water to form the conjugate
weak acid:
RNH2 + H2O <=> RNH3+ + OHR2NH + H2O <=> R2NH2+ + OH-
Kb = [R2NH2+][OH-] / [R2NH]
R3N + H2O <=> R3NH+ + OH
Kb = [RNH3+][OH-] / [RNH2]
Kb = [R3NH+][OH-] / [R3N]
For the specific case of methylamine:
164
The conjugate acid reacts with water according to the reaction:
Relationship Between Ka and Kb of a conjugated acid-base pair
Consider a weak acid HA in aqueous solution. Two reactions occur:
Dissociation reaction: HA <=> H+ + AKa = [H+][A-] / [HA]
Hydrolysis reaction: A- + H2O <=> HA + OH- Kb = [HA][OH-] / [A- ]
If we add the two reactions:
HA <=> H+ + A+
A- + H2O <=> HA + OHH2O
<=> H+ + OHKw = Ka x Kb
The relationship between a Ka and Kb for a conjugated pair is:
Kw = Ka x Kb
165
Ka and Kb values measure the strength of a weak acid and a weak base,
respectively
8-C. Ask Yourself:
Which is a stronger acid, A or B? Write the Ka reaction for each.
A: Cl2HCCOOH <=> Cl2HCCOO-+ H+ Ka = [Cl2HCCOO-][H+] / [Cl2HCCOOH]
B: ClH2CCOOH <=> ClH2CCOO- + H+ Ka = [ClH2CCOO-][H+] / [ClH2CCOOH]
The ratio [Dissociated Species] / [ Un-dissociated Species] is larger for A
=> For any given concentration of acid ([A] = [B]), A will produce a larger
concentration of H+
=> The stronger acid is A because it has a larger value of Ka
166
Which is the stronger base, C or D? Write the Kb reaction for each.
C: H2NNH2 + H2O <=> H2NNH3+ + OH-
Kb = [H2NNH3+][OH-] / [H2NNH2]
D: H2NCONH2 + H2O <=> H2NCONH3+ + OH-
Kb = [H2NCONH3+][OH-] / [H2NCONH2]
The ratio [Dissociated species] / [Un-dissociated species] is larger for C
For any given concentration of C and D ([C] = [D]), C will produce a
larger concentration of OH-
The stronger base is C because it has a larger value of Kb
167
Using Appendix B: Acid Dissociation Constants
Each compound is shown in its fully protonated form:
Name
Unprotonated form Fully Protonated form
Acetic acid
CH3CO2CH3CO2H
Methylamine
CH3NH2
CH3NH2H+ or CH3NH3+
Both, the pKa and the Ka value are given:
Recall that: pKa = -logKa
For a weak acid such as acetic acid, the information available in the table is:
CH3CO2H
pKa = 4.736 Ka = 1.75 x 10-5
If one wants to know the hydrolysis constant (Kb) for the weak base of the conjugate
pair:
CH3CO2H / CH3CO2CH3CO2- + H2O <=> CH3CO2H + OHWe know that: Kb x Ka = Kw
=> Kb = 1.0 x 10-14 / 1.75 x 10-5 =
For a weak base such as methylamine, the information available in the table is:
CH3NH3+
pKa = 10.645 Ka: 2.26 x 10-11
If one wants to know the hydrolysis constant (Kb) for the weak base of the conjugate
pair:
CH3NH3+ / CH3NH2
CH3NH2 + H2O <=> CH3NH3+ + OHWe know that: Kb x Ka = Kw
=> Kb = 1.0 x 10-14 / 2.26 x 10-11 = 4.42 x 10-4
168
For polyprotic acids, i.e. acids that release more than one proton:
H3PO4 <=> H+ + H2PO4- Ka1
H2PO4- <=> H+ + HPO42- Ka2
HPO42- <=> H+ + PO43-
Ka3
The table reports several Ka values for the fully protonated form. The first
Ka value belongs to the most acidic group
Example: Pyridoxal phosphate
pKa
Group
Ka
1.4
POH
0.04
3.44
OH
3.6 x 10-4
6.01
POH
9.8 x 10-7
8.45
NH
3.5 x 10-9
169
pH of Weak Acids: Weak-Acid Equilibrium
Assume a monoprotic weak acid with the general formula HA. In aqueous
solution:
HA <=>
IC
F
H+ +
0
A-
Ka = [H+][A-] / [HA]
0
Assuming that [H+]H2O zero
FC
F-x
x
x
Substituting these values in the Ka expression:
Ka = x2 / F-x => x2 = Ka (F-x) => x2 = KaF Kax
=> x2 + Kax - KaF = 0
Because x = [H+]:
[H+]2 + Ka[H+] - KaF = 0
Note: This equation can be used for any monoprotic weak acid
170
[H+]2 + Ka[H+] - KaF = 0
This equation is equivalent to a quadratic equation of the general form
ax2 + bx + c = 0, where:
a = 1; b = Ka and c = - KaF
A quadratic equation (ax2 + bx + c = 0) has two possible solutions:
x1= -b + (b2 4ac)1/2
2a
and
x2 = -b - (b2 4ac)1/2
2a
Because the concentration of H+ can not be negative (it doesnt make
sense!), we reject x2
=> [H+] = -Ka + [Ka2 4(1)(-KaF)]1/2
2(1)
or
[H+] = -Ka + (Ka2 + 4KaF)1/2
2
171
Example: What is the pH of a solution containing 0.0200M of benzoic acid?
[H+] = -Ka + (Ka2 + 4KaF)1/2
2
[H+] = - 6.28x10-5 + [(6.28x10-5)2 + 4(6.28x10-5)(0.0200)]1/2
2
[H+] = 1.09 x 10-3M
172
Checking our assumption (or approximation):
> We know that:
[H+]Total = [H+]H2O + [H+]HA
> Our approximation was that [H+]HA >> [H+]H2O or [H+]H2O = 0
> Substituting [H+] = 1.09 x 10-3M in Kw:
[OH-] = Kw / [H+] = 1.0 x 10-14 / 1.09 x 10-3 = 9.20 x 10-12M
> Recall that:
H2O <=> H+ + OH[OH-] = [H+]H2O = 9.20 x 10-12M
=> our approximation is correct because:
[H+]HA = 1.09 x 10-3M >> [H+]H2O = 9.20 x 10-12M
173
Fraction of Dissociation
Fraction of dissociation of a weak monoprotic acid is calculated as:
[A-]
[A-] + [HA]
We know that: [A-] = [H+]
From the acid dissociation:
HA <=>
FC
F-x
H+ +
x
A-
x
=> [A-] + [HA] = F-x + x = F
So, the fraction of dissociation will be: [A-] / F
In the case of benzoic acid:
[A-] = 1.09 x 10-3M
F = 0.0200M
[A-] / F = 1.09 x 10-3M / 0.0200M = 0.054
174
pH of Weak Bases: Weak-Base Equilibrium
The treatment of weak bases is similar to the treatment of weak acids
Hydrolysis reaction for a weak base:
B + H2O <=> BH+ + OH- Kb = [BH+][OH-] / [B]
The total concentration of OH- is given by:
[OH-]Total = [OH-]H2O + [OH-]B
Assuming that [OH-]B >> [OH-]H2O => [OH-]Total [OH-]B
From the hydrolysis reaction:
B + H2O <=>
BH+ + OHIC
F
00
FC
F-x
x
x
=> [BH+] = [OH-] = x
[B] = F-x
Substituting these values in Kb:
x2 / F-x = Kb => x2 = (F-x) Kb => x2 = FKb Kbx
=> x2 + Kbx FKb = 0
Because x = [OH-]:
[OH-]2 + Kb [OH-] - FKb = 0
=> [OH-] = -Kb + [Kb2 + 4KbF]1/2
2
175
Example: Find the pH of a 0.0372M solution of the commonly encountered
weak base cocaine:
[OH-] = -Kb + [Kb2 + 4KbF]1/2
2
Kb = 2.6 x 10-6 and F = 0.0372M
=> [OH-] = -2.6 x 10-6 + [(2.6 x 10-6 )2 + 4(2.6 x 10-6 )(0.0372)]1/2 = 3.10 x
10-4M
2
[H+] = Kw / [OH-] = 1.0 x 10-14 / 3.10 x 10-4 = 3.22 x 10-11M => pH = 10.49
176
Its always good to check our assumption:
From the auto-dissociation of water:
[OH-] H2O = [H+] => [OH-] H2O = 3.22 x 10-11M
In fact:
[OH-] H2O = 3.22 x 10-11 << [OH-]B = 3.10 x 10-4M
=> Our assumption is correct
Fraction of Association of a Base
From the hydrolysis reaction:
B + H2O <=> BH+ + OHThe fraction of B that reacts with water is given by:
[BH+] / [B] + [BH+] ; where [B] + [BH+] = Concentration of base (F or M)
Example: former case of cocaine
[B] + [BH+] = 0.0372M
[BH+] = [OH-] = 3.10 x 10-4M
Fraction of Association = 3.10 x 10-4M / 0.0372M = 0.0083 = 0.83%
=> Only a very small fraction of cocaine dissolves in water
177
Finding the pH of Salts
Salts can be separated in the following groups:
a) Salt of a strong acid and a strong base
b) Salt of a weak acid and a strong base
c) Salt of a strong acid and a weak base
d) Salt of a weak acid and a weak base
pH of Salts of Strong Acids and Strong Bases
Examples:
HCl + NaOH NaCl + H2O
HI + KOH KI + H2O
HCl + KOH KCl + H2O
or any combination between an acid and a base
shown in Table 8-1
When this type of salts dissolve in water, the [H+] and the [OH-] are not
affected
Example: KI in water
KI K+ + ClH2O <=> H+ + OH- Kw = 1.0 x 10-14
=> [H+] = [OH-] = x
x2 = Kw => x = (1.0 x 10-14)1/2 = 1.0 x 10-7M
178
=> pH = 7
The pH of any salt of strong acid and strong base is equal to 7!
pH of a Salt of a Weak Acid and a Strong Base
Examples:
CH3COOH + NaOH CH3COONa +
Acetic Acid
C7H5O2H
Benzoic Acid
H2O
Sodium Acetate
+
KOH
C7H5O2K
+
H2O
Potassium Benzoate
or any combination of any weak acid from Appendix B and any strong base
from Table 8-1
When a salt of this type dissolves in water, a hydrolysis reaction occurs
between the anion of the weak acid and water that increases the
concentration of OH- in aqueous solution
Example: Potassium Benzoate in water
C7H5O2K C7H5O2- + K+
C7H5O2- + H2O <=> C7H5O2H + OH- Hydrolysis reaction
Kb = Kw / Ka, where Ka is the dissociation constant of Benzoic Acid
For any given salt of a weak acid (HA) and a strong base:
Hydrolysis reaction: A- + H2O HA + OHKb = Kw / Ka
where Ka is the dissociation constant of the weak acid
179
The anion of a weak acid behaves as a weak base in aqueous solution
Its pH is calculated in the same manner the pH of a weak base is calculated
The pH of any salt of a weak acid and a strong base is higher than 7
For any given salt of a weak acid (HA) and a strong base:
Hydrolysis reaction:
A- + H2O HA + OH-
Kb = Kw / Ka
where Ka is the dissociation constant of the weak acid
Example:
Find the pH of a 0.01M solution of sodium acetate
CH3COONa or NaC2H3O2
180
pH of salts of Strong Acids and Weak Bases
Examples:
HCl + NH3 NH4Cl
Any combination of an amine (primary, secondary or tertiary) with a strong
acid from Table 8-1
In aqueous solution:
NH4Cl NH4+ + ClNH4+ + H2O
<=> NH3 + H3O+ Hydrolysis reaction: Ka = [NH3][H3O+]
[NH4+]
Ka = Kw / Kb; where Kb is the dissociation constant of the weak conjugate
base (NH3)
For any given salt (BHA) of a strong acid (HA) and a weak base (B):
Hydrolysis reaction: BH+ + H2O <=> B + H3O+
Ka = [B][H3O+] = Kw /
Kb
[BH+]
where Kb is the association constant of the weak base
Note: Because Appendix B correlates the dissociation constants of the fully
protonated forms, you can obtain the Ka value directly from the table
181
Example: Calculate the pH of a 0.25M solution of ammonium chloride
NH4Cl NH4+ + ClNH4+ + H2O <=> NH3 + H3O+
FC
0.26-x
x
Ka = 5.69 x 10-10
x
NH4+ behaves as a weak acid:
Directly from Appendix D
[H+] = -Ka + [Ka2 + 4KaF]1/2
2
[H+] = -5.69 x 10-10 + [(5.69 x 10-10)2 + 4(5.69 x 10-10)(0.25)]1/2
2
[H+] = 1.2 X 10-5M => pH = 4.92
The cation of a salt of a weak base behaves as a weak acid in aqueous
solution
Its pH is calculated in the same manner the pH of a weak acid is calculated
The pH of any salt of a weak base and a strong acid is lower than 7
For any given salt of a weak acid (BHA) and a strong base:
Hydrolysis reaction: BH+ + H2O B + H3O+ Ka = Kw / Kb
where Kb is the dissociation constant of the weak base
Appendix D provides the value of Ka directly
182
Chapter 9: Buffers
Buffers are formed when the following mixtures in aqueous solutions occur:
> Weak acid + salt of its conjugate base
> Weak base + salt of it conjugate acid
Examples of acidic buffers: weak acid + salt of its conjugate base:
Acetic acid + sodium acetate => CH3CO2H + CH3CO2Na
Benzoic acid + potassium benzoate => C7H5O2H + C7H5O2K
Any acidic buffer can be represented as the following generic mixture:
HA + NaA
It could be any other cation
The relevant reactions in aqueous solution are the following:
Dissociation of weak acid:
HA <=> H+ + AKa = [H+ ][A-] / [HA]
Complete dissociation of salt:
NaA Na+ + AHydrolysis reaction of conjugate base: A- + H2O <=> HA + OHKb = [HA][OH-] / [A-]
Because the cation of the salt plays no role on the chemistry of the solution,
we can also say that any acidic buffer is made of a weak acid and its
183
conjugate base: HA + A-
Re-arranging Ka:
Ka x [HA]/[A-] = [H+]
Applying log to both sides:
log[H+] = logKa + log [HA] / [A-]
Multiplying both sides by -1:
- log[H+] = - logKa - log [HA] / [A-]
Or
pH = pKa + log [A-]
Henderson-Hasselbalch Equation for an acidic buffer
[HA]
Example of basic buffer:
Ammonia + ammonium chloride: NH3/NH4Cl
Any basic buffer can be represented as the following generic mixture:
It could be any other anion
B / BHCl
The relevant reactions in aqueous solution are the following:
Association reaction of weak base :
Dissociation of salt
:
B + H2O <=> BH+ + OH-
Kb = [BH+][OH-] / [B]
BHCl BH+ + Cl-
Hydrolysis reaction of conjugate acid:
BH+ + H2O <=> B + H3O+ Ka = [B][H3O+] / [BH+]
184
Re-arranging Ka:
Ka x [BH+] / [B] = [H3O+]
Applying log to both sides:
log[H+] = logKa + log([BH+] / [B])
Multiplying both sides by -1:
- log[H+] = - logKa - log([BH+] / [B])
Or
pH = pKa + log [B]
[BH+]
Henderson-Hasselbalch Equation for a basic buffer
pKa applies to BH+
Another version of the HHE for a basic buffer
It is possible to obtain an equation based on the dissociation of the base.
Rearranging Kb:
Kb x [B]/[BH+] = [OH-]
Substituting [OH-]:
Kb x [B]/[BH+] = Kw/[H+]
185
Re-arranging previous equation:
[H+] = Kw/Kb x [BH+]/[B]
Applying log to both sides:
log[H+] = log Kw/Kb + log [BH+]/[B]
Multiplying both sides by -1:
- log[H+] = - log Kw/Kb - log [BH+]/[B]
Or
pH = log Kb/Kw + log [B]/[BH+]
pH = logKb logKw + log [B]/[BH+]
pH = -log(1.0 x 10-14) ( -logKb) + log [B]/[BH+]
Or
pH = 14 pKb + log [B]/[BH+] Another version of HHE for a basic buffer
Kb applies to B
186
For an acidic buffer, when [A-] = [HA]:
This is the case where the initial concentrations of the weak acid and the
salt of the weak acid are the same
From the HHE:
pH = pKa + log [A-]
[HA]
pH = pKa + log 1
pH = pKa
For a basic buffer, when [BH+] = [B]:
This is the case where the initial concentration of weak base and the salt
concentration of the weak base are the same
From the HHE:
pH = pKa + log [B]
[BH+]
pH = pKa + log 1
pH = pKa (Ka applies to BH+)
187
From the 2nd version of HHE:
pH = 14 pKb + log [B]/[BH+]
pH = 14 pKb + log 1
pH = 14 pKb (Kb applies to B)
FYI:
Recall that Kw = Kb x Ka where Ka applies to the weak conjugate acid of
the weak base
Substituting in the equation above:
pH = 14 (-logKw/Ka)
pH = 14 + logKw/Ka
pH = 14 + logKw logKa
pH = 14 + log(1.0x10-14) logKa
pH = 14 14 + pKa
pH = pKa (Ka applies to BH+)
188
Examples:
1) Calculate the pH of a buffer prepared by adding 10mL of 0.10M acetic acid to 20mL of 0.10M
sodium acetate
Acidic buffer: HA/ApH = pKa +log[A-]/[HA]
> Initial acetic acid concentration in solution is:
10mL x 0.10M / 30mL = 0.033M
> Initial sodium acetate concentration in solution is:
20mL x 0.10M / 30mL = 0.067M
> Equilibria:
HA <=> H+ + AKa = 1.75 x 10-5
NaA Na+ + AA- + H2O <=> HA + OH> Charge Balance: [+] = [-]
[Na+] + [H+] = [A-] + [OH-]
(1)
> Mass Balance:
[Na+] = 0.067M
(2)
[A-] + [HA] = 0.033M + 0.067M
(3)
> Approximations:
Because the concentration of salt is large in comparison to Ka, we can make the following
approximations in the charge balance equation:
[Na+] >> [H+] and [A-] >> [OH-]
=> [Na+] [A-]
From equation 2:
[Na+] = 0.067M = [A-]
From equation 3:
189
0.066M + [HA] = 0.033M + 0.067M => [HA] = 0.033M
Note:
What these approximations are telling us is that in calculating the pH of an
acidic buffer one can neglect the dissociation of the acid and the hydrolysis
reaction of the base:
=> [HA] = initial concentration of the weak acid
=> [A-] = initial concentration of the salt of conjugate base
We can apply the formula directly:
pH = pKa + log[A-]/[HA] or pH = pKa + log [salt of conjugate base]
[weak acid]
pH = -log (1.75 x 10-5) + log 0.067 / 0.033
pH = 4.76 + log 2
pH = 5.06
2) Find the pH of a solution prepared by dissolving 12.43g of tris (FM 121.14)
plus 4.67g of tris hydrochloride (FM 157.60) in 1.00L of water
190
2) Calculate the pH of a buffer prepared by adding 25mL of 0.20M ammonia to 50mL of 0.10M
ammonium chloride
> Initial concentration of weak base: 25mLx0.20M/75mL = 0.067M
> Initial concentration of salt of conjugate weak acid: 50mLx0.10M/75mL = 0.067M
> Equilibria:
NH3 + H2O <=> NH4+ + OH-
Kb = 1.8 x 10-5
NH4Cl NH4+ + ClNH4+ + H2O <=> H+ +NH3
> Charge balance:
[NH4+] + [H+] = [Cl-] + [OH-] (1)
> Mass balance:
[Cl-] = 0.067M
(2)
[NH3] + [NH4+] = 0.067M + 0.067M
(3)
> Approximations:
Because the concentration of salt is large in comparison to Kb, we can make the following
approximations in the charge balance equation:
[NH4+] >> [H+] and [Cl-] >> [OH-] => [NH4+] [Cl-]
From equation 2:
[Cl-] = 0.067M = [NH4+]
From equation 3:
[NH3] + [NH4+] = 0.067M + 0.067M => [NH3] = 0.134M 0.067M = 0.067M
191
Note:
What these approximations are telling us is that in calculating the pH of a
basic buffer one can neglect the association reaction of the weak base and
the hydrolysis reaction of conjugate weak acid:
=> [B] = initial concentration of the weak base
=> [BH+] = initial concentration of the salt of conjugate acid
We can apply the formula directly:
pH = pKa + log [B]
or pH = pKa + log [weak base]
[BH+]
[salt of conjugate acid]
pH = -log(5.69x10-10) + log0.067M/0.067M = 9.24 + log 1 = 9.24
Directly from appendix D
We can also use the formula:
pH = 14 pKb + log [B]/[BH+]
In this case, Kb is obtained from Kb=Kw/Ka = 1.0x10-14/5.69x10-10 = 1.75x10192
5
From the formula:
pH = 14 (-log1.75x10-5) + log 0.067M/0.067M
pH = 14 4.76 + log 1
pH = 9.24
3) Find the pH of a solution prepared by dissolving 12.43g of tris (FM 121.14)
plus 4.67g of tris hydrochloride (FM 157.60) in 1.00L of water.
FYI:
Tris is a very common buffer:
This is a basic buffer: pH = pKa + log[B]/[BH+]
[B] = 12.43 g/L/ 121.14 g/mol = 0.1026M
[BH+] = 4.67 g/L / 157.60g/mol = 0.0296M
pH = 8.07 + log (0.1026 / 0.0296) = 8.61
193
Preparing Buffers
Buffers are usually prepared by starting with a measured amount of either a
weak acid (HA) or a weak base (B). Then OH- is added to HA to make a
mixture of HA and A- (a buffer) or H+ is added to B to make of B and BH+ (a
buffer):
Acidic Buffer
Basic Buffer
HA
B
OHHA + A-
H+
B + BH+
194
How many milliliters of 0.500M NaOH should be added to10.0g of tris
hydrochloride to give a pH of 7.60 in a final volume of 250mL?
Tris hydrochloride = BH+ = weak acid
Initial number of moles = 10g/157.60g/mol = 0.0635
Reaction with OH-:
BH+
Initial number of moles
Final moles
0.0635-x
+
OH- B
0.0635
--- x
x
+
H2O
---
---
---
Number of moles added = number of moles formed
From the HHE equation: pH = pKa + log [B]/[BH+]
Because: [B] = number of moles of B / final volume of solution (250mL)
[BH+] = number of moles of BH+ / final volume of solution (250mL)
The final volume doesnt matter in the ratio: [B]/[BH+]
pH = pKa + log (x / 0.0635-x)
195
Since we know the final pH (i.e. the desired pH) and the pKa:
7.60 8.07 + log (x/0.0635-x)
-0.47 = log(x/0.0635-x)
If we raise 1o to the power in both sides:
10-0.47 = 10log(x/0.0635-x)
Remembering that 10logz = z:
10-0.47 = x / 0.0635 x
0.339 = x / 0.0635 x => 0.339 (0.0635 x ) = x
=> (0.339)(0.0635) 0.339x = x => 0.02153 = 1.339x
=> x = 0.02153 / 1.339 => x = 0.0161 mol
The volume of 0.500M NaOH will be:
M = n/V => V = n/M = 0.0161 mol/0.500 mol/L = 0.0322L = 32.2mL
Note:
Although the volume of NaOH can be theoretically calculated for a desired
pH, it is always convenient to monitor the pH of the buffer with a pH electrode
and adjust the final NaOH volume to the desired pH measured in the scale of
the pH electrode. The same is true when adding an strong acid to a weak base
196
Buffer Capacity
It is the ability of a buffer to resist to
changes in pH when acid or base is
added
The greater the buffer capacity, the
less the pH changes
The figure shows typical curves on the
effect of adding 0.01mol of H+ or OHto a buffer containing HA and A- (HA +
A- = 1 mol)
The Ka of the acid is 10-5
The minimum change in pH takes
place when the initial pH of the buffer
equals the pH for Ka
The buffer capacity is maximum when
pH = pKa for the buffer
197
The useful pH range of a buffer is usually considered to be pKa 1
=> Choose a buffer whose pKa is close to the desired pH
198
199
How does a buffer work?
The buffering mechanism for a mixture of a weak acid and its salt can be explained
as follows:
> The pH is governed by the HHE:
pH = pKa + log [A-]/[HA]
Resistance to dilution:
If the solution is diluted, the ratio [A-]/[HA] remains constant, and so the pH of the
solution does not change
Resistance to changes in pH:
HA <=> H+ + A(1)
A- + H2O <=> HA + OH(2)
Adding small amounts of strong acid:
> H+ from the strong acid will combine with A- to form HA according to the
equilibrium:
HA <=> H+ + A> So, adding a strong acid shifts the equilibrium to the left favoring the formation of
HA
> The formation of HA shifts the second equilibrium to the left favoring the formation
of A-. The amount of A- consumed in the first equilibrium is replaced by the second
equilibrium
=> [A-] constant
> The amount of HA formed in the first equilibrium is consumed by the second
equilibrium
=> [HA] constant
200
=> log [A-]/[HA] constant => pH constant
Ask Yourself
9-D. (a) look up pKa for each of the following acids and decide which one
would be best for preparing a buffer with pH 3.10
(i) hydroxybenzene
(ii) propanoic acid
(iii) cyanoacetic acid
(iv) sulfuric acid
From Appendix D:
(i) hydroxybenzene; pKa = 9.98;
(ii) propanoic acid; pKa = 4.87
(iii) cyanoacetic acid; pKa = 2.47
(iv) sulfuric acid; pKa = 1.99
(b) Why does a buffer capacity increase as the concentration of buffer
increases?
A buffer resists changes in pH because the added acid or base is
consumed by the buffer. The higher the concentration of the buffer, the higher
the resistance to pH change
201
Chapter 10: Acid-Base Titrations
For every type of titration discussed in this chapter our goal is to construct a
graph showing how the pH changes as titrant is added
Titration of strong acid with strong base
Example: titration of 50.00mL of 0.02000M KOH with 0.1000M HBr (titrant)
1st step: calculate the volume of HBr needed to reach the equivalence point
> Titration Reaction:
KOH + HBr NaBr + H2O
Or
OH- + H+ H2O
> At the equivalence point:
number of moles of KOH = number of moles of HBr
MKOH x VKOH = MHBr x VHBr
Or
number of equivalents of KOH = number of equivalents of HBr
NKOH x VKOH = NKOH x VKOH
202
1.
2.
3.
.
Ve (mL) x 0.1000M = 50.00mL x 0.02000M => Ve = 10.00mL
When 10.00mL of HBr have been added, the titration is complete.
Prior to Ve, unreacted OH- is present
After Ve, there is excess H+ in the solution
In the titration of a strong base with a strong acid, three regions of the titration
curve require different kinds of calculations:
Before the EP, the pH is determined by excess OH- in the solution
At the equivalence point, added H+ is just sufficient to react with all the OH- to
make H2O. The pH is determined by the dissociation of water
After the equivalence point, pH is determined by excess H+ in the solution
Before the equivalence point:
[OH-] remaining = [OH-] initial [OH-] consumed by HBr
Example:
If we add 3.00mL of HBr, we add:
3.00mL x 0.1000M = 0.300mmol of H+
number of moles of OH- consumed = 0.300mmol of OHOH- remaining = 1.000mmol 0.300mmol = 0.700mmol
[OH-]remaining = 0.700mmol / 50mL + 3mL = 0.0132M
From the Kw: [H+] = Kw / [OH-] = 1.0 x 10-14 / 0.0132 = 7.58 x 10-13M => pH =
12.12
203
At the equivalence point:
The pH is determined by the auto-dissociation of water:
H2O <=> H+ + OH-
Kw = [H+][OH-] = 1.0 x 10-14
[H+] = [OH-] => [H+]2 = Kw => [H+] = 10-7M => pH = 7
pH = 7 at the equivalence point of any titration between a strong acid and a
strong base
After the equivalence point:
Beyond the equivalence point, excess HBr is present. The pH is given by
excess H+
[H+] excess = [H+] HBr added [OH-]initial
Example:
10.50mL of HBr have been added:
H+ excess = 10.50 x 0.1000 50.00 x 0.02000 = 1.05 1.0 = 0. 05mmol
H+
[H+] excess = 0.05mmol / 50 + 10.50 mL = 8.26 x 10-4M => pH = 3.08
204
205
Titration of a weak acid with a strong base
The titration calculations for this case
are shown in the figure:
1. Initial point, i.e. before any
base is added
2. From the first addition of strong
base until immediately before the
equivalence point
3. At the equivalence point
4. Beyond the equivalence point
206
Before any base is added:
> All there is in solution is the weak acid (HA) and water:
HA <=> H+ + A- Ka = [H+][A-] / [HA] = x2 / F-x
F-x
x
x
Or
[H+] = -Ka + [Ka2 + 4FKa]1/2
2
From the first addition of base until immediately before the equivalence
point:
> The titration reaction is:
HA + OH- A- + H2O
=> Main species in solution: HA / A- =>
Buffer
> pH = pKa + log [A-] / [HA]
At the equivalence point:
> All weak acid was consumed; only A- remains in solution
> A- undergoes hydrolysis reaction with water:
A- + H2O <=> HA + OH- Kb = [HA][OH-] / [A-] = x2 / F-x
F-x
x
x
Or
A- is a weak base and its pH can be also calculated as follows:
[OH-] = -Kb + [Kb2 + 4FKb]1/2 and [H+] = Kw / [OH-]
207
2
Beyond the equivalence point:
> There is an excess of strong base and the pH is calculated as:
[H+] = Kw / [OH-] excess => pH = - log {Kw / [OH-] excess}
Example:
Titration of 50.00mL of 0.02000M MES with 0.1000M NaOH
MES = 2-(N-morpholino)ethane-sulfonic acid
The titration reaction is:
Remember that is always helpful to calculate the volume of base (Ve) needed to
reach the equivalence point (EP):
number of moles of acid = number of moles of base
(stoichiometry: 1:1)
MMES x VMES = MNaOH x Ve => 0.02000M x 50.00mL = 0.1000M x Ve
=> Ve = 0.02000M x 50.00mL / 0.1000M = 10.00mL
208
Before any base is added: weak acid (HA)
[H+] = -Ka + [Ka2 + 4FKa]1/2
2
pKa = 6.72 => Ka = 10-6.72 = 5.37 x 10-7
[H+] = - 5.37 x 10-7 + [ (5.37 x 10-7 )2 + 4 (0.02000)(5.37 x 10-7 )]1/2
2
[H+] = 1.03 x 10-4 => pH = 3.99
Before the equivalence point (0mL < VNaOH < 10mL): buffer (HA/A-)
> pH = pKa + log [A-] / [HA] = pKa + log number of m-moles Anumber of m-moles HA
> Titration reaction is:
HA + OH- A- + H2O
(stoichiometry 1:1:1)
number of m-moles of A- = number of m-moles of OHnumber of m-moles of HA = (initial number of m-moles of HA) (number of m-moles of OH-)
209
Consider where 3.00mL of OH- have been added:
number of m-moles of A- = number of m-moles of OH- = 3.00mL x 0.1000M =
0.300
number of m-moles of HA = (initial number of m-moles of HA) (number of mmoles of OH-)
number of m-moles of HA = (50.00mL x 0.02000M) 0.300
number of m-moles of HA= 1.000 0.300 = 0.700 m-mol
pH = pKa + log m-moles A- / m-moles HA = 6.27 + log 0.300 / 0.700 = 5.90
Consider where 5.00mL of OH- have been added:
number of m-moles of A- = number of m-moles of OH- = 5.00mL x 0.1000M =
0.500
number of m-moles of HA = (50.00mL x 0.02000M) 0.500 =0.500
pH = pKa + log m-moles A- / m-moles HA = 6.27 + log 0.500 / 0.500 =6.27
Note: 5.00mL = 1/2Ve
Always: VBase = 1/2Ve => pH = pKa
At the equivalence point: weak base (A-)
A- + H2O <=> HA + OHKb
[OH-] = -Kb + [Kb2 + 4FKb]1/2
2
F = Concentration of A- = initial number of m-moles of HA / 50mL +
10mL = 1.000mmol / 60mL = 0.01667M
210
Kb = Kw / Ka = 1.0 x 10-14 / 5.37 x 10-7 = 1.86 x 10-8
[OH-] = - 1.86 x 10-8 + [(1.86 x 10-8 )2 + 4 x 0.01667 x 1.86 x 10-8 ]1/2
2
[OH-] = 1.76x10-5M
pH = -log Kw / 1.78x10-5 = 9.25
After the equivalence point: excess of titrant (OH-)
HA + OH- A- + H2O
number of mmoles OH- excess = (mmoles of Base added) (initial number of mmoles HA)
Consider the addition of 10.10mL of NaOH:
number of mmoles OH- excess = 10.10mLx0.1000M 50.00mLx0.02000M = 0.01mmol
[OH-] excess = 0.01mmol / 50mL + 10.10mL = 1.66 x 10-4M
[H+] = Kw / [OH-] => pH = -log 1.0x10-14/1.66x10-4 = 10.22
211
212
Titration of a weak base with a strong acid
The titration reaction is:
B + H+ BH+
1.
2.
There are four distinct regions for
the titration curve:
Before the acid is added
Between the initial point and the
equivalence point (0 < V < Ve)
3.
At the equivalence point
4.
After the equivalence point
213
Before any acid is added:
There is only a weak base in solution:
B
+
H2O <=> BH+ + OH- Kb = [BH+][OH-] / [B]
Fx
x
x
Kb = x2 / F-x
Or
[OH-] = -Kb + [Kb2 + 4FKb]1/2
2
Between the initial point and the equivalence point:
0 < V < Ve, a buffer is formed in solution:
B + H+ BH+
Buffer: B / BH+
pH = pka + log [B] / [BH+] Note: pKa for BH+
Because the volume is the same for both species:
pH = pKa + log {# of moles of B / # of moles of BH+}
At the special point where [B] = [BH+] or # of moles of B = # of moles of
BH+
pH = pKa + log 1 => pH = pKa
214
At the equivalence point:
B has been completely consumed. The main species in solution is BH+
BH+ is a weak acid and dissociates in water according to the equilibrium:
BH+ + H2O <=> B + H3O+
F-x
x
x
Ka = [B][H3O+] / [BH+]
Ka = x2 / F-x
Or
[H+] = -Ka + [Ka2 + 4FKa]1/2
2
At the equivalence point, the pH must be below 7
After the equivalence point:
There is an excess of strong acid. The pH is given by the excess of strong
acid:
pH = - log [H+]
215
Example: Titration of 25mL of 0.08364M pyridine with 0.1067M HCl:
It is always a good idea to find out the equivalence volume:
# mmoles of pyridine = # of mmoles of HCl
M pyridine x V pyridine (mL) = M HCl x Ve (mL)
0.08364M x 25mL = 0.1067M x Ve (mL) => Ve = 19.60mL
a)
Find the pH when VHCl = 4.63mL
b)
Find the pH at the equivalence point
216
VHCl = 4.63mL
This is before the equivalence point (0 mL < VHCl < Ve = 19.60mL):
There is a buffer in solution: B / BH+
pH = pKa + log mmoles B /mmoles BH+
pKa refers to BH+ => Ka = Kw / Kb = 1.0 x 10-14 / 1.6 x 10-9 = 6.3 x 10-6
Initial # of mmoles of B is: 0.08364M x 25mL = 2.091
# of mmoles of HCl added: 0.1067M v 4.63mL = 0.494
Titration reaction is:
B
+
H+ BH+
2.091
0.494
2.091 0.494
1.597
0
0
0.494
0.494
pH = -log 6.3 x 10-6 + log 1.597 / 0.494 = 5.71
At the equivalence point:
All B is gone. There is only BH+. Because the reaction is 1:1:1, the number of
mmoles of BH+ formed is equal to the initial number of mmoles of B:
BH+
<=> B
+
H+
F x
x
x
Ka = [B][H+] / [BH+] = x2 / F-x
F = 2.091 mmol / 25 + 19.60 mL = 0.04688M
x2 / 0.04688 x = Ka = 6.3 x 10-6 => x = [H+] = 5.40 x 10-4M
217
pH = -log 5.40 x 10-4 = 3.27
Or:
[H+] = -Ka + [Ka2 + 4FKa]1/2 = - 6.3 x 10-6 + [(6.3 x 10-6 )2 + 4 (0.04688)(6.3 x 10-6)]1/2
2
2
[H+] = 5.40 x 10-4M => pH = 3.27
218
Chapter 11: Polyprotic Acids and Bases
Section 11-1: Amino Acids are Polyprotic
Amino-acids from which proteins are built have an acidic carboxylic group, a basic
amino group and a substituent group:
Because the amino group is more basic than the carboxyl group, the acidic proton
resides on nitrogen of the amino group instead of on oxygen of carboxylic group
The resulting structure, with positive and negative sites, is called a zwitterion
At low pH, both the amino group and the carboxylic group are protonated
At high pH, neither group is protonated
The substituent (R) may also have acidic or basic properties
Acid dissociation constant of the 20 common amino acids are given in Table 11-1
219
220
Example: Cysteine
Cysteine has three acidic protons:
H3A+ <=> H2A + H+ Ka1 = 0.020
H2A <=> HA- + H+
HA- <=> A2- + H+Ka3 = 1.8 x 10-11
Ka2 = 4.4 x 10-9
221
Calculating pH of diprotic acids and their salts
In general, a diprotic acid has two acid dissociation constants:
H2A <=> HA- + H+
Ka1 or K1 = [HA-][H+] / [H2A]
HA- <=> A2- + H+
Ka2 or K2 = [A2-][H+] / [HA-]
Always: Ka1 > Ka2
The equilibrium for the total dissociation is:
H2A <=> HA- + H+
+
HA- <=> A2- + H+
H2A <=> A2- + 2 H+ K = Ka1 x Ka2 = [HA-][H+] / [H2A] x [A2-][H+] /
[HA-]
K = [H+]2[A-] / [H2A]
Example: H2CO3
H2CO3 <=> HCO3- + H+ Ka1 = [H+][HCO3-] / [H2CO3] ; pKa1 = 6.34
HCO3- <=> CO32- + H+
10.36
Total dissociation:
Ka2 = [H+][CO32-] / [HCO3-] ; pKa2 =
222
When H2A dissociates in aqueous solution, it forms two conjugate bases:
HA- and A2The conjugate pairs are: H2A/HA- and HA-/A2The association constants and association reactions of the conjugate bases
are:
A2- + H2O <=> HA- + OH- Kb1= [HA-][OH-] / [A2-]
HA- + H2O <=> H2A + OH-
Kb2 = [H2A][OH-] / [HA-]
Always: Kb1> Kb2
Note that:
Kb1= [HA-][OH-] = Kw / Ka2 = [H+][OH-] x
[A2-]
[A2-][H+]
Kw
[HA-]
= [HA-][OH-]
[A2-]
1/Ka2
Kb2 = [H2A][OH-] = Kw / Ka1 = [H+][OH-] x [H2A] = [H2A][OH-]
[HA-]
[HA-][H+]
Kw
[HA-]
1/Ka1
223
This is always true. For a diprotic acid:
Ka1 x Kb2 = Kw
Ka2 x Kb1 = Kw
In a solution of a diprotic acid all three species (H2A, HA- and A2-) are
present to some extent
The questions are:
1. Which one is the predominant species?
2. How do we calculate the pH due to the predominant species?
pH of an H2A solution
Example: Suppose you have a 0.10M solution of a diprotic acid H2A where
Ka1 = 1.0 x 10-3 and Ka2 = 1.0 x 10-7:
H2A <=> HA- + H+
Ka1 = 1.0 x 10-3
HA- <=> A2- + H+
Ka2 = 1.0 x 10-7
[H+]TOTAL = [H+]Ka1 + [H+]Ka2
Note: we neglect [OH-] H2O because Kw << Ka1 and Ka2
Because Ka1 >> Ka2, we can assume that [H+]Ka1 >> [H+]Ka2
=> [H+]TOTAL [H+]Ka1
Considering only the first equilibrium:
H2A
<=>
HA+ H+ Ka1 = 1.0 x 10-3
0.10 x
x
x
x2 / 0.10 x = 1.0 x 10-3
Or
[H+] = -Ka1 + [Ka12 + 4FKa1]1/2
224
[H+] = - 1.0x10-3 + [(1.0x10-3)2 + 4x0.10x 1.0x10-3)1/2 = 0.0095M
2
From the equilibrium reaction:
[H+] = [HA-] = 0.0095M
[H2A] = 0.10 0.0095 0.10M
[A2-] is obtained from the Ka2:
Ka2 =[A2-][H+] / [HA-] => [A2-] = Ka2 x [HA-] / [H+] = 1.0x10-7 x
0.0095/0.0095
=> [A2-] = 1.0x10-7M
Note that [H+]Ka2 = [A2-] = 1.0x10-7M. So our assumption that [H+]Ka1 >>
[H+]Ka2 is correct
The pH of H2A (Ka1/Ka2 > 10) can be calculated as the pH of a monoprotic
weak acid
225
pH of a Na2A solution
Example: Lets calculate the pH of a 0.10M solution of Na2A and the
concentrations of H2A, HA- and A2-.
The main species in solution is A2-:
Na2A 2Na+ + A20.10M = Initial Concentration
A2- is a weak base:
A2- + H2O <=> HA- + OH-
Kb1 = Kw / Ka2 = 1.0x10-7
HA- + H2O <=> H2A + OH-
Kb2 = Kw / Ka1 = 1.0 x 10-11
[OH-] Total = [OH-] Kb1 + [OH-] Kb2
Note: we neglect [OH-]H2O because Kw << Kb1 and Kb2
Comparison of the two base constants tells us that A2- is a much stronger
base than HA-:
[OH-] Total [OH-] Kb1
=> [OH-] = [HA-]
and [A2- ] = 0.10 [OH-] 0.10M
226
Substituting in Kb1:
[HA-][OH-] / [A2-] = Kb1
[OH-]2 / 0.10 = 1.0x10-7
[OH-] = 1.0x10-4M => pOH = 4
pH + pOH = 14 => pH = 10
Since [OH-] = [HA-] => [HA-] = 1.0x10-4M
The concentration of H2A is obtained from Kb2:
[H2A][OH-] / [HA-] = Kb2
[H2A] = Kb2 x [HA-] / [OH-] = 1.0x10-11 x 1.0x10-4 / 1.0x10-4 = 1.0x10-
11M
Note that [H2A] = [OH-] Kb2 = 1.0x10-11M << [OH-] Kb1 = 1.0x10-4M; so
our assumption was correct
The pH of N2A (Kb1/Kb2 > 10) can be calculated as the pH of a monoprotic
weak base
227
pH of a NaHA solution
Do not panic! We will derive a formula that can be applied to all NaHA cases
The main species in a NaHA solution is:
NaHA Na+ + HA-
HA- is an amphoteric species, i.e. it can act as a weak acid and as a weak
base:
Salt:
NaHA Na+ + HA-
Weak acid:
HA- <=> H+ + A2-
Ka2 = 1.0x10-7
Weak base: HA- + H2O <=> H2A + OH-
Kb2 = Kw/Ka1 = 1.0x10-
11
Charge Balance:
[Na+] + [H+] = [HA-] + 2[A2-] + [OH-]
(1)
Mass Balance:
[HA-] + [A2-] + [H2A] = Initial Concentration of Salt (Ci)
[Na+] = Ci
(2)
(3)
Adding equations (1) and (2) and noting that [Na+] = Ci
[Na+] + [H+] + [HA-] + [A2-] + [H2A] = [HA-] + 2[A2-] + [OH-] + Ci
228
We obtain:
[H+] + [H2A] = [A2-] + [OH-]
(4)
From Ka1:
Ka1 = [H+][HA-] / [H2A] => [H2A] = [H+][HA-] / Ka1
From Ka2:
Ka2 = [A2-] [H+] / [HA-] => [A2-] = Ka2 x [HA-] / [H+]
From Kw:
[OH-] = Kw / [H+]
Substituting these three in equation (4):
[H+] + [H+][HA-] = Ka2 x [HA-] + Kw
Ka1
[H+]
[H+]
Multiplying by [H+]:
[H+]2 + [H+]2[HA-] = Ka2 x [HA-] + Kw
Ka1
Factoring out [H+]2:
[H+]2 {1 + [HA-]} = Ka2 x [HA-] + Kw
Ka1
[H+]2 {Ka1+ [HA-]} = Ka2 x [HA-] + Kw
Ka1
229
[H+]2 = [HA-]Ka2 + Kw
Ka1 + [HA-]
Ka1
Or
[H+]2 = Ka1Ka2[HA-] + Ka1 Kw
Ka1 + [HA-]
1/2
=>[H+] = Ka1Ka2[HA-] + Ka1 Kw
Ka1 + [HA-]
Since [HA-] is the main species in solution, from equation (2):
[HA-] Ci
1/2
=>[H+] = Ka1Ka2Ci + Ka1 Kw
Ka1 + Ci
230
Example: Calculate the pH of a 0.10M solution of NaHA and the
concentration of the various species in solution
1/2
[H+]
= (1.0 x 10-3)(1.0x10-7)(0.10) + (1.0 x 10-3)(1.0x10-14)
1.0x10-3 + 0.10
[H+] = 9.95x10-6 1.0x10-5M => pH = 5
From Kw:
[OH-] = 1.0x10-14/1.0x10-5 = 1.0x10-9M
From Ka2:
(1.0x10-5)[A2-] = 1.0x10-11 => [A2-] = 1.0x10-7M
0.10
From Ka1:
(1.0x10-5)(0.10) = 1.0x10-3 => [H2A] = 1.0x10-3M
[H2A]
231
Note that the approximation [HA-] Ci made from equation (2) is good:
[HA-] + [A2-] + [H2A] = Initial Concentration of Salt (Ci) = 0.10M
1.0x10-3 + 0.10 + 1.0x10-7 0.10
For any salt NaHA, where Ci >> Ka1 and Ka2:
1/2
=>[H+] = Ka1Ka2Ci + Ka1 Kw
Ka1 + Ci
If:
Ci >> Ka1 => Ka1 + Ci Ci
And
Ka1Ka2Ci >> Ka1 Kw => Ka1Ka2Ci + Ka1 Kw Ka1Ka2Ci
1/2
=> [H+] = Ka1Ka2Ci
Ci
=> [H+] = (Ka1Ka2)1/2 => log[H+] = log(Ka1Ka2)
-log[H+] = [-logKa1-logKa2 ] => pH = (pKa1 + pKa2)
232
In the previous NaHA case, where: Ci = 0.10M; Ka1 = 1.0x10-3 and Ka2=
1.0x10-7:
pH = (3 + 7) = 5, which is the same result obtained without using the
approximation
Section 11-3: Which is the Principal Species?
Consider a weak monoprotic acid, HA. When HA dissolves in water, it
dissociates according to the well-know equilibrium:
HA + H2O <=> H3O+ + A-
Or
HA <=> H+ + A
The predominant species in solution depends on the pH of the solution
The ratio between [HA] and [A-] is given by the HH equation:
pH = pKa + log [A-] / [HA]
Notes:
1. Do not get confused! The pH of HA is not calculated with this equation
2. However, you should remember that this equation is obtained from the
Ka of the weak acid (see your notes) and, therefore, it can be used to predict
the ratio between [A-] and [HA] of a solution of a weak acid at any given pH
233
Consider benzoic acid:
=> pH = 4.20 + log [A-] / [HA]
Setting the pH of the solution at 4.20:
log [A-] / [HA] = 0 => [A-] / [HA] = 100 = 1 => [A-] = [HA]
Setting the pH of the solution to pKa + 1 = 5.20:
5.20 = 4.20 + log [A-] / [HA] => log [A-] / [HA] = 1 => [A-] / [HA] = 101
=> [A-] = 10 [HA]
Setting the pH of the solution to pKa + 2 = 6.20:
6.20 = 4.20 + log [A-] / [HA] => log [A-] / [HA] = 2 => [A-] / [HA] = 102
=> [A-] = 100 [HA]
Setting the pH of the solution to pKa -1 = 3.20:
3.20 = 4.20 + log [A-] / [HA] => log [A-] / [HA] = -1=> [A-] / [HA] = 10-1
=> [A-] = 0.1[HA] or [HA] = 1 / 0.1[A-] => [HA] = 10 [A-]
234
Setting the pH of the solution to pKa 2 = 2.20:
2.20 = 4.20 + log [A-] / [HA] => log [A-] / [HA] = -2=> [A-] / [HA] = 10-2
[A-] = 0.01[HA] or [HA] = 1 / 0.01[A-] => [HA] = 100 [A-]
In summary, for monoprotic systems we can say that:
At pH values below pKa, HA is predominant
At pH values above pKa, A- is predominant
When pH = pKa, the concentrations of HA and A- are the same
235
For diprotic systems, the reasoning is similar, but there are two pKa values:
H2A <=> HA- + H+
Ka1
HA- <=> A2- + H+Ka2
From Ka1, the following equation is obtained:
pH = pKa1 + log [HA-] / [H2A]
From Ka2, another equation is obtained:
pH = pKa2 + log[A2-] / [HA-]
At pH = pKa1:
pKa1 = pKa1 + log [HA-] / [H2A] => 0 = log [HA-] / [H2A] => [HA-] / [H2A] =
100
=> [HA-] / [H2A] = 1 => [HA-] = [H2A]
At pH = pKa2:
pKa2= pKa2 + log[A2-] / [HA-] => 0 = log[A2-] / [HA-] => [A2-] / [HA-] = 100
=> [A2-] / [HA-] = 100 => [A2-] = [HA-]
236
The chart below shows the predominant species in each pH region for a
diprotic system:
At pH values above pKa2, A2- is predominant
At pH values below pKa1, H2A is predominant
At pH values between pKa1 and pKa2, HA- is predominant
237
For a triprotic system:
H3A <=> H2A- + H+ Ka1 => pH = pKa1 + log [H2A-] / [H3A]
H2A- <=> HA2- + H+ Ka2 => pH = pKa2 + log[HA2-] / [H2A-]
HA2- <=> A3- + H+ Ka3 => pH = pKa3 + log[A3-] / [HA2-]
The chart below shows the predominant speciesin each pH region for a
triprotic system:
1/2
1/2
At pH below pKa1, H3A is the predominant species
H2A- is the predominant species between pH = pKa1and pH = pKa2
HA2- is the predominant species between pH = pKa2 and pH = pKa3
238
At pH above pKa3, A3- is the predominant species
To calculate the pH of any triprotic system you should:
Treat H3A as a monoprotic weak acid:
Ka1 = x2 / Ci x; where x = [H+]
Treat A3- as monoprotic weak base:
Kb1 = Kw/ Ka3 = x2 / Ci x; where x = [OH-]
Treat H2A- as intermediate:
pH = (pKa1 + pKa2)
Treat HA- as an intermediate:
pH = (pKa2 + pKa3)
239
Fractional Composition Diagrams (FCD)
Consider a biprotic system (H2A):
CH2A = [H2A] + [HA-] + [A2-]
Where:
CH2A = analytical concentration of acid or initial total concentration of acid
[H2A], [HA-] and [A2-] are the equilibrium concentrations of the individual
species
The fractions or concentrations of individual species at equilibrium are
defined as:
H2A = [H2A] / CH2A
HA- = [HA-] / CH2A
A2- = [A2-] / CH2A
Note:
H2A + HA- + A2- = 1
240
The value of each fraction depends on the pH of the solution:
H2A <=> HA- + H+ => [HA-] = Ka1 x [H2A]/[H+] (1)
HA- <=> A2- + H+ => [A2-] = Ka2 x [HA-]/[H+]
(2)
Substituting (1) in (2):
[A2-] = Ka1x Ka2 [H2A]/[H+]2
(3)
Recall that:
CH2A = [H2A] + [HA-] + [A2-]
(4)
Substituting (1) and (3) in (4):
CH2A = [H2A] + Ka1 x [H2A] + Ka1x Ka2 [H2A]
[H+]
[H+]2
Dividing each side of the expression by [H2A]:
CH2A / [H2A] = [H2A] / [H2A] + Ka1 x [H2A] + Ka1x Ka2 [H2A]
[H+] [H2A]
[H+]2 [H2A]
1/ H2A = 1 + Ka1/[H+] + Ka1 x Ka2 / [H+]2
Or
1/ H2A = [H+]2 + Ka1[H+] + Ka1 x Ka2
[H+]2
241
Or:
H2A =
[H+]2
[H+]2 + Ka1[H+] + Ka1 x Ka2
A similar approach gives the following fractions for the other species:
HA-=
Ka1[H+]
[H+]2 + Ka1[H+] + Ka1 x Ka2
A2- = Ka1 x Ka2
[H+]2 + Ka1[H+] + Ka1 x Ka2
242
243
Chapter 13: EDTA Titrations
Ethylenediaminetetraacetic acid (EDTA) is the most widely used chelator in
analytical chemistry
A chelate is a multidentate ligand
Multidentate ligand: binds to a metal ion through more than one ligand
atom:
EDTA forms strong 1:1 complexes with most metal ions, binding through
four oxygen and two nitrogen atoms
244
Some ligands are monodentate
ligand because only one atom of
the ligand binds to the metal
Example: Cyanide
Metal ligand reactions can be
seen
as
Lewis
Acid-Base
reactions:
Lewis
acceptors
acids:
electron
pair
Lewis
donators
bases:
electron
pair
Monodentate ligands
chelate ligans
are
not
245
In addition to EDTA, other examples of aminocarboxylic acids whose
nitrogen and carboxylate oxygen atoms can lose protons and bind to metal
ions are:
The chelating agents form strong 1:1 complexes with all metal ions, except
univalent ions such as Li+, Na+ and K+
246
The stoichiometry is 1:1 regardless of the charge on the ion
EDTA
EDTA is a hexaprotic system represented as H6Y2+:
The blue hydrogen atoms are the ones lost on metal-complex formation
Neutral EDTA is tetraprotic, with the formula H4Y
A common EDTA reagent is the disodium salt Na2H2Y.2H2O
247
The equilibrium constant for the
reaction of a metal with a ligand is
called the formation constant, Kf,
or the stability constant:
Mn+ + Y4- <=> MYn-4
Kf = [MYn-4] / [Mn+][Y4-]
Formation constant for EDTA are
large and tend to be larger for
more positively charged ions
Note that stability constants are
defined for reactions of the
species Y4- with the metal ion
At low pH, most EDTA is in one of
its protonated forms, not Y4At low pH H+ competes with the
metal ion for EDTA and metalEDTA
complexes
become
unstable
At high pH, OH- competes with
EDTA for the metal ion and EDTA
complexes become unstable
248
There is an optimum pH for metal-EDTA complex formation
Formation Constants
Complexes are formed step-wise. Step-wise formation constants (Ki) are
defined as follows:
M + X <=> MX
K1 = [MX] / [M][X]
MX + X <=> MX2 K2 = [MX2] / [MX][X]
MX2 + X <=> MX3
.
.
.
.
.
.
.
.
K3 = [MX3] / [MX2][X]
.
MXn-1 + X <=> MXn Kn = [MXn] / [MXn-1][X]
Where:
M is the metal ion and X is the ligand
249