Chemistry, Acid-Base Ka/Kb, pH,
Buffers
February 7, 2012
Chemistry Joke of the Day
Why do chemistry professors always teach a little bit
about ammonia?
Its very basic material.
Getting Started
Opening Prayer
Reminder: Im in the Tutorial lab 2-3 T, 12-1 and 2-3 Th,
and 1-3 F. Its a great place to come in and get help. Im
also available in the mornings (9-10 except for Th) if you
e-mail me to set up an appointment (do it!).
Practice problem suggestion: Mastering Chemistry has a
study area which will link you to a site related to the
textbookitll have practice quizzes and other problems
you can work for practice.
What are acids and bases?
Two definitions (so far):
Arrhenius acids/bases
Brnstead-Lowry acids/bases
Arrhenius definitions
Acid: a compound that produces H+ in water
Base: a compound that produces OH- in water
Brnstead-Lowry definitions
Examples of Acids and Bases
Acids
Bases
Strong (memorize these!
Youll need to recognize
them if mentioned)
HCl, H2SO4, HNO3, HI,
HBr, HClO4, HClO3
LiOH, NaOH, KOH,
RbOH, CsOH, Ca(OH)2,
Sr(OH)2, Ba(OH)2
Weak (lots more than
this)
NH4+, CH3COOH, HF
NH3, CH3COO-, F-
This is not a comprehensive list anything that meets the definition of an acid or base
given in the previous slide would fall into these categories.
For most of this class, well be working problems with Arrhenius acids and bases
these would also be Brnstead-Lowry acids and bases, but well be worrying about
the concentrations of H+ and OH- in water.
Conjugate Acid/Base pairs
Consider the following 2 reactions:
CH3COOH + H2O D CH3COO- + H3O+
NH3 + H2O D NH4+ + OH-
In each reaction, which reactant is the acid and which is
the base?
In the first reaction, CH3COOH is the acid and water is the
base.
Key point: water is
amphiprotic
In the second reaction, NH3 is the base and water is the
acid.
Conjugate Acid/Base Pairs, cont.
Youll notice that in each reaction, there are 2 pairs
of compounds which differ from each other by
exactly one H+ ion.
CH3COOH + H2O D CH3COO- + H3O+
NH3 + H2O D NH4+ + OH-
Acetic acid/acetate ion
Water/hydronium ion
These areion
conjugate pairs!
Water/hydroxide
Ammonia/ammonium ion
Relationship between conjugate
acid/base pairs
As acid strength increases, conjugate base strength
decreases.
Describe the strength of
the conjugate bases of:
H2SO4
HF
H2O
CH4
Clarifying acid/base comparison
Remember, a strong acid (or strong base) means that it COMPLETELY
dissociates in water. Consider HCl it will entirely break into hydrogen and
chlorine ions. So the conjugate base, Cl-, would never act as a base at all a
base would be accepting a proton (reforming HCl if that happened, HCl
wouldnt be a strong acid!)
In a weak acid or base, there is partial dissociation and thus at equilibrium, you
will have some of the compound protonated and some not (e.g., CH3COOH in
solution will exist as CH3COO- and CH3COOH both the acid and the base are
WEAK because the species partially exist as acetate and hydrogen ions and
partially as acetic acid).
Acid
Base
Strong
Negligible
Weak
Weak
Negligible
Strong
Autoionization of Water
Water molecules are constantly moving around and
the H2O molecules are constantly participating in
collisions and forming new bonds. While the effect is
very small, a few of these collisions result in water
molecules autoionizing: breaking into H3O+ and
OH-.
2H2O (l) D H3O+ + OH-
This reaction proceeds (at room temp, 25 oC)
according to the equilibrium expression:
Kw = [H+][OH-] = 1.0x10-14
Neutral Solutions
Strong Acids and Bases
We wont spend a lot of time on strong acids and
bases in this class because the calculations are very
straightforward but make sure you can do it (even
though we dont talk about it a lot, itll probably still
come up on a few questions on the test).
Remember, we use the term strong to designate
complete dissociation. That means that [HA] = [H+]
or that [BOH] = [OH-]
From here, you can calculate pH, pOH, [H+], or
Weak Acids and Bases
When you have a weak acid or base, you mean that it
DOES NOT completely dissociate. The extent to which it
dissociates is based on its EQUILIBRIUM CONSTANT
(which will not change no matter what you start with)!
HA D H+ + AA- + H2O (l) D HA + OH-
Ka = [H+][A-]/[HA]
Kb = [HA][OH-]/[A-]
By multiplying these expressions together
Predictions about Equilibrium
In acid/base chemistry, the equilibrium will favor
formation of the weaker acid and base.
This means that the stronger base will be taking a proton
from the stronger acid and the weaker acid and base will
be formed. The easiest way to compare this is by looking
at a table (either giving Ka values or relative acidity).
ICE tables in Acid-Base Chemistry
Lets say you put 0.30 M HSO3- into a liter of water.
Bisulfite ion is a WEAK acid, so [H+]
0.30 M.
What do we do???
Set up an ICE table!
(But first, you have to know Ka. In this case, its
6.4x10-8.)
Illustration: Bisulfite Ion
HSO3-
H+
SO32-
I
0.30 M
none
None
C
-x
+x
+x
E
0.30-x
x
x
x=1.4x10-4
Checking Assumptions and Making
Calculations
First, lets check our assumption. Was x less than
5% of our initial concentration?
1.4x10-4 = .00014
.00014/.30 = .00047*100=.047%
So were good there. x=.00014 M (and youll recall
from the ICE table, x=[H+])
Now lets find the pH and pOH of the solution:
This example holds true for any monoprotic weak acid or base plug your data into an ice table,
-log(.00014)=3.85 find out pH or pOH. Alternatively, you can be given pH or
calculate [H+] or [OH-], and you canpOH=14-pH=10.15
pOH and asked to calculate Ka, but the principle is the same. Acid/base equilibria are no different
So 0.30 M bisulfite ion in water gives15. solution with
than those we worked with in chapter a
A word on Polyprotic acids
If Ka1 is more than 1000 times (a factor of 103)
greater than Ka2, the second dissociation will not
affect pH significantly.
If not, follow the same principles as before work
the ICE table for the first dissociation (H2A D HA- +
H+) and then use the concentrations calculated from
that ICE table in another ICE table for the second
dissociation (HA- D A2- + H+).
Acid-Base properties of salt solutions
Remember that weak acids and bases act according
to an equilibrium expression. For example, HF D H+
+ F- in solution with a Ka of 6.8x10-4. Whether you
start with products (hydrogen and fluoride ions) or
reactants (hydrofluoric acid) the reaction will still go
to equilibrium.
This example explains the acid-base properties of
salt solutions, which are as follows:
Acid-Base Salt Properties
The conjugate base of a weak acid will increase the
[OH-] making a solution more basic.
A- + H2O (l) D HA + OH-
The conjugate base of a strong acid will not affect
the pH (if it were to pull an H+ away from water, it
would become a strong acid and immediately
dissociate!) Neither will the 1A cations or heavy 2A
cations (because they are the cations of the strong
bases!)
The conjugate acid of a weak base (e.g., NH4+) will
increase [H+], making the solution more acidic.
Factors affecting acid strength
Bond strength (inversely related to bond lengththink of
the size of atoms)
HF vs. HI
Polarity (electronegativity)
HCl vs. H2S
The net effect of these two effects is that acidity increases
going DOWN and to the RIGHT when it comes to
element effects.
Stability of the conjugate base
CH3COOH vs CH3CH2OH (RESONANCE!)
Oxyacids
You wont be asked to compare oxyacids to regular acids,
but you do need to know two trends about them:
Electronegativity: the more electronegative the central
atom, the stronger the acid.
Number of oxygen atoms: increasing the number of
oxygens increases the acidity because the negative charge
can be spread out to more oxygens.
We already know about several strong oxyacids: H2SO4,
HNO3, HClO3, HClO4.
Quick definition: Lewis acids/bases
The definition a of Lewis acid and base:
Lewis acid an electron pair acceptor
Lewis base an electron pair donor
These are simply definitions explaining the same concept
of Bronstead-Lowry acid/base chemistry, but now were
talking about electrons. This makes our definition more
extensive because anything exchanging electrons can be
discussed in terms of acidity or basicity.
Common-Ion Effect
Think of the equilibrium expression for a weak acid: HA
D A- + H+
Any time you have any of these things already present in
solution it will affect the equilibrium.
For example, if you have a sodium fluoride solution, when
you add in hydrofluoric acid not so much will dissociate
because there is already significant A- in the solution
vessel.
Buffers
Buffer a solution in which the weak acid and its
conjugate base (or vice versa) are both present. The
purpose of a buffer is to resist change in pH. A good
buffer has an acid to base ratio smaller than 10:1.
One good example of buffers importance is in your
blood. A carbonic acid-bicarbonate buffer in the blood
keeps the pH of your blood at a normal level and makes
life possible.
Buffer capacity the amount of acid or base the buffer
Henderson-Hasselbach Equation
An easy way to calculate the pH of a buffered solution is
to use the Henderson-Hasselbach equation. This formula
only works for a buffered solution, but it is much easier
than making an ICE table in these cases to calculate the
pH.
pH = pKa + log ([A-]/[HA])
Moving into problems
Once again, come see me if you need help!
Dont forget to read so you can get your i-clicker points!
Were not going to use the worksheet this week because a
fair amount of the material is not conducive to recitation
and problem 2 is slightly beyond the scope of this class.
Practice problems!
Kb of ammonia (NH3) is 1.8x10-5. Find:
The Ka of NH4+
The following for a 0.50 M solution of NH4Br:
pH
pOH
[H+]
[OH-]
Solution
Kb of ammonia = 1.8x10-5.
Ka of the conjugate acid = Kw/Kb = 1x10-14/1.8x10-
5 = 5.6x10-10.
ICE table:
NH4+
H+
NH3
I
0.50 M
none
none
C
-x
+x
+x
E
0.50-x
x
x
5.6x10-10=(x2)/(0.50-x) ASSUME x is small
0.5(5.6x10-10)=x2
Solution cont.
x=1.7x10-5
x=[H+]=1.7x10-5
pH=-log(1.7x10-5)=4.77
NOTE: sig figs in logs. When you take the log of a
number, the number of places after the decimal point
should equal the number of sig figs you had initially.
pOH=14-pH=9.23
[OH-]=10-pOH=10-9.23=5.9x10-10
Practice Problem
1. Designate the Brnsted-Lowry acid and the
Brnsted-Lowry base on the left side of each
equation, and also designate the conjugate acid and
conjugate base on the right side.
a. HBrO(aq) + H2O(l) D H3O+(aq) + BrO-(aq)
b. HSO4-(aq) + HCO3-(aq) D SO4-2(aq) +
H2CO3(aq)
c. HSO3-(aq) + H3O+(aq) D H2SO3(aq) + H2O(l)
Solution
a. HBrO(aq) + H2O(l) D H3O+(aq) + BrO-(aq)
Acid
base
conj acid
conj base
b. HSO4-(aq) + HCO3-(aq) D SO4-2(aq) +
H2CO3(aq)
acid
base
conj base
conj acid
c. HSO3-(aq) + H3O+(aq) D H2SO3(aq) + H2O(l)
Base
acid
conj acid
conj base
Problem
Predict whether aqueous solutions of the following
substances are acidic, basic, or neutral.
CrBr3
LiI
K3PO4
[CH3NH3]Cl
KHSO4
Solution
CrBr3
Dissociation: Cr3+ and BrBr- has no effect (related to strong acid); Cr3+ is metal =
acidic
LiI
Dissociation: Li+ and ILi+ and I- = no effect (related to strong base and acid)
K3PO4
Dissociation: K+ and PO43K+ has no effect (related to strong base); PO43- is
conjugate base of weak acid = basic
Question
What is the [H+] and pH of a 0.10 M solution of
hypochlorous acid (HOCl), Ka=3.5x10-8?
Solution:
ICE table draw it out for practice
3.5x10-8 = (x2)/(0.10-x)
ASSUMPTION
x=5.9x10-5
CHECK (5.9x105/0.1)*100=0.059%
x=[H+]=5.9x10-5
pH= -log (5.9x10-5) = 4.23
Question
A man decided to conduct an experiment. Suppose he has
an alkaline buffer consisting of 0.20 M aqueous ammonia
(NH3) and 0.10 M ammonium chloride (NH4Cl). What is
the pH of the solution?
Kb for ammonia is 1.8x10-5. Use the reaction
NH3 + H2O (l) D NH4+ + OH-
Solution to Problem
NH3 + H2O (l) D NH4+ + OH-
I
0.20 M |
0.10 M 0
C
-x
+x
E
0.20-x M |
|
0.10+x
+x
x
KbNH3 = 1.8x10-5
1.8x10-5=(0.10+x)(x)/(0.20-x)
ASSUMPTION
1.8x10-5(0.20)/(0.10)=x
x=3.6x10-5
[OH-]=x=3.6x10-5
pH=14-pOH
14-[-log(3.6x10-5)]=9.56
Question 5
Someone is considering two solutions, solution A and
solution B. [H+] in solution A is 343 times greater than
that in solution B. What is the difference in pH values of
the two solutions that he is considering?
Question 5 Solution
The easiest way to solve a problem like this (in my
opinion) is to pick two values of [H+] that fit the
requirements and solve for pH.
A: [H+] = 0.001 M pH=-log(0.001)=3
B: [H+] = 0.001*343 = .343 M pH=-log (.343)=0.465
The difference in pH values is 3-0.465=2.535.
To check this, use different values of [H+] for A and B.
A: [H+] = 0.000451 M pH=-log(.000451)=3.346
B: [H+] = 0.000451*343= .155 M pH=-log(.155)=0.810
The worksheet we skipped
1. How does the acid strength of an oxyacid depend on
a) The electronegativity of the central atom.
b) The number of non-protonated oxygen atoms in the
molecule.
A-more electronegative central atom=stronger acid
(negative charge on conjugate base is stabilized)
B-more oxygen atoms=stronger acid (oxygen is
electronegative-stabilizes conjugate base)
2. Predict which member of each pair produces the more
acidic aqueous solution:
a) K+ or Cu2+
b) Al3+ or Ga3+
Beyond the scope of this classCu is more acidic in
solution because K is the cation of a strong acid
(KOH). Im not positive about the answer to B and so
you dont need to know that either.
3. Identify Lewis acid and Lewis base among the reactants
in each of the following reactions:
a) CN- (aq) + H2O(l) --- HCN(aq) + OHb) SO2(g) + H2O(l) --- H2SO3(aq)
A) lewis acid is H2O, lewis base is CNB) lewis base is H2O, lewis acid is SO2 (visualizing this
is slightly beyond the scope of this classwhat is
happening is that the electrons on the oxygen are donated
to the sulfur atom of SO2)
4. Why is ammonia (NH3) a stronger base than water?
And, why is NH3 a stronger base than CH4?
Basicity is ability to accept a proton. NH3 is a stronger
base than water because acidity increases going DOWN
and TO THE RIGHT of the periodic table. Because H2O
is more acidic than NH3 we know NH3 is a stronger base.
By this logic, we would expect CH4 to be a stronger base
than NH3. However, carbon doesnt have any lone pairs in
CH4 to form a bond (remember the proton doesnt have
any electrons of its own).
5. What is common ion effect? Using Appendix D in your
book, calculate the pH of
a) A solution that is 0.060 M in potassium propionate
(C2H5COOK) and 0.085 M in propionic
acid(C2H5COOH). [Ka=1.3x10-5]
b) A solution that is 0.250 M in sodium formate
(HCOONa) and 0.100 M in formic acid
(HCOOH). [Ka=1.8x10-4]
A = 4.73 B = 4.14 common-ion effect just describes what
6. A) Calculate percent ionization of 0.0075M butanoic
acid (Ka = 1.5 x 10-5). B) Calculate the percent ionization
of 0.0075 M butanoic acid in a solution containing 0.085
M sodium butanoate.
A) 4.47% ionization
B) 0.0176% ionization
7. A buffer contains 0.10 mol of acetic acid (pKa=4.74)
and 0.13 mol of sodium acetate in 1.00 L. a) What is the
pH of this buffer. b) What is the pH of the buffer after the
addition of the 0.02 mol of KOH. C) What is the pH after
the addition of 0.02mol of HNO3
A) 4.85
B) 5.01
C) 4.78 (this is assuming you are adding HNO3 to the
initial mixture, not part B)
8. How many milliliters of 0.0850 M NaOH are required
to titrate each of the following solutions to
the equivalence point:
a) 40.0 ml of 0.0900 M HNO3
b) 35ml of 0.0850 M acetic acid.
A) .0036 mol HNO3 = 42.3 mL of .0850 M NaOH
B) equivalence point: moles acid = moles of base. Since
the concentrations are the same you need 35 mL of NaOH
(note: equivalence point is NOT when pH=7)