Unformatted text preview: Module 30 — Weak Acids and Bases Calculations In Chemistry
*****
Module 30: Weak Acids and Bases
Module 31: BrønstedLowry Definitions
Module 30 – Weak Acids and Bases ........................................................................... 923
Lesson 30A:
Lesson 30B:
Lesson 30C:
Lesson 30D:
Lesson 30E:
Lesson 30F:
Lesson 30G: Ka Math and Approximation Equations ........................................................ 923
Weak Acids and Ka Expressions ..................................................................... 927
Ka Calculations .................................................................................................. 934
Percent Dissociation and Shortcuts................................................................. 943
Solving Ka Using the Quadratic Formula ...................................................... 947
Weak Bases and Kb Calculations..................................................................... 950
Polyprotic Acids ................................................................................................ 960 Module 31 – BrønstedLowry Definitions................................................................. 966
Lesson 31A:
Lesson 31B: BrønstedLowry Acids and Bases ................................................................... 966
Which Acids and Bases Will React?................................................................ 970
For additional modules, visit www.ChemReview.Net © 2009 ChemReview.Net v. 6w Page i Module 30 — Weak Acids and Bases Module 30 — Weak Acids and Bases
Timing: Begin this module when you are assigned calculations involving Ka.
Prerequisites: Do Lessons 28C and 28D on equilibrium and Module 29 on acidbase
fundamentals before starting this module.
Pretests: In this module, if you think you are familiar with a lesson topic, try the last
problem in each problem set of the lesson. If you get those right, move to the next lesson.
***** Lesson 30A: Ka Math and Approximation Equations
Pretest: This lesson has two parts. The first part is a quick review of the math of powers
and roots covered in Lessons 19A and 28B. The second part is new material. Even if you
can do the Practice A problems quickly, devote extra care to the section after Practice A.
***** Squares and Square Roots
The calculations in this module require solving the square and square root of numbers in
exponential notation. Rules for these calculations were covered in Lesson 19A. The
following is a brief reminder of those rules (which you need to be able to apply from
memory).
Estimating exponential calculations is one way to catch mistakes in calculator use. For this
reason, we will review squares and square roots both with and without the calculator.
Rules for Squares and Square Roots in Exponential Notation
1. To take an exponential term to a power, multiply the exponentials.
2. “Taking the square root” has the same meaning as taking a quantity to the 1/2 power.
3. When taking exponential notation to a power, handle numbers by number rules and
exponents by exponential rules.
4. A square root can be calculated without entering the exponent on the calculator by
• making the exponent even. If the exponent is odd, make the significand 10 times
larger, and the exponent ten times (one number) smaller. Then, • take the square root of the significand either on the calculator or by estimation. To
take the square root of the exponential term, multiply the exponent by 1/2. Practice A: Do the last part of each numbered problem, then more if you need more
practice. Convert final answers to scientific notation. Check answers at the end of this
lesson. If you cannot solve these problems easily, review Lesson 19A.
Complete these first three numbered problems without a calculator.
1. a. (103)2 = © 2011 ChemReview.Net v. 7f b. (10―4)2 = c. (10―4)1/2 = (Rule 1) Page 923 Module 30 — Weak Acids and Bases 2. The square root of a. 1014 = b. 10―6 = c. 10―16 = (Rule 2) 3. Without a calculator:
a. (7.0 x 10―5)2 = b. (36 x 10―6)1/2 = c. Square root of 25 x 10―12 = d. 0.0064 = 4. Do Problem 3 using the square and square root function on your calculator.
a. (7.0 x 10―5)2 = b. (36 x 10―6)1/2 = c. Square root of 25 x 10―12 = d. 0.0064 = Compare Problem 3 to 4. Which was easier: with or without the calculator?
5. Use a calculator for the number part, but do the exponential part by inspection.
a. (2.5 x 10―5)2 = b. (9.5 x 10―6)2 = c. (2.56 x 10―4)1/2 = d. 6. Estimate the square root of 1.44 x 10  6 = a. 45 b. 18 c. 95 d. 7 7. Use a calculator to take a square
root and compare to above a. 45 b. 18 c. 95 d. 7 8. Estimate the square root.
a. (4.2 x 10+5)1/2 = b. (3.0 x 10―11)1/2 = c. (8.1 x 10―7)1/2 = d. Square root of 7.2 x 10―3 = 9. Calculate the square root. Either adjust to make the exponent even, then take the
significand root on the calculator and the exponential root by inspection, or do the
entire calculation on the calculator. Compare answers to Problem 7.
a. (4.2 x 10+5)1/2 = b. (3.0 x 10―11)1/2 = c. (8.1 x 10―7)1/2 = d. Square root of 7.2 x 10―3 = Bottom Line On Calculator Use
With practice, you can quickly do squares and square roots of exponential notation on the
calculator, but first write a quick estimate of the answer without the calculator. Check that
the estimate and the calculator answer are close. © 2011 ChemReview.Net v. 7f Page 924 Module 30 — Weak Acids and Bases Approximation Equations
If equations involve both large and small numbers, they can often be simplified to
approximations that are easier to solve.
For K approximations, the rules are
• You cannot ignore small numbers that are by themselves as terms in equations, but
• You can ignore small numbers that are added or subtracted from much larger numbers.
In the following questions, ≈ means approximately equals, A is any large number, B is any
different large number, and x is any number much smaller than A and B. Write the
approximation equation by applying the rule above.
Q. ( A + x )( x )
(B ─ x) ≈? *****
A. ( A + x )( x )
(B─ x) ≈ (A)( x )
B Let’s test this rule using numbers. For this term, calculate an exact numeric answer:
(18.0 + 0.10)(0.10) =
(9.0 ─ 0.10)
Now simplify that equation by applying the approximation rules. Cross out the small
terms added or subtracted from larger terms, calculate a numeric answer, then compare to
the answer above.
(18.0 + 0.10)(0.10) ≈
(9.0 ─ 0.10)
*****
A. Exact: (18.1)(0.10)
(8.9)
A. Approximation: = 0.204 (18.0 + 0.10)(0.10)
(9.0 ─ 0.10) ≈ (18.0)(0.10) = 0.200
(9.0) If the small numbers that are added or subtracted from the large numbers are removed, the
difference between the two answers is 2%. In some science experiments and procedures,
that would be more error than we would like to see. However, because K values often
involve significant inherent uncertainty, a change in an answer of up to 5% due to the use
of an approximation is considered acceptable.
In the acid and base equilibrium calculations that we are about to encounter,
approximation equations will often be used to solve. © 2011 ChemReview.Net v. 7f Page 925 Module 30 — Weak Acids and Bases Practice B. Using the notation and rules above, simplify these equations using
approximations. Save a few for your next practice session.
1. (A + x )( x )
(A ─ x ) ≈ 2. (x)(x) ≈
(B ─ x ) 3. (B + x )( x )
(A ─ x ) ≈ 4. (A + x )
(A ─ x )( x ) ≈ 5. Assuming x is much smaller than the other numbers in these equations, apply the
approximation rules to simplify these terms.
a. (0.050 + x) (x) b. ≈ (2x) (x) 0.020 ─ x
c. (x) (x) ≈ 0.10 ─ x
d. ≈ 0.020 ─ x (x + x) ≈ [WB] + x 6. Assuming x is very small compared to the other numbers, apply the approximation
rules to these equations, then solve for x. (Try these first without a calculator).
a. (x) (x) = 8.0 x 10―12 b. (2x + x) 2.0 ─ x
c. = 1.8 x 10―5 0.50 + x (0.60 + x) (x) = 3.6 x 10―9 d. 0.20 ─ x = 1.25 x 10―11 (2x) (x)
4.0 ─ x ANSWERS
Practice A
1a. 106 b. 3,4: a. 4.9 x 10―9
5. a. 6.2 x 10―10 10―8 c. 10―2 b. 6.0 x 10―3
b. 2a. 107
c. 5.0 x 10―6 9.0 x 10―11 c. 6. a. 45 62 = 36 and 72= 49, estimate ≈ 6.7
c. 95
7. a. 6.71 10―3 c. 10―8 d. (64 x 10―4)1/2 = 8.0 x 10―2 1.60 x 10―2
b. 18 92 =81 and 102= 100, est. ≈ 9.7 b. d. 1.20 x 10―3 42 = 16 and 52= 25, est. ≈ 4.2 d. 7 22 =4 and 32= 9 est. ≈ 2.7 b. 4.24 c. 9.7 d. 2.65 When estimating to check answers, close is a good sign. 8. To estimate the square root, first convert so that the power is divisible by 2.
8,9. a. 6.5 x 102 b. 5.5 x 10―6 c. 9.0 x 10―4 d. 8.5 x 10―2 Practice B
1. (A + x )( x ) ≈ (A)( x ) ≈ x
(A ─ x )
(A) © 2011 ChemReview.Net v. 7f 2. ( x )( x ) ≈
(B─ x) x2
B Page 926 Module 30 — Weak Acids and Bases 3 . ( B + x )( x )
(A ─ x )
5. a.
c. ≈ (B)( x )
A (0.050 + x) (x)
0.020 ─ x
(x) (x)
0.020 ─ x ≈
≈ 4.
(0.050)(x) = 2.5x
0.020 x2
0.20 ≈ ( x )― 1 (A + x )
(A ─ x )( x )
(2x) (x)
0.10 ─ x 5b.
d. (x + x)
[WB] + x ≈
≈ 2x2
0.10 20x2 = 2x
[WB] 6. On these, you may do the steps differently, but you must get the same answers.
a. (x) (x)
2.0 ─ x
x2
2.0
x2
x c. = 8.0 x 10―12
= 8.0 x 10―12 = 3.6 x 10―9
= 3.6 x 10―9 3x = 3.6 x 10―9
x = 1.2 x 10―9 (2x + x)
0.50 + x
3x
0.50 = 16 x 10―12
= 4.0 x 10―6 (0.60 + x) (x)
0.20 ─ x
(0.60) (x)
0.20 b. 3x
x
d. = 1.8 x 10―5
= 9.0 x 10―6
= 3.0 x 10―6 (2x) (x)
4.0 ─ x
2x2
4.0
x2
x = 1.8 x 10―5 = 1.25 x 10―11
= 1.25 x 10―11 = 25.0 x 10―12
= 5.0 x 10―6 ***** Lesson 30B: Weak Acids and Ka Expressions
So far, our acidbase calculations have all involved either strong monoprotic acids (HCl and
HNO3) or strong bases (NaOH and KOH). Those four compounds are widely used in
laboratories and industrial processes when strong acids and bases are needed.
However, far more compounds are weak acids and bases than strong. An understanding of
weak acids and bases is important in fields as interesting and diverse as the biological
sciences and the culinary arts. Weak Acid Solutions
Strong acids dissociate (ionize) 100% to form H+ ions when dissolved in water.
Weak acids are defined as substances that dissociate to form H+ ions when dissolved in
water, but do so only slightly.
The loss of a proton by an acid is reversible: the proton can return to the base conjugate to
reform the acid. In a weak acid solution at equilibrium, the gain and loss of protons occurs
continuously, but because both reactions proceed at the same rate, no net change occurs.
A familiar weak acid solution is vinegar, which can be formed by the action of bacteria and
oxygen on fruit juice. The oxidation of the sugar in fruit juice produces a mixture that is
© 2011 ChemReview.Net v. 7f Page 927 Module 30 — Weak Acids and Bases composed primarily acetic acid, a weak acid that is a liquid at room temperature, and water.
Most vinegar solutions are about one volume of acetic acid per 10 to 20 volumes of water.
Acetic acid is a weak acid because in water, it ionizes slightly to produce H+ ions. As the
products form, the reverse reaction of the proton returning to the acetate ion also occurs,
and the solution will quickly reach an equilibrium state where no further net change takes
place.
The behavior of acetic acid in water can be represented as either a separation into ions
(which can be termed an ionization or a dissociation):
1 H+(aq) + 1 CH3COO─ (aq) 1 CH3COOH(aq)
In water: ~99% unionized ~ 1% ionized (goes ~ 1%) (~ means approximately) or as a hydrolysis (a reaction with water):
1 H3O+(aq) + 1 CH3COO─(aq) 1 CH3COOH(l) + 1 H2O(l) (goes ~ 1%) Compare the two equations. Both represent a weak acid that is 100% dissolved in water
but ionizes only slightly. Recall that H+ and H3O+ are equivalent ways of representing the
proton released by acids. These two equations are equivalent ways of representing the same
reaction. [H+] In Weak versus Strong Acids
In water, strong acids dissociate completely, but when weak acids dissolve in water,
equilibrium favors the undissociated (unionized) species. A weak acid solution therefore
has fewer H+ ions, and a higher pH (is less acidic), than a strong acid solution with the same
concentration.
For example:
In 0.10 M HCl, a solution of a strong acid, the concentrations and pH include
[HCl] = 0 M, [H+] = 0.10 M = 10─1 M; pH =1 . In a 0.10 M acetic acid solution, roughly one acetic acid particles per 100 is ionized.
This means that in this solution,
[CH3COOH] ≈ 0.099 M ≈ 0.10 M, [H+] ≈ 0.001 M ≈ 10─3 M, pH ≈ 3.
Compare the [original acid], the [H+], and pH in the two solutions. At the same
mixed concentrations of the acids, the hydrochloric acid solution has about 100
times more protons than the acetic acid.
Vinegar can be used in cooking because acetic acid solutions do not contain high
concentrations of the reactive, corrosive particles in acids: H+. However, even at low
concentrations, the [H+] in aqueous solutions can have a major impact on reactions in
chemistry and biology. © 2011 ChemReview.Net v. 7f Page 928 Module 30 — Weak Acids and Bases Base Conjugates
An acid can be defined as any particle that can lose a proton. Acids can be positive or
negative ions as well as electrically neutral particles.
Examples: HCl, NH4+, and HPO42─ can all act as an acid by losing a proton.
The ionization of an acid produces a proton and a base conjugate. The chemical formula
for the base conjugate of an acid is simply the acid formula with one less H atom and one less
positive charge.
For the weak acid HF reacting with water:
The general equation is: 1 Weak acid 1 proton + 1 base conjugate In the ionization format: HF(aq) H+(aq) + F─(aq) In the hydrolysis format: HF(g) + H2O(l) H3O+(aq) + F─(aq) Use the definition of a base conjugate to answer these two questions.
Q. Assume the first particle in each reaction is acting as an acid. Complete the reaction
by writing chemical formulas for the products. Circle the base conjugate of the acid.
1. HCN(aq)
2. NH4+ + H2O(l)
(For particles in a solution, if no state is shown, assume (aq).)
*****
1. HCN(aq) H+(aq) + CN─(aq) An acid ionizes to produce a proton (H+). The other particle is the base conjugate:
the acid particle with one less H atom and one less positive charge.
2. NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq) When a weak acid ionization is written in the hydrolysis format (losing a proton by
donating the proton to water), one of the product particles is the always the
hydronium ion (H3O+); the other is the base conjugate.
All reaction equations must be balanced for atoms and charge. Check the balancing
in each of the answers above. Practice A: In the reactions below, assume that the first particle is acting as a weak acid
in an aqueous solution. Write the formulas for the products. Circle the base conjugate. 1. HI 2. HCO3─ © 2011 ChemReview.Net v. 7f Page 929 Module 30 — Weak Acids and Bases 3. HS─ + H2O 4. HPO42─ + H2O The Ka Expression
For the reaction of an acid losing one proton, the equilibrium constant K is given a special
name, the aciddissociation constant, and a special symbol, Ka . For any reaction
equation in which one acid particle ionizes to form one H+ or reacts with water to form one
H3O+, the symbol for the equilibrium constant is Ka.
By convention, an acid ionization or hydrolysis reaction is always written with one unionized acid particle on the left and one H+ (or H3O+) on the right.
Example: The loss of a proton by acetic acid can be represented by either:
H+(aq) + CH3COO─(aq) CH3COOH(aq)
or H3O+(aq) + CH3COO─(aq) CH3COOH(aq) + H2O(l) Write the K expression for these two reactions.
*****
Because each reaction consists of a weak acid releasing one proton, the K is a Ka .
A Ka expression always has [H+]1 or [H3O+]1 on top, and [weak acid]1 on the bottom.
For the above two reactions,
+
─
Ka = [H ]eq. [CH3COO ]eq.
[CH3COOH]at eq. or +
─
Ka = [H3O ] [CH3COO ]
[CH3COOH] For all Ka expressions,
• as in standard K expressions: product concentrations are on top and reactant
concentrations on the bottom. • In K expressions, concentrations are assumed to be measured at equilibrium. • Since [H+] = [H3O+], the two reaction formats and the two Ka expressions are
equivalent. One format may be substituted for the other. • [H2O] is not included in the Ka expression for the hydrolysis reaction. • The powers of all of the concentrations are 1 and are omitted as understood. © 2011 ChemReview.Net v. 7f Page 930 Module 30 — Weak Acids and Bases • [H2O] is left out of Ka hydrolysis expressions for the same reason that it is left out of
the Kw expression. During acid hydrolysis, the concentration of liquid water
remains high and close to constant, at about 55 M, except in solutions that have a
high concentration of the acid. By convention, if a particle concentration in a
reaction cannot be significantly changed by a reaction, its term is represented by a 1
in the K expression for the reaction. Acetic acid (CH3COOH) is also a liquid at room temperature. Because acetic acid is a polar
molecule, it soluble in water: these two liquids mix to form one layer, unlike the
immiscible “oil and water.” In the dilute acetic acid solutions used in most experiments, its
concentration is low compared to the [water], and the state of the acetic acid is better
described as aqueous (dissolved in a comparably large amount of water) rather than liquid.
Since, depending on how much acetic acid is mixed into the water, the [CH3COOH(aq)]
can vary substantially, the [CH3COOH] term is included in the K expression.
In general,
A [dissolved liquid] term is included in a K expression, but the [solvent] is not.
In Ka expressions, if the state of a particle is left out, it is assumed to be aqueous. Since a K
must be calculated based on concentrations at equilibrium, if a label after and below a
concentration term is omitted, the subscript [ ]at equilibrium is understood.
A polyprotic acid can ionize to lose more than one proton, but in all polyprotic acid
solutions, each successive proton ionizes less than the one before. These successive
ionizations are written separately, with the base conjugate of the first ionization becoming
the weak acid in the second, etc. A Ka value always refers to the ionization of a weak acid
to produce one H+ or one H3O+. Practice B: In the reactions below, assume that the first particle is acting as a weak acid
in an aqueous solution. Complete the reaction by writing the formulas for the products.
Then write the Ka expression for each reaction.
1. H2CO3 2. HCO3─ 3. H2S + H2O(l) 4. NH4+ + H2O(l) © 2011 ChemReview.Net v. 7f Page 931 Module 30 — Weak Acids and Bases Ka Values
Acids have characteristic Ka values at a given temperature.
Sample Ka Values at 25ºC:
Hydrochloric acid (strong) HCl H+ + Cl─ Ka = very large Hydrofluoric acid (weak) HF H+ + F─ Ka = 6.8 x 10―4 Acetic acid (weak) CH3COOH Hydrocyanic acid (weak) HCN H+ + CH3COO─
H+ + CN─ Ka = 1.8 x 10―5
Ka = 6.2 x 10―10 For strong acids, ionization strongly favors the products. The Ka values for strong acids will
be very large: a number much greater than one. Since strong acid ionization is considered
to go to completion rather than equilibrium, the K value is not needed for calculations.
Weak acids have Ka values between 1.0 and 10―16, usually written in scientific notation
with a negative power of 10. The weaker is the acid, the less it will ionize and the smaller
will be its Ka. In the table above, the weakest acid is ? ? ?
*****
HCN.
As the temperature of a weak acid solution rises, more bonds to H break, more protons
form, and Ka values increase. However, unless otherwise noted, you should assume that
Ka calculations are based on reactions at 25ºC (77ºF).
For each H atom in a compound, a Ka value can be measured that represents the tendency
of the H to ionize. However, if a Ka value is smaller than 10―16, the H will not react as an
acid except with bases that are very strong: stronger than hydroxides. Under most
circumstances, an H with a Ka smaller than 10―16 is considered to be a nonreactive rather
than an acidic hydrogen (see Lesson 14A).
Because K values can be difficult to measure precisely, reported K values can vary slightly
among textbooks. When doing textbook calculations, use K values in that text. Practice C:
1. For the weak acid NH4+, the Ka value is 5.6 x 10―10.
a. Write the reaction that this is this a K value for.
b. Write the Ka expression for this reaction.
c. What term do Ka expressions always have in their numerator? © 2011 ChemReview.Net v. 7f Page 932 Module 30 — Weak Acids and Bases 2. The weak acid HSO3─ has a Ka value of 6.2 x 10―8.
a. What reaction is this a K value for?
b. Write the Ka expression for this reaction.
3. Which particle above is the weaker acid: NH4+ or HSO3─ ? ANSWERS
Practice A
H+ + I ─ 1. HI 3. HS─ + H2O H+ + CO32─
4. HPO42─ + H2O
H3O+ + PO43─ 2. HCO3─
H3O+ + S2─ Practice B: In the K expressions below, all concentrations are measured at equilibrium.
1. H+ + HCO3─ H2CO3 Ka = [ H+ ][ HCO3─ ]
[ H2CO3 ]
3. H2S + H2O(l) 2. HCO3─ H+ + CO32─ Ka = [ H+ ][ CO32─ ]
[ HCO ─ ]
3 H3O+ + HS─ Ka = [ H3O+ ] [ HS─ ]
[ H2S ] 4. NH4+ + H2O (l) H3O+ + NH3 Ka = [ H3O+ ][ NH3 ]
[ NH + ]
4 For 3 and 4, in aqueous solutions [H2O] is essentially constant and is given a value of 1 in K expressions. Practice C:
1a. NH4+
1b. H+ + NH3 Ka = [ H+ ][ NH3 ]
[ NH4+ ] A Ka is always a K for the reaction where the reactant loses an H+ ion. In
the equation for a Ka reaction, an H+ ion will always be on the right side.
1c. A Ka expression always has [H+] (or [H3O+] ) in the numerator. 2b. Ka = [ H+ ][ SO32─ ]
[ HSO3─ ]
3. The weaker acid is the one with the lower Ka value: NH4+ 2a. HSO3─ H+ + SO32─ ***** © 2011 ChemReview.Net v. 7f Page 933 Module 30 — Weak Acids and Bases Lesson 30C: Ka Calculations
Ka Equations
The Ka expression and the Ka value together form the Ka equation.
For the ionization of acetic acid: 1 CH3COOH
the Ka equation is: 1 H+ + 1 CH3COO─ +
─
Ka = [H ][CH3COO ] = 1.8 x 10―5 at 25oC [CH3COOH]
Calculations using Ka follow the same rules as other K calculations:
• units are not attached to K values, • terms in the equation must have consistent units, but • units may be omitted when substituting into the K expression, and • when substituted concentrations are in mol/L, and a concentration is WANTED, the
unit moles/liter (M) must be added to the answer. When To Use Ka
In solutions of strong acids, a Ka value is not needed to calculate particle concentrations at
equilibrium. Since HCl and HNO3 ionize completely in a 1 to 1 to 1 ratio, the original
[HCl and HNO3]mixed equals the [H+] and [Cl─ or NO3─] that form in the solution. These
simple ratios are used to solve calculations.
Weak acids ionize to form one proton, but the reaction goes slightly: to equilibrium instead
of completion. The mixed original [weak acid] does not directly convey how much of the
weak acid ionizes, nor how much of the products form. For reactions that go to
equilibrium, a K equation is needed to calculate the particle concentrations at equilibrium.
This we will call weak acidbase Rule
1. In calculations involving acids and [particles],
• For 100% ionization (as in HCl or HNO3 solutions), use the REC steps.
• For slight ionization (as in weak acid solutions), use WRECK steps. The WRECK Steps For Weak Acid Ionization
For all weak acid ionizations, the general R and E (Reaction and Extent) can be written
1 WA 1 H+ + 1 base conjugate (goes slightly) Since the reaction does not go to completion, the WRECK steps are needed to solve
calculations. Write the K expression for this general reaction, then check your answer
below.
***** © 2011 ChemReview.Net v. 7f Page 934 Module 30 — Weak Acids and Bases Ka ≡ [H+]at eq. • [base conjugate]at eq. (Definition) [WA]at equilibrium
Because this ionization forms H+, the K is a Ka . The [ ]at equilibrium subscript can be
omitted as understood in K expression terms, but in this case it will be helpful to emphasize
in the discussion that follows.
In Lesson 28F, to solve K calculations, we used a rice table. Let’s apply the rice table method
to the general reaction above. The Rice Table For Weak Acid Ionization.
There are several types of reactions in which a compound reacts slightly by separating into
ions. To solve calculations in these reactions, we will define the small moles per liter of the
original, mixed compound that does ionize as x.
For compounds that ionize slightly: mol/L that does ionize = small = x By this definition, for reactions in which compounds ionize slightly ,
[compound]remaining at equilibrium = [compound]initially, as mixed ─ x
The ionization of a weak acid in water is one example of this type of reaction. In weak acid
ionization, the [ion] of interest is usually the [H+] ion. In this reaction, for every one weak
acid particle that reacts (ionizes and is used up), one H+ particle forms. This one to one
ratio applies to measurements in particles and moles. In addition, since the particles are all
in the same volume, the 1 to 1 ratio applies to moles per liter of particles.
We can therefore write this key relationship:
Mol/L weak acid that ionizes = small = x = [H+]formed
The above relationships are the same for all weak acids that ionize. That similarity will
simplify our problem solving.
Based on the general weak acid reaction
1 WA 1 H+ + 1 base conjugate (goes slightly) and the definition that x is the small amount of initial weak acid that reacts, fill in the
blanks and empty boxes in the rice table below.
Reaction
Initial
Change (+,―) __ WA __ H+ __ base conjugate [WA]mixed 0 0 ―x At Equilibrium
***** © 2011 ChemReview.Net v. 7f Page 935 Module 30 — Weak Acids and Bases 1 WA At Equilibrium 0 0 +x +x [WA]mixed ― x Change 1 base conjugate ―x Initial 1 H+ [WA]mixed Reaction +x +x As in all rice tables (Lesson 28F), in the Change row, quantities used up must be negative
and formed positive. The coefficients in front of the x terms in rows 3 and 4 are determined
by (and must be the same as) the coefficients in the balanced equation in Row 1.
To solve K calculations, we substitute the concentrations in the bottom row of the rice table
into the K definition equation. Do that step for the above reaction.
Ka ≡ [H+]at eq. • [base conjugate]at eq. ≡
[WA]at equilibrium _____________________ *****
Ka ≡ [H+]at eq. [BC]at eq.
[WA]at eq. x2 ≡ [WA]mixed ─ x ^ Definition ^ Exact The two terms on top both have a value of x; multiplied they are x2 . Writing the Equilibrium Concentrations By Inspection
Compared to general K calculations, Ka calculations are simplified because
• there is only one reactant, and it is always a weak acid. • One product is always [H+] or [H3O+] and the other is the base conjugate. • The initial quantities of the products are zero. • The coefficients are always 1 weak acid particle used up equals 1 H+ ion formed
and 1 base conjugate formed, and • the powers of the [ ] terms in a Ka expression are always 1. This means:
The general rice moles table is the same for all weak acid ionization or hydrolysis.
The purpose of the rice table was to determine the “Concentrations at equilibrium”: the
bottom row of the rice table. A rice table can be written with each Ka calculation. However,
because all weak acid ionization calculations are so similar, instead of writing a rice table
for each problem, it is quicker to instead write the key first and last rice rows by inspection,
as part of the WRECK steps. © 2011 ChemReview.Net v. 7f Page 936 Module 30 — Weak Acids and Bases Let’s walk through the logic of abbreviating the rice table with the WRECK steps.
1. Use x.
Since the coefficients in a weak acid ionization are all one,
1 WA 1 H+ + 1 base conjugate (goes slightly) the ratio of moles of weak acid used up to moles H+ formed to moles of base conjugate
(BC) formed is always 1 to 1 to 1.
Since all of the particles in an ionization reaction are dissolved in the same volume of
solution, the moles per liter ratios are also 1 to 1 to 1. This means that all three of the
following concentrations can be represented by an x representing a relatively small
concentration.
x = small = [WA]that ionizes = [H+]formed = [BC]formed
2. The REC steps. In a solution, the [WA] at equilibrium, after ionization, is usually difficult
to measure directly. However, the mol/L of weak acid that we originally added to mix
the solution is usually a quantity that we know.
Substituting x from the equality above simplifies the math. By definition:
[WA]at eq. ≡ [WA]mixed ─ small [WA]that ionizes ≡ [WA]mixed ─ x
If we can solve for x, we solve for four quantities:
x = [H+]formed = [BC]formed = [WA]that ionizes plus [WA]at eq. ( ≡ [WA]mixed ─ x)
Substitute into the blanks in the C step below a symbol for each concentration. Write a
term that includes x and uses the symbols for measurable concentrations in the
equalities above.
1 H+
^
___ Reaction and Extent: 1 WA
^
Conc. At Equilib.: _______________ + 1 CB
^
___ (goes slightly) *****
Reaction and Extent:
Conc. At Equilib.: 1 H+ + 1 CB
^
^
x
x 1 WA
^
[WA]mixed ─ x (goes slightly) The C row is the same as the bottom row of the rice table above, and it will be the same
for all weak acid ionization reactions.
3. The K step. Substitute into the K definition, for each of the three concentration terms
below, the symbol for that term that includes x from the C row above.
Ka ≡ [H+]at eq. • [BC]at eq.
[WA]at eq. ≡ _____________________ ***** © 2011 ChemReview.Net v. 7f Page 937 Module 30 — Weak Acids and Bases Ka ≡ [H+]at eq. [BC]at eq. x2 ≡ [WA]at eq. Ka ≡ [WA]mixed ─ x ^ Definition based on R row ^ Exact based on C row These two equations are the same as those that were written based on the rice table.
Both of the Ka equations above are true at equilibrium, but the exact form of the
equation has an advantage. Our goal is usually to solve for x, and the exact equation
can be solved for x if the Ka value and the [WA as mixed] are known, as they usually are.
However, because the exact equation has both x2 and x terms, it is a quadratic equation.
To solve exactly, you can solve for a, b, and c, and then use the quadratic formula.
x = ─b ± (b2 ─4ac)1/2
2a The arithmetic is not difficult,
but it is timeconsuming. For most weak acid calculations, the following approximation may be used which
solves for x more quickly.
4. The K approximation. In a weak acid solution, the value of x is small compared to the
value of the [WA] as mixed. Using the approximation rule from the previous lesson,
write the simplified, approximate term in the blank below.
[WA]at equilibrium ≡ [WA]mixed ─ x ≈ __________ *****
[WA]at equilibrium ≡ [WA]mixed ─ x ≈ [WA]mixed
When a small quantity is added to or subtracted from a larger quantity, the larger
quantity remains approximately the same.
Now fill in the box below: write an approximation based on the exact equation.
Ka ≡ [H+]eq. [BC]eq.
[WA]at eq. ≡ x•x
≈
[WA]mixed ─ x *****
The Ka expression can be simplified to this approximation:
Ka ≡ [H+]eq. [BC]eq.
[WA]at eq.
^ Definition ≡ x2
[WA]mixed ─ x
^Exact ≈ x2
[WA]mixed ≈ Ka ^ Approximation The small ( ─ x ) difference between the exact and the approximation equation means
that solving the exact equation for x requires the quadratic formula, while solving the
approximation requires only a square root. The approximation can therefore be solved
more quickly. For Ka calculations, we will first solve the approximation equation, and in
most cases the approximation will be an acceptable answer. © 2011 ChemReview.Net v. 7f Page 938 Module 30 — Weak Acids and Bases Summary
For weak acid dissociation (ionization) reactions, these are always the same:
• The general reaction: WA H+ + CB • The rice table.
• The WRECK steps.
• The C step (which is the same as the bottom row of the rice table).
• The K step: the definition, exact, and approximate expressions.
In weak acid ionization (Ka) calculations, all that will vary are the specific formulas for WA
and CB, the values for [WA] and [BC], and the value of Ka . By representing each weak
acid ionizations using the general equation, we simplify problemsolving. Practice A: Check your answers as you go. 1. For the ionization of the weak acid WA to form protons and CB,
a. Write the REC steps. In the C row, each term should include x.
b. Write the Ka definition.
c. Write the Ka exact equation using C row terms that include x.
d. Write the Ka approximation equation.
2. For the ionization of the weak acid NH4+ :
a. Write the Reaction and its Extent, using the particle formulas in the reaction.
b. Below the Reaction and Extent, write the Concentrations at equilibrium, defined so
that each term includes x .
c. Write the Ka definition equation, using the particle formulas.
d. Write the Ka exact equation based on the terms in the C step.
e. Write the Ka approximation equation, based on the answer in part d. Ka Calculations
We will solve Ka calculations in the same way we solved K calculations: by writing the
WRECK steps. To simplify Ka calculations, we will add a step: we will write both the
reaction showing the particles for the specific weak acid and then the general reaction
equation for all weak acids. These two R steps will define the terms in the equations. © 2011 ChemReview.Net v. 7f Page 939 Module 30 — Weak Acids and Bases We will then solve all weak acid calculations using the same general equations. Our rule
will be weak acid rule
2. The Ka Prompt: If a problem involves a weak acid or Ka and [ions], write the
WRRECK steps.
• WANTED: Write the general and specific symbol. • Write the specific Reaction using the symbol for the particles in the problem,
then write this general Reaction and extent:
Rxn. and Extent:
WA
H+ + base conjugate (goes slightly) • Conc. at Eq: • Ka ≡ [H+]eq. [BC]eq. ≡
[WA]at eq. ^
[WA]mixed ─ x ^Definition ^
x ^
x x2 x2 ≈ [WA]mixed ─ x ≈ Ka [WA]mixed ^ Exact ^Approximation c. SOLVE the Ka approximation for the WANTED symbol.
In short: See weak acid or Ka and [ions]? Write WRRECK, solve the approximation. Use the rules above on this example, and then check your answer below.
Q. In a 2.0 M acetic acid (CH3COOH) solution, calculate the [H+] (Ka = 1.8 x 10―5)
*****
Answer
WANT: [H+] = x Specific R: 1 CH3COOH R & E: 1 WA
^
[WA]mixed ─ x Conc. at Eq: Ka ≡ [H+]eq. [BC]eq. ≡
[WA]at eq. 1 H+ + 1 CH3COO─
1 H+ + 1 CB
^
^
x
x
x2 [WA]mixed ─ x ^Definition ^ Exact ≈ (goes slightly) x2
≈
[WA]mixed Ka ^Approximation Solve the boxed approximation. Start with a DATA TABLE using the equation symbols.
*****
DATA:
x = [H+] = ?
[WA]mixed = [CH3COOH]mixed = 2.0 M
Ka = 1.8 x 10―5
Use those values to solve the approximation for the WANTED symbol. © 2011 ChemReview.Net v. 7f Page 940 Module 30 — Weak Acids and Bases *****
Since x2 Ka ≈ and ? = [H+] = x, solve for x2 then x . [WA]mixed
*****
x2 = (Ka ) x ( [WA]mixed ) = (1.8 x 10―5 ) x ( 2.0 ) = 3.6 x 10―5 = 36 x 10―6
x = 6.0 x 10―3 M = [H+] Practice B. (When solving a K for [ ], add M as the unit.) Do both problems. If in doubt, check your answers after each part. 1. Hydrogen cyanide (HCN) is both a weak acid and a notorious poison.
a. Calculate the [H+] in a 0.50 M HCN solution. Use Ka = 6.2 x 10―10 for HCN .
b. Calculate the pH of the HCN solution.
2. The weak acid hydrogen fluoride (HF) is used to etch glass. The [F―] in a 0.25 Molar
HF solution is measured to be 0.012 Molar.
a. Find the Ka value for HF at the temperature of this experiment.
b. What is the pH of the solution?
c. What is the [OH─] in this HF solution? ANSWERS
Practice A
1a. R&E:
Cat eq.: Ka ≡ [H+]eq [BC]eq
[WA]eq.
1b,c,d.
2a. R&E: 2b. Cat eq.:
2d. Ka ≡ 1 H+ + 1 CB
^
^
x
x 1 WA
^
[WA]mixed ─ x
≡ ^ Definition
1 NH4+
^
+]
[NH4 mixed ─ x
x•x
[NH4+]mixed ─ x © 2011 ChemReview.Net v. 7f x2
[WA]mixed ─ x ≈ ^Exact (goes slightly) x2
[WA]mixed
^Approximation 1 H+ + 1 NH3 (goes slightly)
^
x ^
x
1e. Ka ≈ 2c. Ka ≡ [H+]eq [NH3]eq
[NH4+]at eq x2
[NH4+]as mixed Page 941 Module 30 — Weak Acids and Bases Practice B
1a. If Ka and [ions] are mentioned, write the WRRECK’s, solve the approximation.
[H+] = ? WANTED: Specific Reaction: 1 HCN 1 H+ + 1 CN─ R&E: 1 H+
^
x + 1 WA
^
[WA]mixed ─ x Conc. at Eq: Ka ≡ [H+]eq [BC]eq
[WA]at eq. 1 CB
^
x x2
[WA]mxd ─ x ≡ ^ Definition (goes slightly) ≈ ^Exact (goes slightly) ≈ Ka x2
[WA]mixed
^Approximation Solve the Ka approximation for the WANTED symbol.
x2 ≈ (Ka ) ( [WA]mixed ) = (6.2 x 10―10 ) ( 0.50 ) = 3.1 x 10―10
x = [H+] = ? ≈ (estimate: 12 x 10―5) ≈ 1.8 x 10―5 M = [H+]
1b. See pH? Write pH ≡ ─ log [H+] [H+] ≡ 10─pH and WANT pH = ─ log (1.8 x 10─5) = estimate 4.? = 4.74
2a. (4.74 rounded up = 5: Check.) If Ka and [ions] are mentioned, write the WRRECK’s, solve the approximation.
Ka = ?
Specific Reaction: WANTED:
R&E: 1 HF
1 WA
^
[WA]mixed ─ x Conc. at Eq: Ka ≡ [H+]eq [BC]eq
[WA]at eq. ≡ ^ Definition 1 H+ + 1 F─
1 H+ + 1 CB
^
^
x
x x2
[WA]mxd ─ x ≈ ^Exact (goes slightly)
(goes slightly) x2
[WA]mixed ≈ Ka ^Approximation Solve the Ka approximation for the WANTED symbol.
? = Ka ≈ x2
≈
[WA]mixed (1.2 x 10─2)2
0.25 ─4 = 5.8 x 10─4
≈ 1.44 x 10
0.25 A K value has units, but by convention units are not listed with a K value.
b. pH? pH ≡ ─ log [H+] and [H+] ≡ 10─pH WANT: pH = ─ log [H+] = ─ log (1.2 x 10─2) = 1.? = 1.92 (estimate, then calculate.) c. In an acid solution, find [H+] using acid rules, then [OH─] using Kw.
DATA: [H+] = x = 0.012 M = 1.2 x 10─2 M © 2011 ChemReview.Net v. 7f (from Part a) Page 942 Module 30 — Weak Acids and Bases SOLVE: Kw = [H+][OH─] = 1.0 x 10─14 [OH─] = 1.0 x 10─14 = 1.0 x 10─14
[H+]
1.2 x 10─2 = 0.83 x 10─12 = 8.3 x 10─13 M (Kw check: [H+] x [OH─] (circled) must estimate to 10.0 x 10─15 or 1.0 x 10─14).
***** Lesson 30D: Percent Dissociation – and Shortcuts
Moving Numbers Instead of Symbols
For past calculations that use equations, we have solved in symbols before plugging in
numbers. We did so because when using algebra, symbols can be written more quickly
than numbers with their units. In K calculations, however, you may choose to plug
numbers into the K equation and then do the algebra. Why? Numbers without units often
move more quickly than symbols.
The downside is that in Ka as in other K calculations, you lose unit cancellation as a check
on your algebra. This means that in K calculations, you must double check your math. Solving the Approximation Directly
It is important to be able to solve Ka calculations by writing out the methodical WRECK
steps. For upcoming problems that are more complex, we will need those methodical
steps. However, once you master the WRRECK steps, you may solve Ka calculations by
writing these WA quick steps (WASS): W, Approximation, Substitute, Solve.
• Write the WANTED symbol. • Write the Ka approximation equation: Ka ≈ x2
[WA]mixed • Substitute the DATA and solve the approximation for the WANTED symbol. Try those steps on this problem.
Q. Hypochlorous acid ionizes by this reaction.
HOCl H+ + OCl─ Ka = 3.5 x 10―8 Calculate the [H+] in a 0.40 M HOCl solution.
***** © 2011 ChemReview.Net v. 7f Page 943 Module 30 — Weak Acids and Bases Answer
WANTED: [H+] = x
Since hypochlorous acid has a small Ka , it is a weak acid.
Using the WA quick steps (WASS), write the Ka approximation, substitute, solve.
Ka ≈ x2 Substituting: 3.5 x 10―8 = [WA]mixed x2
0.40 SOLVE: To find x WANTED, first solve for x2.
x2 = ( 3.5 x 10―8 ) x ( 0.40 ) = 1.4 x 10―8
x = 1.2 x 10―4 M = [H+] (When solving a K for [ ], add M as the unit.) Percent Dissociation
The percent dissociation is a measure of the strength of a weak acid solution. Percent
ionization, percent dissociation, and percent hydrolysis are all terms that have the same
meaning.
The percent dissociation is simply the percentage of the original weak acid concentration
that ionizes. Since [WA that ionizes] = x , for equation consistency we will write as weak
acid Rule
3. % Dissociation ≡ x ● 100% [WA]ionized ● 100%
[WA]mixed ≡ [WA]mixed
The terms and the definition equations must be memorized. The equations for percent dissociation and the K approximation both have x terms in the
numerator and the same denominator. Note how they differ: The K approximation has x2
on top, while percent dissociation has x on top and multiplies by 100%.
Apply the definition above to the following problem, and then check your answer below.
Q. Benzoic acid ionizes to form the benzoate ion.
C6H5COOH H+ + C6H5COO─ Ka = 6.3 x 10―5 a. Calculate the [H+] in a 0.040 M benzoic acid solution.
(Use the quick steps: solve the approximation equation directly.)
b. Calculate the percent dissociation in the solution.
***** © 2011 ChemReview.Net v. 7f Page 944 Module 30 — Weak Acids and Bases Answer
a. WANTED: [H+] = x = ? To solve using the WA quick steps (WASS),
Ka ≈ x2 Substituting: 6.3 x 10―5 [WA]mixed ≈ x2
0.040 M SOLVE: To find the x WANTED, first solve for x2.
x2 ≈ ( 6.3 x 10―5 ) ( 0.040 ) = 2.5 x 10―6
? = x ≈ (estimate 12 x 10―3) ≈ 1.6 x 10―3 M = [H+] If needed, adjust your work and then try part b.
*****
b. Watch for the answer from one part to be used as DATA for later parts.
Since the dissociation equation is simple, you may solve without listing the symbols
and values in a DATA table.
SOLVE: % Dissoc. = ● 100% x
[WA]mixed = 1.6 x 10―3 M ● 102 % = 4.0 %
4.0 x 10―2 M In 0.040 M C6H5COOH at 25ºC, 4.0% of the weak acid is ionized. Practice
1. A 2.0 Molar solution of a monoprotic weak acid has a pH of 4.49 .
a. Calculate the [H+] in the solution.
b. Calculate the percentage of ionization in this solution.
2. If a weak acid in a 0.50 M solution is 4% dissociated, what is its Ka ? ANSWERS
1a. pH Prompt: pH ≡ ─ log [H+] and [H+] ≡ 10─pH WANT: [H+] = ? = x
For Ka and [ions], write the WRRECK’s, but Ka is not mentioned in this problem.
Want [H+] and know pH. The relationship between the two that solves for [H+] is:
[H+] = 10─pH © 2011 ChemReview.Net v. 7f = 10─4.49 = 3.2 x 10─5 M = [H+] (check: 4.49 up = 5) Page 945 Module 30 — Weak Acids and Bases b. For weak acids, x = [H+] from part a.
% Ionization = % Dissoc. =
x ● 100% = 3.2 x 10─5 M x 102 % = 1.6 x 10─3 % = 0.0016 %
[WA]mixed
2.0 M
2. WANT: Ka Write needed equations based on the terms used in the problem. % Dissociation =
Ka
DATA: ≈ ● 100%
x
[WA]mixed x2
[WA]mixed (Approximation) [WA]mixed = 0.50 M % Dissociation = 4%
In this problem, the % dissociation equation above has one unknown value, and the K approximation has
two. Begin by substituting into the top equation to find x.
*****
4% =
x
● 100% ; x = (4.0%/100%) ● 0.50 M = 0.02 M = 2 x 10─2 M
0.50 M
Solve for the WANTED symbol: Ka ≈ x2
[WA]mixed ≈ (2 x 10─2)2 ≈ 8 x 10─4
0.50 ***** © 2011 ChemReview.Net v. 7f Page 946 Module 30 — Weak Acids and Bases Lesson 30E: Solving Ka Using the Quadratic Formula
Timing: Do this lesson if you are asked to solve Ka calculations using the 5% test and/or
the quadratic formula.
***** The 5% Test
So far, we have solved Ka calculations using the Ka approximation. For weak acid ionization,
the only difference between the Ka exact equation and the Ka approximation equation is the
─ x term in the denominator.
Ka ≡ [H+][conjugate] ≈ x2 [WA]mixed ─ x [WA]mixed ^Exact ^Approximation If x is small, the ─ x term does not substantially change the answer, and the above
approximation may be used. However, if a weak acid solution is very dilute, and/or the
weak acid is close to being a strong acid and ionizes to a significant extent, x may not be
relatively small, and the approximation can introduce significant error in Ka calculations.
When can you safely use the Ka approximation and solve using squares and square roots?
When must you use exact Equation 2 and solve the quadratic formula?
The generally accepted rule is that if the percent dissociation (ionization) is less than 5%,
using the approximation is permitted. In most experiments involving Ka, a 5% variation is
within the range of experimental error.
When solving a Ka equation, the approximation equation is solved first. The 5% test is then
applied to the answer by calculating the % dissociation. If the % dissociation is 5% or less,
the approximation result is accepted. If the percent dissociation is more than 5%, the
calculation is done again, with the original problem data substituted into the exact equation.
The exact equation is then converted to the general quadratic format. The quadratic formula
is then solved to find x.
We will add this rule to our steps for solving Ka calculations:
In Ka calculations,
• SOLVE first for the WANTED symbol using the approximation equation. • Calculate % dissociation. IF greater than 5%, solve the exact Ka quadratic. Solving Ka with the Quadratic Formula
Weak acid calculations in which the dissociation is more than 5% error generally involve
weak acids that either are not very weak (those with a Ka of 10―6 or larger) or are very
dilute solutions. © 2011 ChemReview.Net v. 7f Page 947 Module 30 — Weak Acids and Bases Try this example in your notebook, then check the answer below.
Q. In a 0.010 M HF solution at 25oC ( Ka HF = 6.8 x 10―4 ),
a. find [H+] using the approximation equation.
b. Does the HF in this solution ionize more than 5%?
*****
Answer
WANT: [H+] = x Part a: Ka x2 ≈ [WA]mixed
x2 ≈ (6.8 x 10―4 ) ( 0.010 ) = 6.8 x 10―6 x = ? ≈ (estimate 23 x 10―3) ≈
**** 2.6 x 10―3 M ≈ [H+] Now try Part b. * Part b:
WANT: % Dissociation = ● 100% x
[WA]mixed = 2.6 x 10―3 ● 102 % = 26 %
1.0 x 10―2 Applying the 5% test, because the % dissociation is 26%, the x value is too high to use the
approximation equation. In such cases, the exact Ka equation should be solved. Write the
exact form of the Ka equation, then substitute the values for Ka and [WA].
*****
Ka ≡ x2 Substituting: 6.8 x 10―4 = [WA]mixed ─ x x2
0.010 ─ x The substituted equation is a quadratic because it has x2 and x terms. Solve the substituted
equation above for x by applying the quadratic formula. Use an online or handheld
calculator if permitted for solving quadratics in your course. Refer to Lesson 28J if needed.
*****
Begin by converting the numeric equation into the quadratic format: ax2 + bx + c = 0,
6.8 x 10―4 = x2
0.010 ─ x 6.8 x 10―4 (0.010 ─ x) = x2
6.8 x 10―6 ─ (6.8 x 10―4) x = x2
x2 + (6.8 x 10―4)(x) ─ 6.8 x 10―6 = 0
To solve for x, substitute values for a, b, and c into the quadratic formula.
***** © 2011 ChemReview.Net v. 7f Page 948 Module 30 — Weak Acids and Bases a = 1 , b = 6.8 x 10―4 , c = ─6.8 x 10―6 . For substituting into an online quadratic
calculator, you may need to convert to a = 1, b = 0.000 68, c = ─0.000 006 8 .
Solving without the online calculator by substituting into the quadratic formula:
x = ─b ± (b2 ─4ac)1/2 =
2a ─( 6.8 x 10―4) ± {(―6.8 x 10―4)2 ─ 4(1)( ─6.8 x 10―6)}1/2
2(1) x= (─ 6.8 x 10―4) ± {(+ 46.2 x 10―8) + (27.2 x 10―6)}1/2
2 x= (─ 6.8 x 10―4) ± {(+ 0.462 x 10―6) + (27.2 x 10―6)}1/2
2 x= (─ 6.8 x 10―4) ± (+ 27.7 x 10―6)1/2
2 x= (─ 0.68 x 10―3) + (5.26 x 10―3)
2 x = (+ 4.58 x 10―3)
2 OR x = =
= (─ 6.8 x 10―4) ± (+ 5.26 x 10―3) =
2 = OR = x= (─ 0.68 x 10―3) ─ (+ 5.26 x 10―3))
2 ─ 5.94 x 10―3 ; x = + 2.29 x 10―3 OR x = ─ 2.97 x 10―3
2 Since x is a concentration, and concentrations cannot be negative, x = 2.3 x 10―3 M Using the approximation equation in Part a, x was 0.0026 M. Using the exact equation, x =
0.0023 M. The difference between those answers is roughly 12%. For a weak acid that is
not very weak, the exact formula provides significantly more accurate answers. Practice
Use the exact Ka equation to solve. Use an online quadratic formula calculator or handheld
calculator for assistance if it is allowed in your course.
1. In a 0.010 M CH3COOH solution, find the [H+]. (Ka = 1.8 x 10―5) ANSWERS
1. WANTED = [H+] = x . The exact Ka equation based on the original [WA] as mixed, is
x2 Ka ≡ Substituting: 1.8 x 10―5 = [WA]mixed ─ x x2
0.010 M ─ x Since the equation includes x2 and x terms, it is a quadratic.
To solve, convert to the quadratic format, then solve the quadratic formula.
x2 = ( 1.8 x 10―5 ) ( 0.010 M ─ x)
x2 = ( 1.8 x 10―7 ) ─ ( 1.8 x 10―5 )x
x2 + ( 1.8 x 10―5 )x ─ ( 1.8 x 10―7 ) = 0
© 2011 ChemReview.Net v. 7f Page 949 Module 30 — Weak Acids and Bases a = 1 , b = 1.8 x 10―5 , c = ─1.8 x 10―7 For a quadratic calculator, you may need to input a = 1 , b = 0.000 018 , c = ─ 0.000 000 18
Substituting these values into a quadratic equation calculator to solve:
x = + 4.15 x 10―4 or x = ─ 4.33 x 10―4 Since x is a concentration and must be positive, x = 4.2 x 10―4 M = [H+]
***** Lesson 30F: Weak Bases and Kb Calculations
Timing: Do this lesson if you are assigned calculations involving Kb.
***** Weak Bases
A base reacts with water to produce OH─ ions and the acid conjugate of the base.
The strong bases NaOH and KOH will dissociate (ionize) ~100% when dissolved in water.
For a weak base (WB), only a small percentage of particles will react with water.
An example of the reaction of a weak base and water occurs in the mixture of ammonia
(NH3) and water, a solution used in many glasscleaning products. When ammonia gas is
dissolved in water, a small percentage of the NH3 molecules react by removing a proton
from a water molecule. This hydrolysis (reaction with water) of a base can be represented
as
Specific: NH3(aq) + H2O(l) General: WB + H2O OH─ + + NH4+(aq) (goes slightly) acid conjugate OH─(aq) (goes slightly) The formation of OH─ ions in the reaction with water makes ammonia a base. The small
percentage of ammonia molecules that react means that ammonia is a weak base.
The reaction of a weak base with water is reversible, and in the closed system of an
aqueous solution the reaction will go to equilibrium. The equilibrium constant expression
for the hydrolysis of a weak base is given the symbol Kb.
Write the K expressions for the general and specific reactions above.
*****
General: Kb = [OH─] [acid conjugate]
[WB] Specific: Kb = [OH─] [NH4+]
[NH3] with all concentrations measured at equilibrium.
A Kb expression always has [OH─] on top and [WB] on the bottom. © 2011 ChemReview.Net v. 7f Page 950 Module 30 — Weak Acids and Bases As in the case of Ka and Kw expressions, the concentration of liquid water is not included in
the Kb expression. In most weak base solutions, the concentration of water remains very
high and essentially constant during hydrolysis reactions.
One difference between weak acids and weak bases is that a weak acid nearly always
contains an H atom and loses an H+ ion when it ionizes or hydrolyzes (“Lewis acids” are an
exception). In contrast, as in the case of NH3 above, most weak bases do not contain OH─
ions, so that they do not dissociate (ionize) to lose them. Instead, most weak bases create an
OH─ by hydrolysis: by reacting with water to remove an H+ from H─OH.
For example, the weak acid behavior of acetic acid can be represented as either
H+(aq) + CH3COO─(aq) CH3COOH(aq) ionization:
or hydrolysis: H3O+(aq) + CH3COO─(aq) CH3COOH(l) + H2O(l) but the weak base behavior of ammonia (NH3) can only be represented by hydrolysis:
NH3(aq) + H─OH(l) NH4+(aq) + OH─(aq) As with weak acids, the values of Kb for the hydrolysis of weak bases will be between 1.0
and 10―16. The equilibrium constant equation for the hydrolysis of ammonia is
Kb = [NH4+][OH─] = 1.8 x 10―5 at 25oC. [NH3] Kb Calculation Rules
The rules for calculations involving weak bases and Kb are similar to those for weak acids
and Ka. Compare:
1. Acids hydrolyze (react with water) to form H3O+ ions.
Bases hydrolyze (react with water) to form OH─ ions.
2. Weak acids react with water (ionize slightly in water) to form H3O+ (or H+) ions.
Weak bases react with water slightly to form OH─ ions.
3. In acidbase reactions, acid particles lose an H+ to form the base conjugate particle.
In acidbase reactions, base particles gain an H+ to form the acid conjugate particle.
Using those definitions, cover the answer below and then try this problem.
Q. Fluoride ion (F─) is a weak base. When F─ is mixed with water, which particles
will form in the hydrolysis reaction?
F─ + ***** © 2011 ChemReview.Net v. 7f H2O Page 951 Module 30 — Weak Acids and Bases Answer
Weak base
F─ Acid Conjugate H2O + HF + OH─ ~1% of the F─ reacts to form these at equilibrium ~99% unreacted The base particle F─ gains an H+ when it reacts, and becomes the acid conjugate of the
HF. When a base reacts with water, one of the products is always OH─ ion.
Check that this equation is balanced for atoms and charge. Practice A: Check your answers at the end of the lesson. 1. Assuming the first particle is acting as a base, complete the reaction.
a. HS─ + H O
2 b. CH3COO─ + H2O
c. H2PO4─ + H2O
2. Write the expression for Kb for each reaction above.
3. If each particle below is acting as a base, write the formula for its acid conjugate.
a. CO32─ b. HSO4─ c. HPO42─ More Ka and Kb Comparisons
4. The general reaction for the hydrolysis of a weak acid is
WA + H O
H O+ + base conjugate
2 3 The general reaction for the hydrolysis of a weak base is
OH─ + acid conjugate
WB + H O
2 (goes slightly)
(goes slightly) The acid conjugate will have one more H and one more + charge than the base.
5. When a weak acid is mixed with water,
[WA]that ionizes = x = [H3O+ or H+]formed = [base conjugate]formed
When a weak base is mixed with water, x is the small mol/L of weak base that reacts.
[WB]that hydrolyzes = x = [OH─]formed = [acid conjugate]formed
6. The K for the ionization of an acid is termed the aciddissociation constant: Ka.
The K for the hydrolysis of a base is termed the basehydrolysis constant: Kb .
Since most bases do not dissociate by losing an OH─ ion, basehydrolysis is the
preferred term. © 2011 ChemReview.Net v. 7f Page 952 Module 30 — Weak Acids and Bases 7. Ka is between 1 and 10─16 for a weak acid. Kb is between 1 and 10─16 for a weak base.
The higher the Ka, the stronger the acid. The higher the Kb, the stronger the base.
8. Ka ≡ [H+]eq. [base conjugate]eq. ≡
[WA]at eq. x2 ≈ [WA]mixed ─ x Kb ≡ [OH─]eq. [acid conjugate]eq. ≡
x2
[WB]at eq.
[WB]mixed ─ x
^ Definition ^Exact ≈ x2
[WA]mixed
x2
[WB]mixed
^ Approximation 9. Ka expressions have [H+] or [H3O+] on top. Kb expressions have [OH─] on top.
***** Percent Hydrolysis for Bases
In Kb expressions, x measures the small [WB] used up in the hydrolysis reaction. The
percent hydrolysis is the percentage of the original base that reacts.
The percent hydrolysis of a weak base can be calculated in the same way as the percent
dissociation of a weak acid. We will therefore modify Rule 3 to read
3: % Dissociation/Ionization/Hydrolysis/5% Test:
% Dissociation = x
● 100%
[ WA or WB ]mixed = [ WA or WB ]hydrolyzed ● 100%
[ WA or WB ]mixed The 5% Test For Weak Bases
If you are asked to use the 5% dissociation test for weak bases, use the same steps as for
weak acids. First solve with approximation equation, then do the 5% test. If the percent
hydrolysis of the weak base is greater than 5%, solve for x using exact equation (with the
─ x term) and a procedure for solving quadratic equations. © 2011 ChemReview.Net v. 7f Page 953 Module 30 — Weak Acids and Bases Summary: Weak Base Calculations.
4. Kb Prompt. If a problem involves a weak base and/or Kb and its [ions], write the
WRRECK steps.
• WANTED: Write the general and specific symbol. • Write the specific reaction using the symbol for the particles in the problem, then
write this general reaction for weak base hydrolysis and its extent:
WB + H O
OH─ + acid conjugate
(goes slightly)
R & E:
2 ^
[WB]mixed ─ x • Conc. • Kb ≡ [OH─]eq.[conjugate]eq.
[WB]at eq. ^
x
≡ ^
x
x2 ≈ [WB]mixed ─ x x2
[WB]mixed where: x = [OH─] = [acid conjugate] = [WB]hydrolyzed
5. In Ka and Kb calculations, first solve the approximation for the WANTED unit or
symbol, then calculate the percent dissociation. IF >5%, solve the exact equation
using the quadratic formula.
6. Ka expressions have [H+] or [H3O+] on top. Kb expressions have [OH─] on top.
Ka solves for x = [H+] , Kb solves for x = [OH―].
In short: For Ka and Kb calculations, write the WRRECK’s, solve the approximation, then
calculate % dissociation. If greater than 5%, solve the exact quadratic.
After mastering the WRRECK steps, you may solve Ka and Kb calculations using the WASS
quick steps: Write the WANTED symbol, the Approximation equation, Substitute, Solve,
and then apply the 5% test. Quick 5% Test
Here’s a quick way to apply the 5% test to see if a Ka or Kb approximation may be used.
• Using scientific notation, compare [WA] to the calculated [H+] or [WB] to [OH─] . • If the exponents differ by 3 or more, the ionization must be less than 5% and the
approximation is acceptable to use. • If the exponents differ by 2 or less, solve the % dissociation equation for the 5% test. Let’s test this rule.
Q. For a weak acid solution, [H+] is calculated using the Ka approximation to be
9.9 x 10─4 M at [WA] = 0.10 M. Does this calculation pass the 5% test?
***** © 2011 ChemReview.Net v. 7f Page 954 Module 30 — Weak Acids and Bases [WA] = 0.10 M = 1.0 x 10─1 M , [H+] = x = 9.9 x 10─4 M.
Since the difference between ─1 and ─4 is 3 or greater, this ionization must be less
than 5%, so the approximation gives acceptable results, but let’s calculate to be sure.
% Dissoc. = x
● 100% = 9.9 x 10―4 M ● 102 % = 0.99 %
[WA or WB]mixed
1.0 x 10─1 M This example is contrived to show the highest percent dissociation possible when
the exponents differ by 3 or more, and it passes the 5% test, so the quick rule works. Practice B: Try both. 1. Codeine (C18H21NO3) is a physically addictive opiate which is an ingredient in some
prescription cough suppressants. Codeine is a weak base with a Kb = 9.2 x 10―7. Use
the full WRRECK steps to calculate the [OH─] in a 0.010 M codeine solution.
2. [OH─] is 3.6 x 10―5 M in a 0.12 M solution of the weak base hydroxylamine (HONH2).
a. Using the WASS quick steps, calculate the Kb value for HONH2 .
b. Apply the 5% test to the Kb approximation calculation. Calculating Kb from Ka
In general,
• The conjugate of a strong acid or base (K = very large) is pH neutral. • The conjugate of a weak acid or base is a weak opposite. • The conjugate of a very weak acid or base (K < 10―16) is a strong opposite. The term weak acid or base generally refers to particles with a K between 1 and ~10―16.
For these particles, at a given temperature,
• a weak acid has a characteristic Ka, and its base conjugate has a characteristic Kb. • A weak base has a characteristic Kb, and its acid conjugate has a characteristic Ka. Because weak base solutions are systems at equilibrium, the weak base reacts slightly to
form an acid conjugate, and the acid conjugate can react in the reverse reaction to reform
the weak base. K values take into account both the forward and reverse reactions.
Since reactions for weak acids and bases are reversible, they can be written backwards,
from the perspective of the conjugate. One consequence of this reversibility is that for a
weak acid or a weak base and its conjugate, the Kb and Ka of the particles in the conjugate
pair are related mathematically. Ka and Kb, when multiplied, must equal Kw.
For acidbase conjugate pairs: © 2011 ChemReview.Net v. 7f Kw = Ka x Kb = 1.0 x 10─14 at 25ºC. Page 955 Module 30 — Weak Acids and Bases Note that for any conjugate pair, Kb and Ka are inversely proportional: the larger is one,
the smaller must be the other.
We will summarize these points as rules
7. An acid particle loses an H+ to become the base conjugate of the acid.
A base particle gains an H+ to become the acid conjugate of the base.
8. For acidbase conjugate pairs: Kw = Ka x Kb = 1.0 x 10─14 at 25ºC. Ka values are often listed in tables. Kb values are less often listed in tables. However, if
either a Ka or Kb value is known, the K value of the conjugate can be calculated by
• writing the chemical formula for the conjugate, and • applying the relationship between the K values of acidbase conjugates:
Kw = Ka • Kb = 1.0 x 10─14 . Using that rule, try this problem.
Q. The Ka for HF is 6.8 x 10―4.
a. What is the chemical formula for its base conjugate?
b. What is the Kb of its base conjugate?
*****
Answer
a. The formula for the base conjugate of HF is F─. The base conjugate of an acid has
one fewer H atoms and one fewer positive charges.
b. WANT:
DATA: Kb o f F ─
Ka for HF = 6.8 x 10―4 . To find the Kb of F─, write the chemical formula for its acid conjugate: HF.
Then apply the Ka value of HF and the rule for acidbase conjugates:
Kw = Ka • Kb = 1.0 x 10─14
SOLVE: ? = Kb = 1.0 x 10─14
Ka = 1.0 x 10─14
6.8 x 10―4 = 1.5 x 10─11 = Kb of F─ Check: just as [H+] ● [OH─] must estimate to = Kw = 10.0 x 10─15 or 1.0 x 10─14,
Kb ● Ka (circled) must estimate to = Kw = 10.0 x 10─15 or 1.0 x 10─14. © 2011 ChemReview.Net v. 7f Page 956 Module 30 — Weak Acids and Bases Practice C
Complete the oddnumbered problems, and more if you need more practice.
1. Write Rules 18 for weak acids and bases until you can write them from memory.
2. If the Kb of the weak base methylamine (CH3NH2) is 4.4 x 10─5, what is the Ka value
for CH NH +?
3 3 3. Phosphoric acid is triprotic: it has three
hydrogens that can ionize. Each acid
particle that can lose a proton has a
different Ka value, as shown in this chart.
Based on this data, what would be the Kb
of HPO42─? Acid Ka at 25˚C H3PO4 7.5 x 10─3 H2PO4─ 6.2 x 10─8 HPO42─ 4.2 x 10─13 4. In acetic acid, CH3COOH, the last H in the formula is weakly acidic. The Ka for acetic
acid is 1.8 x 10─5. Write the
a. Chemical formula for the base conjugate of acetic acid.
b. Kb value for the base conjugate.
c. Balanced equation for the reaction that this Kb is the equilibrium constant for.
d. Kb expression for the hydrolysis of the base conjugate of acetic acid.
5. Aniline (C6H5NH2) is a weak base used in synthesizing textile dyes (Kb = 3.8 x 10─10).
a. Find the Ka of C6H5NH3+.
b. Write a balanced equation for aniline reacting as a weak base with water.
c. Find the [OH─] in a 0.020 M aniline solution. Use the full WRRECK steps.
d. What will be the pH of the Part c solution? ANSWERS
Practice A
1. a. HS─ + H2O
c. H2PO4─ + H2O
2. a. Kb = [H2S][OH─]
[HS─] H2S + OH─
H3PO4 + OH─ CH3COOH + OH─ Bases reacting with water form OH─. b. Kb = [CH3COOH][OH─]
[CH3COO─ ] Kb expressions always have [OH─] on top.
3. Acid conjugates: a. CO32─ HCO3─ © 2011 ChemReview.Net v. 7f b. CH3COO─ + H2O b. HSO4─ c. Kb = [H3PO4][OH─]
[H2PO4─ ] H2SO4 c. HPO42─ H2PO4─ Page 957 Module 30 — Weak Acids and Bases Practice B
2a. If Kb and [ions] are mentioned, write the WRRECK’s, solve the approximation, do the 5% test.
WANT: [OH─] = x
1 OH─ + 1 HC18H21NO3+ Specific Reaction: 1 C18H21NO3 + H2O (In this problem, those specific formulas are not needed to solve.)
OH─ + acid conjugate
General R+E:
WB + H O
^
[WB]mixed─ x Conc. at Eq: 2 Kb = [OH─][acid conjugate]
[WB]mixed ─ x Kb : ^
x
≈ x2
[WB]mixed (goes slightly)
(goes slightly) ^
x
≈ Kb Solve the Kb approximation for x:
x2 ≈ (9.2 x 10─7 ) ( 0.010 ) = 9.2 x 10―9 = 92 x 10―10
x ≈ (estimate 910 x 10―5) ≈ 9.6 x 10―5 M = [OH─]
Since the approximation was used, apply the 5% test:
% Hydrolysis = ● 100% = 9.6 x 10―5 ● 102 %
x
[WA or WB]mixed
1.0 x 10―2 = 9.6 x 10─1 % = 0.96 %, which is less than 5%, so approximation is OK.
2a. WANT: Kb
Kb ≈ starting from the approximation equation.
x2
[WB]mixed SOLVE: ? = Kb ≈ b. % Hydrolysis = Remember that in Kb, x solves for [OH―]. = (3.6 x 10─5)2 = 13.0. x 10─10 = 1.1 x 10─8 = Kb
x2
[WB]mixed
0.12
0.12 x
● 100% = 3.6 x 10―5 ● 102 % = 3.0 x 10─2 %
[WA or WB]mixed
0.12
= 0.03% < 5% Practice C
2. WANT:
DATA: SOLVE: Ka of acid conjugate (the acid conjugate has one more H and one more + charge)
Kb of base = 4.4 x 10─5
For acidbase conjugates: Kw = Ka x Kb = 1.0 x 10─14
Ka = Kw = 1.0 x 10─14 = 1.0 x 10─14 = 0.227 x 10─9 = 2.3 x 10─10 = Ka
Kb
Kb
4.4 x 10―5 Quick check: Kb x Ka must estimate to Kw = 10.0 x 10─15 or 1.0 x 10─14. Try it. © 2011 ChemReview.Net v. 7f Page 958 Module 30 — Weak Acids and Bases 3. WANT: Kb of HPO42─
DATA: Kb can be found from the Ka values in the table using Kw, but which Ka is used?
That of the acid conjugate of HPO42─, which is H2PO4─ with a Ka = 6.2 x 10―8
Kw = Ka x Kb = 1.0 x 10─14
SOLVE: Kb = Kw = 1.0 x 10─14 = 1.0 x 10─14 = 0.16 x 10─6 = 1.6 x 10─7 = Kb
Ka
Ka
6.2 x 10―8 Quick check: Kb x Ka must estimate to = Kw = 10.0 x 10─15 or 1.0 x 10─14.
4. a. CH3COO─ . The base conjugate is the acid particle minus an H+.
b. WANT:
DATA: Kb of base conjugate
Ka of acid = 1.8 x 10―5
For acidbase conjugates: Kw = Ka x Kb = 1.0 x 10─14 SOLVE: Kb = Kw = 1.0 x 10─14 = 1.0 x 10─14 = 0.56 x 10─9 = 5.6 x 10─10 = Kb
Ka
Ka
1.8 x 10―5 Quick check: Kb x Ka must estimate to = Kw = 10.0 x 10─15 or 1.0 x 10─14
c. Kb is always the K for this reaction: acid conjugate + OH─ base + H2O Here the acidbase conjugate pair are CH3COOH and CH3COO─.
So the base is the acetate ion: CH3COO─ + H2O CH3COOH + OH─ d. Kb is always the K expression for the reaction of a base with water: the reaction in Part c above:
with all concentrations are measured at equilibrium.
Kb = [OH─] [CH3COOH]
[CH3COO─]
A Kb expression always has [OH─] on top.
5. a. WANT: Ka of C6H5NH3+ (the acid conjugate of aniline  one more H and one more + charge)
DATA: Kb of base = 3.8 x 10─10
For acidbase conjugates: Kw = Ka x Kb = 1.0 x 10─14 Ka = 1.0 x 10─14 = 1.0 x 10─14 = 0.26 x 10─4 = 2.6 x 10─5 = Ka
Kb
3.8 x 10─10
Quick check: Kb x Ka must estimate to = Kw = 10.0 x 10─15 or 1.0 x 10─14. SOLVE: b. Aniline is a weak base. Weak bases take a proton from water:
Specific Reaction: © 2011 ChemReview.Net v. 7f C6H5NH2 + H2O C6H5NH3+ + OH─ (goes slightly) Page 959 Module 30 — Weak Acids and Bases c. WANT: [OH─] = x = ?
General R+E:
Conc. at Eq:
Kb: Kb = For Ka or Kb and [ions] calculations, write the WRRECK’s.
OH─ + acid conjugate
^
^
x
x WB + H2O
^
[WB]mixed─ x
[OH─][conjugate]
[WB]mixed ─ x ≈ x2
[WB]mixed (goes slightly) ≈ Kb Since [OH─] = ? = x , solve the Kb approximation for x:
x2 ≈ (Kb ) ( [WB]mixed ) = ( 3.8 x 10─10 ) ( 0.020 ) = 7.6 x 10―12
x ≈ (estimate: 23 x 10―6) = 2.8 x 10―6 M = [OH─] Since the Kb approximation was used, ask: is the weak base more than 5% dissociated?
By the quick 5% test, since for 2.8 x 10―6 M = [OH─] and Kb = 3.8 x 10─10, the exponents
differ by more than 3, we can assume the dissociation is less than 5%. To be sure:
% Hydrolysis = x
● 100% = 2.8 x 10―6 ● 102 % =
[WA or WB]orig.
2.0 x 10―2 = 1.4 x 10─2 % = 0.014%, which is less than 5%, so the approximation is OK.
d. WANT: pH Write pH ≡ ─ log [H+] and [H+] ≡ 10─pH We don’t know [H+]. We know 2.8 x 10―6 M = [OH─] One way to solve is to find pOH, then pH:
pOH ≡ ─ log[OH─], [OH─] = 10─pOH and pH + pOH = 14.00
pOH ≡ ─ log[OH─] = ─ log(2.8 x 10―6) = 5.? = 5.55 = pOH
then pH + pOH = 14.00 ; pH = 14.00 – pOH = 14.00 – 5.55 = 8.45 = pH
Note that this is a slightly basic pH as would be expected in a weak base solution.
*** ** Lesson 30G: Polyprotic Acids
Timing: Do this lesson if you are assigned problems that involve calculating the ion
concentrations or pH in polyprotic acid solutions.
***** Ka Values For Polyprotic Acids
Acids that have more than one acidic hydrogen are polyprotic acids. Examples of polyprotic
acids include H2SO4 (sulfuric acid), H2CO3 (carbonic acid), H3PO4 (phosphoric acid), and
H3C6H5O7 (citric acid). © 2011 ChemReview.Net v. 7f Page 960 Module 30 — Weak Acids and Bases Polyprotic acids, by definition, lose their first proton to form another acid.
For example, the weak polyprotic acid H2CO3 can lose one proton to form HCO3─.
The HCO3─ can also act as an acid, losing its proton to form CO3─2. These two
reactions can be written as
H2CO3 H+ + HCO3― Ka = 4.3 x 10―7 HCO3― H+ + CO32― Ka = 5.6 x 10―11 For all polyprotic acids,
• the base conjugate formed by the first acid dissociation is the acid in the second; • the second hydrogen ionizes less readily than the first, and, if there are more acidic
protons, the each ionizes less readily than the prior hydrogen; • Ka values are smaller for each successive ionization. When the successive ionizations of a polyprotic acid are written in a series, if there are
middle particles in the series (between the first and last particles), all will be amphoteric:
they can react as an acid when mixed with bases, and a base when mixed with acids.
The Ka values for the successive ionizations are often numbered.
For example, for H2CO3 above, Ka = 4.3 x 10―7 and Ka = 5.6 x 10―11.
1
2
Ka1 is the Ka value for the reaction where H2CO3 loses its first proton. Ka2 is the Ka
for the reaction in which the second proton of the original polyprotic acid is lost:
the proton is lost from HCO ―.
3 Try this problem.
Q. For citric acid (H3C6H5O7),
Ka = 7.1 x 10―4 , Ka = 1.7 x 10―5, and Ka = 4.0 x 10―7.
1
2
3
2―
+ + C H O 3―
H
a. Write the Ka value for HC6H5O7
657
b. For Ka = 1.7 x 10―5, write the reaction that this is a Ka value for.
2
*****
Answer
a. It helps to write the equations for the successive ionizations.
Ka is the K value for this reaction:
1 H3C6H5O7 H+ + H2C6H5O7― Ka is the K value for:
2 H+ + HC6H5O72― Ka is the K value for:
3
For part a: HC6H5O72―
b. H2C6H5O7―
HC6H5O72― H+ + C6H5O73― Ka2 is the K value for: H+ + C6H5O73― Ka = Ka = 4.0 x 10―7.
3
―
+ + HC H O 2―, the reaction in
H2C6H5O7
H
657 which H3C6H5O7, after losing one acidic hydrogen, loses a second. © 2011 ChemReview.Net v. 7f Page 961 Module 30 — Weak Acids and Bases Practice A
1. For phosphoric acid (H3PO4),
Ka = 7.5 x 10―3 , Ka = 6.7 x 10―8, and Ka = 4.2 x 10―13.
2
3
1
H+ + HPO42―
a. Write the Ka value for: H2PO4―
b. For Ka , write the reaction that this is a Ka value for.
3 The [H+] in Polyprotic Acid Solutions
To calculate the [H+] in polyprotic acid solutions, there are three cases to consider.
•
• Acids in which a numeric Ka is 100 or more times larger than Ka ;
1
2
Acids in which Ka is not 100 or more times larger than Ka , and
1
2 • The special case of sulfuric acid, which is both a strong and a weak acid.
For the first case, in which the first two Ka values are widely separated, the [H+] and pH of a
weak polyprotic acid solution is calculated using only Ka1. The [H+] contributed by the
second ionization will be small: so small that it can be ignored when calculating the and
pH due to the larger first ionization. This will be the case for most polyprotic weak acids.
However, in citric acid, as shown in the example above, Ka1 and Ka2 are relatively close.
In the case of sulfuric acid (H2SO4), the first ionization is essentially 100%, and the
dissociation of the second proton is high (Ka2 = 1.0 x 10―2). In both of these cases, the [H+]
from both the first and second ionizations are calculated and added. There are no
polyprotic acids in which Ka3 has a substantial effect on the pH of the weak acid solution.
We will call this Rule
9. For solutions of polyprotic acids, to find [H+] and pH:
• If Ka1 is 100 or more times larger than Ka2, use Ka1 to find [H+] and ignore Ka2 ;
• If Ka1 is not 100 or more times larger than Ka2, add [H+] from Ka1 and Ka2 . Practice B
1. For phosphoric acid (H3PO4), calculate the [H+] in a 3.5 M H3PO4 solution. Use the
approximation equation and the K values in Practice A above.
2. For sulfuric acid (H2SO4), Ka = very large, and Ka = 1.0 x 10―2. Calculate the [H+]
1
2
in a 0.50 M H2SO4 solution. © 2011 ChemReview.Net v. 7f Page 962 Module 30 — Weak Acids and Bases ANSWERS
Practice A
1a. Phosphoric acid has 3 acidic hydrogens. For clarity, write the three ionization equations.
H+ + H2PO4―
Ka is the K value for: H3PO4
1
Ka is the K value for: H2PO4―
H+ + HPO42―
2
Ka is the K value for: HPO42―
H+ + PO43―
3
For the part a reaction, the Ka is Ka = 6.7 x 10―8
2
H+ + PO43―, the reaction where the original
1b. As shown above, Ka is the K value for HPO42―
3
phosphoric acid, after losing two acidic hydrogens, loses its third. Practice B
1. For phosphoric acid, Ka1 is > 100 (102) times larger than Ka2 ; so the second ionization is small compared
to the first and can be ignored. Calculate [H+] based only on the first ionization, using Ka1 .
Ka1 is the K value for: H3PO4
H+ + H2PO4― Ka = 7.5 x 10―3
1
Solve the approximation equation, then apply the 5% test.
WANT: [H+] = x
x2
[WA]mixed Ka ≈ 7.5 x 10―3 = Substituting: x2
3.5 x2 ≈ (7.5 x 10―3) ( 3.5 ) = 2.6 x 10―2
x = [H+] = ? ≈ (estimate: 12 x 10―1) ≈ 0.16 M = [H+]
Since the approximation was used, apply the 5% test:
5% test = % Hydrolysis = x
● 100% = 0.16 M ● 102 %
[WA or WB]mixed
3.5 M = 4.6 %, which is close but less than 5%, so the approximation may be used.
2. Because sulfuric acid is both a strong and a weak acid, the [H+] must be calculated in two parts.
Since the first Ka is very large, the first H+ from H2SO4 is formed essentially 100%, as in strong acids.
For strong acids losing one proton: H2SO4
^
0.50 M used up H+ + HSO4― (goes ~100%)
^
^
0.50 M formed of both The HSO4― formed also ionizes substantially, since it has a relatively high Ka = 1.0 x 10―2
2
―
+ + SO 2―
H
(goes slightly)
which is the Ka for this reaction: HSO4
4
Calculate the [H+] in the solution due to the ionization of the moderately weak acid HSO ― .
4 ***** © 2011 ChemReview.Net v. 7f Page 963 Module 30 — Weak Acids and Bases If Ka and [ions] are mentioned, write the WRRECK’s, solve the approximation, apply the 5% test.
[H+] = ? = x WANTED: HSO4― R& E: H+ ^
[WA]mixed─ x Conc. at Eq: + ^
0.50 + x SO42― (goes slightly, use K) ^
x (rice bottom row) Note the [H+] in the C step above. The [H+] must include the H+ present from the first ionization.
Use the C step results to write the K expression.
K step: [ H+ ][ SO42─ ]
≡ (0.50 + x) (x) ≈ (0.50) (x) = x ≈ Ka
0.50 ─ x
0.50
[HSO4―]mixed ─ x
^ Definition
^ Exact
^ Approximate
x = [H+] = [SO42―] = [HSO4―]that ionized in the second ionization Ka ≡ where Based on the approximation, the result is x ≈ Ka = 1.0 x 10―2 = 0.010 M
Since an approximation was used, apply the 5% test:
% Hydrolysis = ● 100% = 0.010 M ● 100 %
x
[WA or WB]mixed
0.50 M = 2.0 %, which is less than 5%, so using the approximation is OK.
The TOTAL [H+] = the sum from the two ionizations = 0.50 + 0.010 M = 0.51 M H+
When adding sf , round to the highest place with doubt. In this solution, the second ionization occurs,
but does not have a major impact on the [H+] due to the first ionization. In less concentrated sulfuric
acid solutions, the effect of the second ionization is larger.
***** SUMMARY  Weak Acids and Bases
You may want to organize this summary into charts and flashcards.
1. In acid and base solutions, to calculate [particles],
• in strong acids such as HCl or HNO3, or strong bases such as NaOH and KOH,
either write the REC steps for 100% ionization, or use the quick steps to find [ions]. • in weak acid and base solutions, write the WRECK steps. Use x to represent the
small mol/L of weak acid or base that reacts (ionizes or hydrolyzes). 2. Ka Prompt: If a problem has a Ka and [ions], write the WRRECK steps.
• WANTED: Write the general and specific symbol. • Write the specific reaction using the symbol for the particles in the problem, then
write this general reaction and extent: • Rxn. and Extent: • Conc. WA
^
[WA]mixed ─ x © 2011 ChemReview.Net v. 7f ▐ H+ + base conjugate
^
^
x
│
x (goes slightly) Page 964 Module 30 — Weak Acids and Bases Ka ≡ [H+]eq. [BC]eq. ≡
[WA]at eq. • x2
[WA]mixed ─ x ^Definition x2
≈
[WA]mixed ≈ ^ Exact Ka ^Approximate 3: % Dissociation and 5% Test : % Dissociation =
x
● 100%
[WA or WB]mixed 4: Kb Prompt. If a problem is about a Kb and its [ions], write the WRRECK’s:
• WANTED: Write the general and specific symbol. • Write the specific reaction using the symbol for the particles in the problem, then
write this general reaction for weak base hydrolysis and its extent:
R & E:
WB + H O
OH─ + acid conjugate
(goes slightly)
2 ^
[WB]mixed ─ x ^
x • Conc. • Kb ≡ [OH─]eq.[acid conjugate]eq. ≡
[WB]at eq. ^
x
x2 ≈ [WB]mixed ─ x ^Definition ^ Exact x2
[WB]mixed ^Approximate where: x = [OH─] = [conjugate] = [WB]used up/reacting/hydrolyzing
5. In Ka and Kb calculations, first solve the approximation for the WANTED unit or
symbol, then calculate the percent dissociation. IF >5%, solve the exact equation
using the quadratic formula. 6. The Quick 5% Test to see if an approximation equation provides acceptable results.
• Using scientific notation, compare [WA] to [H+] = x or [WB] to [OH─] = x . • If the exponents differ by 3 or more, the ionization is less than 5%. • If the exponents differ by 2 or less, use the % dissociation equation for the 5% test. 7. Ka expressions have [H+] or [H3O+] on top. Kb expressions have [OH─] on top.
Ka solves for x = [H+] , Kb solves for x = [OH―]. 8. An acid particle loses an H+ to become the base conjugate of the acid.
A base particle gains an H+ to become the acid conjugate of the base. 9. For acidbase conjugate pairs: Kw = Ka x Kb = 1.0 x 10─14 10. For polyprotic acid solutions, to find [H+] and pH:
•
• If Ka is 100 or more times larger than Ka , use Ka1 to find [H+] and ignore Ka2 ;
1
2
If Ka is not 100 or more times larger than Ka , add [H+] from Ka1 and Ka2 .
1
2
##### © 2011 ChemReview.Net v. 7f Page 965 Module 31 — BrønstedLowry Definitions Module 31 — BrønstedLowry Definitions
Timing: Do this module when you are assigned problems that involve the BrønstedLowry
definitions or you are asked to predict whether acid and base particles will react.
***** Lesson 31A: BrønstedLowry Acids and Bases
ProtonTransfer
So far, we have defined acids and bases in terms of their reaction with water. Those
definitions are useful in calculations involving particle concentrations.
An important qualitative question is: when acids and bases are combined, which
combinations react, and which do not? To answer these questions, it is helpful to use a
broader definition of acids and bases, termed the BrønstedLowry definitions. These
expanded definitions will help to predict and explain a wide variety of important acidbase
interactions in chemistry and biology.
By the BrønstedLowry definitions:
1. An acidbase reaction is a proton transfer: a proton moves from one particle to another.
2. An acid is a particle that can release an H+. A base is a particle that can accept an H+.
3. Protontransfer reactions are reversible: in a closed system, reactions go to equilibrium.
4. Each side of an acidbase reaction equation has a one acid (the proton donor) and one base
(the proton acceptor). In an acidbase reaction equation, the difference between the
particle formulas on one side and the other is the particle to which the mobile proton is
attached.
5. A particle that is an acid on one side of a reaction loses its proton and becomes a base on
the other. A base particle on one side gains a proton to become the acid on the other.
6. Particles in acidbase reaction behave as two acidbase conjugate pairs. In a conjugate
pair, the particles are identical, except that the acid on one side has an H+ that its
conjugate base on the other does not.
It will help in learning the rules to do a few examples. Apply the rules above to the
questions below. If unsure about an answer, check it before doing the next question.
Q1. Circle the two acids in this reaction: HCO3― + F― CO32― + Q2. Is F― in the above reaction an acid or a base? Explain your answer.
Q3. Complete this reaction with the first reactant behaving as an acid.
H2PO4― + CO32―
Q4. Complete this reaction with the first reactant acting as a base.
***** HSO3― + H2O © 2011 ChemReview.Net v. 7f Page 966 HF Module 31 — BrønstedLowry Definitions Answers
HCO3― + F― Q1. CO32― + HF Each side must have one acid. The HCO3― gives up a proton in going to the right: it
is therefore acting as the acid on the left side. The HF gives away its proton when the
reaction goes to the left, making HF the acid on the right side.
Q2. F― is a base because (a) the other particle on that side is an acid, and each side must
have an acid and a base, and/or (b) F― gains a proton in going to the right, and
gaining a proton is what a base does when it reacts.
Q3. H2PO4― + CO32―
Q4. HSO3― + H2O HPO42― + HCO3―
H2SO3 + OH─ Acids donate an H+ in reactions. Base particles gain a proton in reactions. Check: are each of the above reactions balanced for atoms and charge? Practice A: Identifying Acids and Bases in ProtonTransfer Reactions 1. For these reactions, assume that the charges on the left side are correct. If no charge is
shown on the left, assume the particle is neutral. On the right side, add correct charges
to the particles. If the particle is neutral, leave the charge blank.
a. SO42― + HCl b. HPO42― + HS― c. NH4+ + HSe― HSO4
PO4
NH3 + Cl
+ H2S
+ H2Se 2. In 1c above, which particle is in the acidbase conjugate pair with HS―?
3. Assuming that the first reactant below is acting as an acid, complete these reactions.
a. HNO3 + NH3 b. HS― + SO42― c. OH― + OH― 4. Assuming that the first reactant below is acting as a base, complete these reactions.
a. HCO3― + OH― b. H2O + H2O c. PO43― + H2PO4― © 2011 ChemReview.Net v. 7f Page 967 Module 31 — BrønstedLowry Definitions Stronger and Weaker Acids and Bases
7. A simple protontransfer equilibrium consists of 4 particles: two pairs of acidbase
conjugates. Each of the four particles can be labeled as one of the following.
• Stronger acid (sA): the molecule or ion that releases a proton in the reaction. In
the stronger acid, the proton is more “loosely bound” than in the weaker acid. • Weaker acid (wA): the particle in an equilibrium that has the stronger bond to its
acidic hydrogen. • Stronger base (sB): the particle that more strongly attracts a proton. It will form
the stronger bond to the proton if it is acquired. • Weaker base (wB): the particle that can accept a proton, but does not tend to do so.
If it does acquire a proton, it will bind it weakly. In these lessons, we will use the upper case SA and SB when referring to strong acids
and bases in absolute terms (Ka or Kb = very large), and WA and WB when referring to
particles that are weak acids and bases (Ka or Kb < 1). When comparing two particles,
we will use the lower case s and w.
For example, both HCN and HF are weak acids (WA) because both have a Ka < 1,
but in comparing the two, based on their Ka values, HF (Ka =6.8 x 10―4) is the
stronger acid (sA) and HCN (Ka =6.2 x 10―10) is the weaker acid (wA).
8. In a reaction, the particle that is the stronger acid (sA), upon losing H+ becomes the
weaker base (wB). The stronger base, upon gaining a proton, becomes the weaker acid.
9. In a protontransfer reaction equation, the stronger base (sB) and stronger acid (sA) are
always on the same side and the wA and wB are always on the same side.
10. Equilibrium favors the side with the weaker acid (wA) and weaker base (wB).
An A and B react to form another A and B, which can then react to reform the original
A and B. Which particles win the battle to react more often? The A and B that are
stronger. This means that the wA and wB are formed more often.
When more than one protontransfer reaction is possible, the proton from the strongest
acid will migrate to the strongest base to form the weakest acid.
11. Acidbase behavior is explained by chemical structure.
a. A particle that bonds a proton strongly is a strong base without the proton and a
weak acid with it. b. A particle that can accept a proton, but does not bond it tightly, is a strong acid
when it has the proton, and a weak base when it does not. Practice B: Learn the 11 rules above, then test your knowledge with these. TRUE or FALSE. In protontransfer equilibria,
_____ a. An acid and a base react to form another acid and another base. _____ b. The stronger acid winds up at equilibrium with the proton. © 2011 ChemReview.Net v. 7f Page 968 Module 31 — BrønstedLowry Definitions _____ c. A strong base will bond weakly to the proton it acquires. _____ d. The weaker acid has a stronger bond to H+ than the stronger acid. _____ e. The stronger acid is on the same side of the equation as the weaker base. _____ f. Equilibrium favors the weaker base. _____ g. The weaker base attracts a proton more than the stronger base. _____ h. The stronger acid reacts to become the weaker base. _____ i. The weaker acid becomes the stronger base when it reacts. ANSWERS
Practice A
SO42― + 1. a. HCl b. HPO42― + HS― c. NH4+ + HSe― HSO4―
PO43―
NH3 Cl― +
+
+ H2S
H2S e The acid particle loses an H+ in the reaction. The base particle gains an H+. Reactions must be balanced
for atoms and charge.
2. H2S. The particles in an acidbase conjugate pair differ only by one H+.
3. a.
b. HNO3 + NH3
HS― + SO 2― c. OH― + OH― 4 HCO3― + OH― 4. a.
b. H2O + H2O c. PO43― + H2PO4― NO3― + NH4+
S2― + HSO ―
4 O2― + H2O
H2CO3 + O2―
H3O+ + OH―
2 HPO42― Practice B
True a. An acid and a base react to form another acid and another base. False b. The stronger acid winds up at equilibrium with the proton. False
True c. A strong base will bond weakly to the proton it acquires.
d. The weaker acid has a stronger bond to H+ than the stronger acid. False e. The stronger acid is on the same side of the equation as the weaker base. True f. False g. The weaker base attracts a proton more than the stronger base. True h. The stronger acid reacts to become the weaker base. True i. Equilibrium favors the weaker base. The weaker acid becomes the stronger base when it reacts. ***** © 2011 ChemReview.Net v. 7f Page 969 Module 31 — BrønstedLowry Definitions Lesson 31B: Which Acids and Bases React?
Prerequisites: Lesson 31A.
* * ***
Most chemistry textbooks contain a table of acid strengths. An abbreviated version of such
a table is given below.
Table of Acid Strengths
Acid
HCl
HNO3
H2SO4
H O+
3 HSO4―
H3PO4
HF
C6H5COOH
CH3COOH
H2CO3
H PO ―
2 4 HCN
NH +
4 HCO3―
HPO 2―
4 H2O
*
** Base
H+ + Cl―
H+ + NO ―
3 H+ + HSO4―
H+ + H2O
H+ + SO 2―
4 H+ + H2PO4―
H+ + F―
H+ + C H COO―
65 H+ + CH3COO―
H+ + HCO3―
H+ + HPO 2―
4 H+ + CN―
H+ + NH3
H+ + CO 2―
3 H+ + PO43―
H+ + OH― Ka Values at 25oC
Very Large
Very Large
Very Large
1.0
1.0 x 10―2
7.2 x 10―3
6.8 x 10―4*
6.3 x 10―5
1.8 x 10―5
4.3 x 10―7
6.3 x 10―8
6.2 x 10―10*
5.6 x 10―10
5.6 x 10―11
4.2 x 10―13
1.8 x 10―16** Ka values vary among textbooks. When doing textbook homework, use the textbook’s values for Ka.
For consistency when comparing water’s Ka to other acids, Ka = Kw /55 M is used. In acidstrength tables:
• The strongest acid has the largest Ka ; the weakest acid has the smallest Ka. • The strongest acids are listed at the top left in the table. The acid strength and Ka
values for the acids go down as you go down the left (acid) column. • If base conjugates are listed on the right, the strongest base is at the bottom right of the
table. Why? The weakest acid is at the bottom left. When it loses its proton, it
becomes the strongest base. © 2011 ChemReview.Net v. 7f Page 970 Module 31 — BrønstedLowry Definitions The arrangement of the acid strength table in most textbooks is
Strongest Acid H+ + Weakest Base Ka = very large Weakest Acid H+ + Strongest Base Ka = very small Cover the answer below and, using the above table, try this question.
Q1. Label below the particles in this protontransfer reaction: the stronger acid (sA),
the stronger base (sB), the weaker acid (wA), or the weaker base (wB).
Cl― + CH3COOH HCl + CH3COO― *****
Answer
First, label each particle as acting as an acid or a base (some particles can act as both).
Recall that acids release protons, each side will have one acid and one base, and if a
particle is acting on one side as an acid, on the other side is its conjugate base.
*****
Cl― + CH3COOH
B A HCl + CH3COO―
A B Next, using the table, identify one particle as the stronger of the two acids.
*****
Cl― + CH3COOH
B
A HCl + CH3COO―
sA B If you do the first label correctly, labeling the rest is automatic: the stronger acid must
be on the same side as the stronger base, and the weaker acid and base must also be on
the same side.
Cl― + CH3COOH
wB
wA HCl + CH3COO―
sA sB To check, find the two bases on the right of the arrows in the table. The stronger base
(sB) should be nearer the bottom of the table than the weaker base.
Answer these.
Q2. Which side of a reversible acidbase reaction is favored at equilibrium?
Q3. When will an acid and a base react if they are combined?
Q4. When will an acid and base not react?
*****
A2. The side with the weaker acid and base.
A3. When the reaction can form a weaker acid and base as products.
A4. They will not react if the only possible products are a stronger acid and base. © 2011 ChemReview.Net v. 7f Page 971 Module 31 — BrønstedLowry Definitions Based on your answers to Q1Q4 above, for the reaction in Q1,
Q5. Which side of the reaction will be favored at equilibrium: left or right?
Q6. Will the acid and base on the left react when mixed? If so, what will they form?
Q7. Will the acid and base on the right react when mixed? If so, what will they form?
*****
Answer
A5. The left side is favored. The favored side at equilibrium is the side with the weaker
acid and base.
A6. The acid and base on the left will tend not to react. At equilibrium, all reactants
and products must be present, so a very small amount of the stronger acid and
base in the products will exist, but since the proton is attached to the weaker acid
on the left, the favored side at equilibrium already exists, and very little reaction
going to the right will take place.
A7. The acid and base on the right are the stronger of the two possibilities, so they will
react. At equilibrium, nearly all of the limiting reactant in the stronger pair is
used up, an equal number of moles of the reactant on its side is also used up, and
both of the weaker acid and base in their conjugate pairs in the table have formed. Practice A: Do each numbered problem. Do every other letter on the problems with
parts, and more if you need more practice. Check answers as you go.
1. Write a letter below each particle to label it as an acid (A) or base (B) in these reactions.
a. HCl + F― Cl― + HF b. CH3COO― + HCN
c. SO42― + C6H5COOH CH3COOH + CN― HSO4― + C6H5COO― 2. In the above reactions, label each particle as the stronger acid (sA), the stronger base
(sB), the weaker acid (wA), or the weaker base (wB).
3. In the above reactions, circle the side of the reaction that is favored at equilibrium.
4. Label each of the above reactions as will go to make the products, or won’t go.
5. Complete these reactions, and then label the reaction as will go or won’t go.
a. HF + CN―
b. H2PO4― + SO42―
c. CN― + H2PO4― © 2011 ChemReview.Net v. 7f Page 972 Module 31 — BrønstedLowry Definitions d. F― + Cl―
e. HNO3 + HCl Quick Predictions For AcidBase Reactions
In the previous lesson, to predict whether an acid and base react, we used the
Labeling Rule: Using the acidstrength table, label each particle in the reactants and
products as stronger acid (sA), sB, wA, or wB. Equilibrium favors the side with the
wA and wB: sA and wB react, wA and wB form, and wA and wB will not react.
A quicker way to predict whether an acid and base mixed together will react can be added
to our BrønstedLowry rules as number
12. If the strongest acid or bases in an acidstrength table are listed at the top left and if
particle concentrations are similar, any particle to the left of the arrows will tend to
react with any particle on the right of the arrows BELOW it in the table. The products
will be the conjugates of each particle: the weaker acid and the weaker base. Cover the answer below, then use BrønstedLowry Rule 12 on these examples.
Q. Assuming the particle concentrations are similar, use the acidstrength table to label
each combination as will react or won’t react as acids and bases when mixed.
a. CN― + HCl b. CH3COOH + SO42― c. PO43―+ NO3― d. H2PO4― + PO43― *****
Answers
a. CN― + HCl
Will react. The table lists the strongest acid at the top left.
HCl is an acid at the top left, and CN― is on the base side and below it.
Won’t react. CH3COOH is an acid at the middle left
b. CH3COOH + SO42―
2― is on the base side but above the acid.
of the table; SO4
Won’t react. Both are in the table as bases. To have an
c. PO 3― + NO ―
4 3 acidbase reaction, you need an acid and a base.
Will react. The H2PO4― appears on both sides of the
d. H2PO4― + PO43―
table: it can be an acid or a base. But PO43―, since it does not have a proton, must
be the base, and on the acid (left) side H2PO4― appears above it. Particles on the
left react with particles on the right and lower. © 2011 ChemReview.Net v. 7f Page 973 Module 31 — BrønstedLowry Definitions Practice B: Complete the odd numbers. Save the even for later practice. Use the acidstrength table to label these as will react or won’t react.
1. HSO4― + F― 2. HPO42― + HCN 3. C6H5COOH + CH3COOH A short version of Rule 12 is the
Diagonal Rule: Particles \ diagonal react to form / diagonals; particles / do not react.
Use the diagonal rule and the acidstrength table to answer the following.
Q. Assuming the particle concentrations are similar, label these as will react or won’t react.
1. HNO3 + H2PO4―
2. H2PO4― + HF
*****
Answers
A1. HNO3 + H2PO4―
A2. H2PO4― + HF Will react. HNO3 is \ above H2PO4― on the right.
Won’t react. HF is / to the H2PO4― on the base side. Practice C: Complete the odds; save the even for later. Use the acidstrength table and
diagonal rule to label these as will react or won’t react.
1. HCO3― + HCN 2. NH4+ + PO43― 3. HPO42― + HF Constructing An AcidStrengths Table
In some problems, a complete acidbase strengths table will not be supplied. Instead, you
will be given a list of acids only, with the strongest at the top. If this is the case, you can
make your own acidbase table by writing the base conjugates on the right. You can then
use your table to solve problems.
Cover the answer below and try the following example. © 2011 ChemReview.Net v. 7f Page 974 Module 31 — BrønstedLowry Definitions Q. The following is a list of acids, ordered with the strongest at the top.
HCl
CH3COOH
H2CO3
NH4+
HCO3―
H2O
a. Add the base conjugates in a column on the right. (Do this part, check your
answer, and then try the parts b  d below.)
b. What is the formula for the strongest base in your completed table?
c. What is the formula for the weakest base in your table?
d. Will these react: CH3COOH + OH― If so, what products will form? *****
Answer
a. See table at right. Acid b. Strongest base? OH― The weakest acid is H2O. Its
conjugate is the strongest base. HCl Base
Conjugate
Cl― c. Weakest base? Cl―. The weakest base is formed when a
proton leaves the strongest acid. CH3COOH CH3COO― H2CO3 HCO3― d. They will react. This reaction will go. The acid
CH3COOH on the left will react with the base OH―
below it in the acidstrength table. The two reactants
form conjugates that are listed opposite them in the table. NH4+
HCO3― NH3
CO 2 ― H2O OH― 3 Practice D
1. Use the acidstrength table that you constructed above to answer these questions.
a. Complete this reaction, and then label below each particle: sA, sB, wA, or wB.
NH4+ + OH―
b. Will the reactants in part 1a form the products? Will the acidbase reaction go?
c. In the part 1a reaction, NH4+ is part of an ionic compound. Ionic compounds do not
boil or evaporate to form gases except at very high temperatures. At room
temperature they have no odor because they do not form gas particles that can
travel through the air. Ammonia (NH3) is a covalent compound and a gas at room
temperature that dissolves readily but reversibly in water. A mixture of ammonia © 2011 ChemReview.Net v. 7f Page 975 Module 31 — BrønstedLowry Definitions and water is often used in commercial bluedyed glass cleaning solutions that have
a characteristic, unpleasant “ammonia odor.”
In the 1a reaction above, NH4+ and OH―, before they are mixed, are ionic; they will
have no odor. After those two ions are mixed in the above reaction, will the mixture
have an odor? Why or why not?
2. You spill an aqueous 1.0 M solution of ammonia (NH3) on the floor. The odor is
unpleasant. Using the acidstrength table that you constructed above, state whether
1.0 M solutions of each of these will deodorize the ammonia by changing it to NH +.
4 a. H2O b. CH3COOH c. NaOH d. NaHCO3 3. To the right is a table of acids in order of
strength, with the strongest acid at the top.
a. If bicarbonate ion (HCO ―) is added e. HCl HCl
CH3COOH 3 to a mixture of CH3COOH, C2H5OH, and H2CO3 C6H5OH, which of those three will C6H5OH react with the bicarbonate? C2H5OH b. If the bicarbonate does react, which products will be formed? ANSWERS
Practice A
1, 2, 3, 4. a. HCl + F―
sA
sB Cl― + HF
wB
wA b. CH3COO― + HCN
wB
wA c. SO42― + C6H5COOH
wB
wA 5. a. HF + CN―
sA
sB
b. H2PO4― + CH3COOH + CN―
sA
sB Won’t go  favors wA and wB.
HSO4― + C6H5COO―
sA
sB F― + HCN
wB
wA Won’t go to right. Will go. Equilibrium favors weaker A and B. SO42― HPO42― + HSO4― wB wA sB c. CN― + H2PO4―
sB Will go  equilibrium favors weaker acid and base. sA d. F― + Cl―
B
B © 2011 ChemReview.Net v. 7f sA Won’t go  favors wA and wB. HCN + HPO42―
wA wB Will go. Equilibrium favors weaker A and B. Can’t go. Need an acid and a base. Page 976 Module 31 — BrønstedLowry Definitions e. HNO3 + HCl
A A Can’t go. Need an acid and a base. Practice B
1. HSO4― + F― Will react. HSO4― on acid side is above F― on base side. 2. HPO42― + HCN 3. C6H5COOH + CH3COOH Won’t react. HCN is acid; HPO42― on the base side is above HCN.
Won’t react. Both are in the table as acids. Practice C
1. HCO3― + HCN 2. NH4+ + PO43― 3. HPO42― + HF Won’t react. HCN is an acid, HCO3― on the base side is / .
Will react. NH4+ is an acid, PO43― is a base and they are \ .
Will react. HF is acid, HPO42― can be a base and they are \ . Practice D
1. a. NH4+ + OH―
sA sB NH3 + H2O
wB wA b. The reaction will go to form the products, because a) equilibrium favors the weaker acid and base;
and/or b) the acid is in the table above the base.
c. Yes. The mixture will smell like ammonia, because NH3 is a product of the reaction, and the reaction
goes to form ammonia.
2. a. H2O No. To change the NH3 to NH4+ ions, you need an acid, because the ammonia must act as
a base. H2O can act as an acid, but NH3 as a base is above water as an acid in the table, so the
acidbase reaction will not take place. The odor continues.
b. CH3COOH Yes. This is an acid above the base NH3 in the table, so the reaction will take place.
NH3 gas which can leave the water becomes NH4+ ions which cannot. The ammonia odor
dissipates.
c. NaOH No. Solid NaOH mixed in a water solution is soluble. It will dissolve and form OH― ions.
In the table, OH― ions are a base. NaOH is a strong base. We need an acid to change NH3 to
NH4+, so there will be no acidbase reaction. The odor continues.
d. NaHCO3 No. When Na, an alkali metal, is part of a compound, the compound will be soluble in
water and separate into ions 100%. The HCO3― (bicarbonate) ion that forms upon ionization is in
the table twice, as both an acid and a base. As a base, it can’t convert NH3 to NH4+. As an acid,
bicarbonate is below the base NH3. There will be no reaction. The odor continues.
e. HCl
Yes. HCl is a very strong acid, and the NH3 is below it as a base. Reaction occurs when HCl
is added, and NH3 is converted to NH4+. The ammonia odor dissipates. © 2011 ChemReview.Net v. 7f Page 977 Module 31 — BrønstedLowry Definitions 3. In the stated mixture, only the CH3COOH is
an acid above the HCO3― base in the table. When Acid Base Conjugate HCl CH3COOH is added, the bicarbonate ion will change CH3COOH to H2CO3 which will quickly decompose into H2O and H2CO3 bubbles of CO2 (see Lesson 14E). C6H5OH Acetate ion also forms. C2H5OH HCO3― The reaction is:
CH3COOH + HCO3― [ H2CO3 ] + CH3COO― H2O + CO2↑ + CH3COO― ***** SUMMARY  BrønstedLowry Definitions
1. Using the BrønstedLowry definitions of acids and bases:
a. An acidbase reaction is a proton transfer: H+ moves from one particle to
another.
b. An acid and a base react to form another acid and another base.
c. The acid is the proton donor and the base is the proton acceptor.
d. The particles in the equilibrium are two acidbase conjugate pairs. A particle on
one side has its conjugate on the other side.
e. The stronger acid reacts to become the weaker base. The stronger base reacts to
become the weaker acid.
f. The stronger acid and base are on the same side of the equilibrium equation. g. Equilibrium favors the weakest acid and the weakest base.
2. If the strongest acid or bases in an acidstrength table are listed at the top left, if
particle concentrations are similar, a particle on the left will tend to react with a
particle on the right BELOW it in the table. The products will be the conjugates of
each particle: the weaker acid and the weaker base.
This can be summarized as the
Diagonal Rule: Particles \ diagonal react to form / diagonals; particles / do not
react.
# © 2011 ChemReview.Net v. 7f #### Page 978 ...
View
Full
Document
This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.
 Summer '09
 J
 Bases

Click to edit the document details