CHAPTER 16
Acid-Base Equilibria
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
16.1. Abbreviate the formula of lactic acid as HL. To solve, assemble a table of starting, change, and
equilibrium concentrations.
Conc. (M) HL + H2O H3O+ + L
Starting 0.025 0 0
Change x +x +x
Equilibrium 0.025 x x x
Substituting into the equilibrium-constant equation gives
[H3O ] [L ] (x) 2
Ka = =
[HL] (0.025 x)
The value of x equals the value of the molarity of the H3O+ ion, which can be obtained from the
pH:
[H3O+] = antilog (−pH) = antilog (−2.75) = 0.00178 M
Substitute this value for x into the equation to get
(x) 2 (0.00178)2
Ka = = = 1.36 10−4 = 1.4 10−4
(0.025 x) (0.025 0.00178)
The degree of ionization is
0.00178
Degree of ionization = = 0.071
0.025
16.2. To solve, assemble a table of starting, change, and equilibrium concentrations. Use HAc as the
symbol for acetic acid.
Conc. (M) HAc + H2O H3O+ + Ac−
Starting 0.10 0 0
Change −x +x +x
Equilibrium 0.10 − x x x
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, Chapter 16: Acid-Base Equilibria 573
Now, substitute these concentrations and the value of Ka into the equilibrium-constant equation
for acid ionization:
[H3 O ] [Ac ] (x) 2
Ka = = = 1.7 10−5
[HAc] (0.10 x)
Solve the equation for x, assuming x is much smaller than 0.10, so (0.10 − x) 0.10.
(x) 2
1.7 10−5
(0.10)
x2 = 1.7 10−5 0.10 = 1.7 10−6
x = 0.00130 M
Check to make sure the assumption that (0.10 − x) 0.10 is valid:
0.10 − 0.00130 = 0.0987, = 0.10 (to two significant figures)
The concentrations of hydronium ion and acetate ion are
[H3O+] = [Ac−] = x = 0.0013 = 1.3 10−3 M
The pH of the solution is
pH = − log [H3O+] = − log (0.00130) = 2.884 = 2.88
The degree of ionization is
0.00130
Degree of ionization = = 0.0130 = 0.013
0.10
16.3. Abbreviate the formula for pyruvic acid as HPy. To solve, assemble a table of starting, change,
and equilibrium concentrations:
Conc. (M) HPy + H2O H3O+ + Py−
Starting 0.0030 0 0
Change −x +x +x
Equilibrium 0.0030 − x x x
Substitute the equilibrium concentrations and the value of Ka into the equilibrium-constant
expression to get
[H3 O ] [Py ] (x) 2
Ka = = = 1.4 10−4
[HPy] (0.0030 x)
Note that the concentration of acid divided by Ka is 0.0030/1.4 10−4 = 21, which is considerably
smaller than 100. Thus, you can expect that x cannot be ignored compared with 0.0030. The
quadratic formula must be used. Rearrange the preceding equation to put it into the form
ax2 + bx + c = 0.
x2 + 1.4 10−4 x − 4.20 10−7 = 0
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, 574 Chapter 16: Acid-Base Equilibria
Substitute into the quadratic formula to get
1.4 104 (1.4 104 ) 2 4(4.20 107 ) 1.4 104 1.303 103
x= =
2 2
Using the positive root, = [H3O+] = 5.81 10−4 M. Now you can calculate the pH.
pH = − log [H3O+] = − log (5.81 10−4) = 3.2350 = 3.24
16.4. To solve, note that K a = 1.3 10−2 >> K a = 6.3 10−8, and hence the second ionization and K a
1 2 2
can be ignored. Assemble a table of starting, change, and equilibrium concentrations.
Conc. (M) H2SO3 + H2O H3O+ + HSO3−
Starting 0.25 0 0
Change −x +x +x
Equilibrium 0.25 − x x x
Substitute into the equilibrium-constant expression for the first ionization.
[H3 O ] [HSO3 ] (x) 2
K a1 = = = 1.3 10−2 = 0.013
[H 2SO3 ] (0.25 x)
This gives x2 + 0.013x − 0.00325 = 0.
Note that the concentration of acid divided by Ka is 0.25/0.013 = 19, which is considerably
smaller than 100. Thus, you can expect that x cannot be ignored compared with 0.25. Reorganize
the above equilibrium-constant expression into the form ax2 + bx + c = 0, and substitute for a, b,
and c in the quadratic formula.
0.013 (0.013) 2 4(0.00325) 0.013 0.1147
x= =
2 2
Using the positive root, x = [H3O+] = 0.05087 M.
pH = − log (0.05087) = 1.293 = 1.29
To calculate [SO32−], which will be represented by y, use the second ionization. Assume the
starting concentrations of H3O+ and HSO3− are those from the first equilibrium.
Conc. (M) HSO3− + H2O H3O+ + SO32−
Starting 0.0508 0.0508 0
Change −y +y +y
Equilibrium 0.0508 − y 0.0508 − y y
Now, substitute into the K a expression for the second ionization.
2
[H3 O ] [SO32 ] (0.0508 y )(y )
Ka2 = = = 6.3 10−8
[HSO3 ] (0.0508 y )
Assuming y is much smaller than 0.0508, note that the (0.0508 + y) cancels the (0.0508 − y) term,
leaving y K a , or
2
y = [SO32−] 6.3 10−8 M (note the assumption that y << 0.0508 is valid)
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, Chapter 16: Acid-Base Equilibria 575
16.5. Convert the pH first to pOH and then to [OH−]:
pOH = 14.00 − pH = 14.00 − 9.84 = 4.16
[OH−] = antilog (−4.16) = 6.92 10−5 M
Using the symbol Qu for quinine, assemble a table of starting, change, and equilibrium
concentrations.
Conc. (M) Qu + H2O HQu+ + OH−
Starting 0.0015 0 0
Change −x +x +x
Equilibrium 0.0015 − x x x
Note that x = 6.92 10−5. Substitute into the equilibrium-constant expression to get
[HQ ] [OH ] (x) 2 (6.92 105 ) 2
Kb = = = = 3.346 10−6 = 3.3 10−6
[Qu] (0.0015 x) (0.0015 6.92 105 )
16.6. Assemble a table of starting, change, and equilibrium concentrations.
Conc. (M) NH3 + H2O NH4+ + OH−
Starting 0.20 0 0
Change −x +x +x
Equilibrium 0.20 − x x x
Assume x is small enough to ignore compared with 0.20. Substitute into the equilibrium-constant
expression to get
[NH 4 ] [OH ] (x) 2 (x) 2
Kb = = = 1.8 10−5
[NH3 ] (0.20 x) (0.20)
Solving for x gives
x2 = (0.20) 1.8 10−5 = 3.6 10−6
x = [OH−] 1.89 10−3 M (Note that x is negligible compared to 0.20.)
Now calculate the hydronium-ion concentration.
Kw 1.0 1014
[H3O+] =
= = 5.29 10−12 = 5.3 10−12 M
[OH ] 1.89 103
16.7. a. Acidic. NH4NO3 is the salt of a weak base (NH3) and a strong acid (HNO3), so a solution of
NH4NO3 is acidic because of the hydrolysis of NH4+.
b. Neutral. KNO3 is the salt of a strong base (KOH) and a strong acid (HNO3), so a solution of
NH4NO3 is neutral because none of the ions hydrolyze.
c. Acidic. Al(NO3)3 is the salt of a weak base [Al(OH)3] and a strong acid (HNO3), so a
solution of Al(NO3)3 is acidic because of the hydrolysis of Al3+.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Acid-Base Equilibria
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
16.1. Abbreviate the formula of lactic acid as HL. To solve, assemble a table of starting, change, and
equilibrium concentrations.
Conc. (M) HL + H2O H3O+ + L
Starting 0.025 0 0
Change x +x +x
Equilibrium 0.025 x x x
Substituting into the equilibrium-constant equation gives
[H3O ] [L ] (x) 2
Ka = =
[HL] (0.025 x)
The value of x equals the value of the molarity of the H3O+ ion, which can be obtained from the
pH:
[H3O+] = antilog (−pH) = antilog (−2.75) = 0.00178 M
Substitute this value for x into the equation to get
(x) 2 (0.00178)2
Ka = = = 1.36 10−4 = 1.4 10−4
(0.025 x) (0.025 0.00178)
The degree of ionization is
0.00178
Degree of ionization = = 0.071
0.025
16.2. To solve, assemble a table of starting, change, and equilibrium concentrations. Use HAc as the
symbol for acetic acid.
Conc. (M) HAc + H2O H3O+ + Ac−
Starting 0.10 0 0
Change −x +x +x
Equilibrium 0.10 − x x x
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 16: Acid-Base Equilibria 573
Now, substitute these concentrations and the value of Ka into the equilibrium-constant equation
for acid ionization:
[H3 O ] [Ac ] (x) 2
Ka = = = 1.7 10−5
[HAc] (0.10 x)
Solve the equation for x, assuming x is much smaller than 0.10, so (0.10 − x) 0.10.
(x) 2
1.7 10−5
(0.10)
x2 = 1.7 10−5 0.10 = 1.7 10−6
x = 0.00130 M
Check to make sure the assumption that (0.10 − x) 0.10 is valid:
0.10 − 0.00130 = 0.0987, = 0.10 (to two significant figures)
The concentrations of hydronium ion and acetate ion are
[H3O+] = [Ac−] = x = 0.0013 = 1.3 10−3 M
The pH of the solution is
pH = − log [H3O+] = − log (0.00130) = 2.884 = 2.88
The degree of ionization is
0.00130
Degree of ionization = = 0.0130 = 0.013
0.10
16.3. Abbreviate the formula for pyruvic acid as HPy. To solve, assemble a table of starting, change,
and equilibrium concentrations:
Conc. (M) HPy + H2O H3O+ + Py−
Starting 0.0030 0 0
Change −x +x +x
Equilibrium 0.0030 − x x x
Substitute the equilibrium concentrations and the value of Ka into the equilibrium-constant
expression to get
[H3 O ] [Py ] (x) 2
Ka = = = 1.4 10−4
[HPy] (0.0030 x)
Note that the concentration of acid divided by Ka is 0.0030/1.4 10−4 = 21, which is considerably
smaller than 100. Thus, you can expect that x cannot be ignored compared with 0.0030. The
quadratic formula must be used. Rearrange the preceding equation to put it into the form
ax2 + bx + c = 0.
x2 + 1.4 10−4 x − 4.20 10−7 = 0
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 574 Chapter 16: Acid-Base Equilibria
Substitute into the quadratic formula to get
1.4 104 (1.4 104 ) 2 4(4.20 107 ) 1.4 104 1.303 103
x= =
2 2
Using the positive root, = [H3O+] = 5.81 10−4 M. Now you can calculate the pH.
pH = − log [H3O+] = − log (5.81 10−4) = 3.2350 = 3.24
16.4. To solve, note that K a = 1.3 10−2 >> K a = 6.3 10−8, and hence the second ionization and K a
1 2 2
can be ignored. Assemble a table of starting, change, and equilibrium concentrations.
Conc. (M) H2SO3 + H2O H3O+ + HSO3−
Starting 0.25 0 0
Change −x +x +x
Equilibrium 0.25 − x x x
Substitute into the equilibrium-constant expression for the first ionization.
[H3 O ] [HSO3 ] (x) 2
K a1 = = = 1.3 10−2 = 0.013
[H 2SO3 ] (0.25 x)
This gives x2 + 0.013x − 0.00325 = 0.
Note that the concentration of acid divided by Ka is 0.25/0.013 = 19, which is considerably
smaller than 100. Thus, you can expect that x cannot be ignored compared with 0.25. Reorganize
the above equilibrium-constant expression into the form ax2 + bx + c = 0, and substitute for a, b,
and c in the quadratic formula.
0.013 (0.013) 2 4(0.00325) 0.013 0.1147
x= =
2 2
Using the positive root, x = [H3O+] = 0.05087 M.
pH = − log (0.05087) = 1.293 = 1.29
To calculate [SO32−], which will be represented by y, use the second ionization. Assume the
starting concentrations of H3O+ and HSO3− are those from the first equilibrium.
Conc. (M) HSO3− + H2O H3O+ + SO32−
Starting 0.0508 0.0508 0
Change −y +y +y
Equilibrium 0.0508 − y 0.0508 − y y
Now, substitute into the K a expression for the second ionization.
2
[H3 O ] [SO32 ] (0.0508 y )(y )
Ka2 = = = 6.3 10−8
[HSO3 ] (0.0508 y )
Assuming y is much smaller than 0.0508, note that the (0.0508 + y) cancels the (0.0508 − y) term,
leaving y K a , or
2
y = [SO32−] 6.3 10−8 M (note the assumption that y << 0.0508 is valid)
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, Chapter 16: Acid-Base Equilibria 575
16.5. Convert the pH first to pOH and then to [OH−]:
pOH = 14.00 − pH = 14.00 − 9.84 = 4.16
[OH−] = antilog (−4.16) = 6.92 10−5 M
Using the symbol Qu for quinine, assemble a table of starting, change, and equilibrium
concentrations.
Conc. (M) Qu + H2O HQu+ + OH−
Starting 0.0015 0 0
Change −x +x +x
Equilibrium 0.0015 − x x x
Note that x = 6.92 10−5. Substitute into the equilibrium-constant expression to get
[HQ ] [OH ] (x) 2 (6.92 105 ) 2
Kb = = = = 3.346 10−6 = 3.3 10−6
[Qu] (0.0015 x) (0.0015 6.92 105 )
16.6. Assemble a table of starting, change, and equilibrium concentrations.
Conc. (M) NH3 + H2O NH4+ + OH−
Starting 0.20 0 0
Change −x +x +x
Equilibrium 0.20 − x x x
Assume x is small enough to ignore compared with 0.20. Substitute into the equilibrium-constant
expression to get
[NH 4 ] [OH ] (x) 2 (x) 2
Kb = = = 1.8 10−5
[NH3 ] (0.20 x) (0.20)
Solving for x gives
x2 = (0.20) 1.8 10−5 = 3.6 10−6
x = [OH−] 1.89 10−3 M (Note that x is negligible compared to 0.20.)
Now calculate the hydronium-ion concentration.
Kw 1.0 1014
[H3O+] =
= = 5.29 10−12 = 5.3 10−12 M
[OH ] 1.89 103
16.7. a. Acidic. NH4NO3 is the salt of a weak base (NH3) and a strong acid (HNO3), so a solution of
NH4NO3 is acidic because of the hydrolysis of NH4+.
b. Neutral. KNO3 is the salt of a strong base (KOH) and a strong acid (HNO3), so a solution of
NH4NO3 is neutral because none of the ions hydrolyze.
c. Acidic. Al(NO3)3 is the salt of a weak base [Al(OH)3] and a strong acid (HNO3), so a
solution of Al(NO3)3 is acidic because of the hydrolysis of Al3+.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
