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Chapter 15 – Buffers and Titration
1. Calculate the pH of the each of the following solutions given that for ammonia, Kb= 1.8x10–5.
a. 0.100 M NH3
NH3 + H2O  NH4+ + OH–
0.100
0
0
–x
+x
+x
0.100
x
x
2
–5
x / 0.100 = 1.8 x 10
x = [OH–] = 1.3 x 10–3 M
pH = 11.13
b. 0.100 M NH4Cl
NH4+  NH3 + H+
0.100
0
0
–x
+x
+x
0.100
x
x
2
–14
x
= 1.0 x 10
x = [H+] = 7.5 x 10–6 M
pH = 5.13
–5
0.100
1.8 x 10
c. A solution containing 0.100 M NH3 and 0.100 M NH4Cl
NH3 + H2O  NH4+ + OH–
0.100
0.100
0
–x
+x
+x
0.100
0.100
x
0.100 x / 0.100 = 1.8 x 10–5
x = [OH–] = 1.8 x 10–5 M
pH = 9.26
2. Calculate the pH of solution 1c after the addition of 0.035 moles of NaOH to 1 L of solution.
Is this solution still a buffer at equilibrium?
OH– + NH4+ 
NH3 + H2O
0.035 0.100
0.100
–.035 –.035
+.035
0
0.065
0.135 still has both base and conjugate acid – it’s still a buffer
for pH, do ICE chart:
NH3 +
0.135
H2O 
0.065 x / 0.135 = 1.8 x 10–5
NH4+ + OH–
0.065
x
x = [OH–] = 3.7 x 10–5 M
pH = 9.57
3. Calculate the pH of solution 2 after the addition of 0.200 moles of HCl to 1 L of solution. Is
this solution still a buffer at equilibrium?
H+ + NH3 
NH4+
0.200 0.135
0.065
–.135 –.135
+.135
0.065
0
0.200
has only acid left – not still a buffer
for pH, use –log [H+] = 1.19
4. A 25.0-mL sample of 0.100 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.100 M
NaOH solution. Calculate the pH after the following volumes of NaOH have been added.
a. 0.0 mL
c. 12.5 mL
e. 32.0 mL
b. 8.00 mL
d. 25.0 mL
pKa = 3.86 so Ka = 1.4 x 10–4
a. pH of weak acid: HC3H5O3 ⇄ H+ + C3H5O3–
0.100
0
0
–x
+x
+x
0.100
x
x
x2 / 0.100 = 1.4 x 10–4
x = [H+] = 3.7 x 10–3 M
pH = 2.43
b. 8.00 mL base added: calculate new molarities
[HC3H5O3] = (25.0 mL)(0.100 M) / 33.0 mL = 0.0758 M
[OH–] = (8.00 mL)(0.100 M) / 33.0 mL = 0.0242 M
one thing gets used up:
OH– + HC3H5O3 → C3H5O3– + H2O
0.0242 0.0758
0
–.0242 –.0242
+0.0242
0
0.0516
0.0242
+
–
do an ICE chart:
HC3H5O3 ⇄ H + C3H5O3
0.0516
x
0.0242
0.0242x / 0.0516 = 1.4 x 10–4
x = [H+] = 2.9 x 10–4 M pH = 3.53
c. 12.5 mL base added: use method from part b, or realize that this is the half-equiv. point
pH = pKa = 3.86
d. 25.0 mL base added: equivalence point; [HC3H5O3] = [OH–]
calculate new molarity: [HC3H5O3] = (25.0 mL)(0.100 M) / 50.0 mL = 0.0500 M
both get used up,
OH– + HC3H5O3 → C3H5O3– + H2O
conj. base left
0.0500 0.0500
0
–.0500 –.0500
+0.0500
0
0
0.0500
do an ICE chart:
C3H5O3– + H2O ⇄ HC3H5O3 + OH–
0.0500
x
x
x2 = 1.0 x 10–14
0.0500 1.4 x 10–4
x = [OH–] = 1.9 x 10–6 M
pH = 8.28
e. 32.0 mL base added: calculate new molarities
[HC3H5O3] = (25.0 mL)(0.100 M) / 57.0 mL = 0.0439 M
[OH–] = (32.0 mL)(0.100 M) / 57.0 mL = 0.0561 M
OH– + HC3H5O3 → C3H5O3– + H2O
0.0561 0.0439
0
-.0439 -.0439
+0.0439
0.0122
0
0.0439
pOH = 1.914 so pH = 12.086
one thing gets used up:
[OH–] = 0.0122 M
5. Consider the titration of 80.0 mL of 0.100 M Sr(OH)2 by 0.400 M HCl. Calculate the pH of
the resulting solution after the following volumes of HCl have been added.
a. 0.0 mL
c. 40.0 mL
b. 20.0 mL
d. 80.0 mL
Since Sr(OH)2 → Sr2+ + 2 OH– , the concentration of OH– is 2 x 0.100 = 0.200 M
a. [OH–] = 0.200 M, pOH = 0.699, pH = 13.30
b. 20.0 mL of acid added to base: calculate new molarites
[OH–] = (80.0 mL)(0.200 M) / 100.0 mL = 0.160 M
[H+] = (20.0 mL)(0.400 M) / 100.0 mL = 0.0800 M
the acid reacts with the base:
one thing gets used up:
H+ +
.0800
–.0800
0
OH– → H2O
0.160
–.0800
0.080
pOH = 1.09, pH = 12.90
c. 40.0 mL of acid added: equivalence point; [OH–] = [H+]
pH = 7.000
d. 80.0 mL of acid added: calculate new molarities
[OH–] = (80.0 mL)(0.200 M) / 160.0 mL = 0.100 M
[H+] = (80.0 mL)(0.400 M) / 160.0 mL = 0.200 M
the acid reacts with the base:
one thing gets used up:
H+ +
.200
–.100
0.100
OH– → H2O
0.100
–.100
0
pH = 1.000
6. A sample of an ionic compound NaA, where A– is the anion of a weak acid, was dissolved in
enough water to make 100.0 mL of solution and was then titrated with 0.100 M HCl. After
500.0 mL of HCl was added, the pH was measured to be 5.00. The experimenter found that
1.00 L of 0.100 M HCl was required to reach the equivalence point of the titration.
a. What is the Kb value for A–?
b. Calculate the pH of the solution at the equivalence point of the titration.
a. 1.00 L of HCl is the equivalence point, so 500.0 mL is the half-equivalence point.
At the half-equivalence point, pOH = pKb.
Since pH = 5.00, pOH = 9.00 = pKb
Kb = 1.0 x 10–9
b. equivalence point: calculate molarity of A– using M1V1 = M2V2
(0.100 M)(1.00 L) = M2 (0.1000 L)
M2 = 1.00 M
one solution added to another: calculate new molarities (they are equivalent!)
[A–] = [H+] = (0.1000 L)(1.00 M) / 1.10 L = 0.0909 M
both gets used up,
conj. acid left
H+ + A– → HA
0
0
0.0909
do ICE chart:
HA ⇄ H+ + A–
0.0909 x
x
x2 = 1.0 x 10–14
.0909 1.0 x 10–9
x = [H+] = 9.5 x 10–4 M
pH = 3.02