Homework 4
16. Between the starting point of the titration and the equivalence point, we are dealing
with a buffer solution. The Henderson-Haselback equation can be used to determine
pH:
pH = pKa + log([Base]/[Acid])
At the halfway point to equivalence, enough OH- has been added by the strong base
to convert exactly one-half of the weak acid present initially into its conjugate base.
Therefore, [conjugate base] = [weak acid] so pH = pKa + log1 = pKa
The pKa value can be determined at any point in a titration, from the initial point to
the equivalence point. In Chapter 14, we calculated the Ka from a solution of only the
weak acid. In the buffer region, we can calculate the ratio of base form to the acid
form and use the Henderson-Hasselbach equation to determine the Ka value. The
equivalence point data can be used to calculate the Kb value for the conjugate base
which is related to Ka by the equation Ka = Ka/Kb.
22. NH3(aq) + H+(aq) → NH4+(aq), NH4+(aq) + OH-(aq) → NH3(aq) + H2O(l)
24. a. Weak base problem:
Initial
Change
Equil.
HONH2
+ H2O ↔
HONH3+
+
OHKb = 1.1•10-8
0.100 M
0
0
x mol/L of HONH2 reacts with H2O to reach equilibrium
-x
→
x
+x
(0.100 – x) M
x
x
Kb = 1.1•10-8 = x2/(0.100 – x) ≈ x2/0.100
x = [OH-] = 3.3•10-5 M; pOH = 4.48,; pH = 9.52 Assumption good.
b. Weak acid problem (Cl- has no acidic/basic properties)
Initial
Change
Equil.
HONH3+
↔
H+
+
HONH2
0.100 M
~0
0
+
x mol/L of HONH3 dissociates to reach equilibrium
-x
→
x
+x
(0.100 – x) M
x
x
Ka = Kw/Kb = 9.1•10-7 = x2/(0.100 – x)
x = [H+] = 3.0•10-4 M; pH = 3.52
c. Pure H2O, pH = 7.00
Assumption good.
d. Buffer solution where pKa = -log(9.1•10-7) = 6.04. Using the HendersonHasselbach equation:
pH = pKa + log([Base]/[Acid]) = 6.04 + log([HONH2]/[HONH3+]) = 6.04 +
log(0.100/0.100) = 6.04
28. a. Added H+ reacts completely with HONH2 (the best base present) to form
HONH3+.
Before
Change
After
HONH2
0.100 M
-0.020
0.080
+
H+
↔
0.020 M
-0.020
0
HONH3+
0
+0.020
0.020
Reacts completely
After this reaction, a buffer solution exists, i.e., a weak acid (HONH3+) and its
conjugate base (HONH2) are present at the same time. Using the HendersonHasselbach equation to solve for the pH where pKa = -log (Kw/Kb) = 6.04:
pH = pKa + log([Base]/[Acid]) = 6.04 + log (0.080/0.020) = 6.04 + 0.60 = 6.64
b. We have a weak acid and a strong acid present at the same time. The H+
contribution from the weak acid, HONH3+, will be negligible. So, we have to
consider only the H+ from HCl. [H+] = 0.020 M; pH = 1.70.
c. This is a strong acid in water. [H+] = 0.020 M; pH = 1.70
d. Major species: H2O, Cl-, HONH2, HONH3+, H+
H+ will react completely with HONH2, the best base present.
Before
Change
After
HONH2
0.100 M
-0.020
0.080
+
H+
↔
0.020 M
-0.020
0
HONH3+
0.100
+0.020
0.120
Reacts completely
A buffer solution results after reaction. Using the Henderson-Hasselbach equation:
pH = pKa + log([Base]/[Acid]) = 6.04 + log (0.080/0.120) = 6.04 - 0.18 = 5.86
30. a. We have a weak base and a strong base present at the same time. The OHcontribution from the weak base, HONH2, will be negligible. Consider only the
added strong base as the primary source of OH-.
[OH-] = 0.020 M; pOH = 1.70, pH = 12.30
b. Added strong base will react to completion with the best acid present, HONH3+.
Before
Change
After
HONH3+
0.100 M
-0.020
0.080
OH- ↔
0.020 M
-0.020
0
+
HONH2
0
+0.020
0.020
+ H2O
Reacts completely
A buffer solution results after reaction. Using the Henderson-Hasselbach equation:
pH = pKa + log([Base]/[Acid]) = 6.04 + log (0.020/0.080) = 6.04 - 0.60 = 5.44
c. This is a strong base in water. [OH-] = 0.020 M; pOH = 1.70; pH = 12.30
d. Major species present: H2O, Cl-, Na+, HONH2, HONH3+, OHAgain, the added strong base reacts completely with the best acid present, HONH3+.
Before
Change
After
HONH3+
0.100 M
-0.020
0.080
OH- ↔
0.020 M
-0.020
0
+
HONH2
0.100 M
+0.020
0.120
+ H2O
Reacts completely
A buffer solution results after reaction. Using the Henderson-Hasselbach equation:
pH = pKa + log([Base]/[Acid]) = 6.04 + log (0.120/0.080) = 6.04 + 0.18 = 6.22
40. a. pKb for C6H5NH2 = -log(3.8•10-10) = 9.42; pKa for C6H5NH3+ = 14.00 – 9.42 =
4.58
pH = pKa + log([C6H5NH2]/[C6H5NH3+]), 4.20 = 4.58 + log(0.50 M/[C6H5NH3+])
-0.38 = log(0.50 M/[C6H5NH3+]), [C6H5NH3+] = [C6H5NH3Cl] = 1.2 M
b. 4.0 g NaOH • 1 mol NaOH/40.00 g • 1 mol OH-/mol NaOH = 0.10 mol OH[OH-] = 0.10 mol/1.0 L = 0.10 M
Before
Change
After
C6H5NH3+
1.2 M
-0.10
1.1
+
OH- ↔
0.10 M
-0.10
0
C6H5NH2
0.50 M
+0.10
0.60
+ H2O
A buffer solution exists. pH = 4.58 + log(0.60/1.1) = 4.32
48. a. No. A solution of a strong acid (HNO3) and its conjugate base (NO3-) is not
generally considered a buffered solution.
b. No. Two acids are present (HNO3 and HF), so it is not a buffered solution.
c. H+ reacts completely with F-. Since equal volumes are mixed, the initial
concentrations in the mixture are 0.10 M HNO3 and 0.20 M NaF.
Before
Change
After
H+ +
0.10 M
-0.10
0
F0.20 M
-0.10
0.10
↔
HF
0
+0.10
0.10
Reacts completely
After H+ reacts completely, a buffered solution results, i.e. a weak acid (HF) and its
conjugate base (F-) are both present in large quantities.
d. No. A strong acid (HNO3) and a strong base (NaOH) do not form buffered
solutions. The will neutralize each other to form H2O.
54. This is a strong base, Ba(OH)2, titrated by a strong acid, HCl. The added strong acid
will neutralize the OH- from the strong base. As is always the case when a strong
acid and/or strong base reacts, the reaction is assumed to go to completion.
a. Only a strong base is present, but it breaks up into two mol of OH- ions for every
mol of Ba(OH)2. [OH-] = 2 x 0.100 M = 0.200 M; pOH = 0.699; pH = 13.301
b. mmol OH- present = 80.0 mL x 0.100 mmol Ba(OH)2/ml x 2 mmol OH-/mmol
Ba(OH)2
= 16.0 mmol OHmmol H+ added = 20.0 mL x 0.400 H+/mL = 8.00 mmol H+
Before
Change
After
OH- +
16.0 mmol
-8.00 mmol
8.0 mmol
H+
→
8.00 mmol
-8.00 mmol
0
H2O
Reacts completely
[OH-]excess = 8.0 mmol OH-/(80.0 mL + 20.0 mL) = 0.080 M; pOH = 1.10, pH = 12.90
c. mmol H+ added = 30.0 mL x 0.400 M = 12.0 mmol H+
Before
Change
After
OH- +
H+
→
H2O
16.0 mmol
12.00 mmol
-12.00 mmol -12.00 mmol
4.0 mmol
0
Reacts completely
[OH-]excess = 4.0 mmol OH-/(80.0 mL + 30.0 mL) = 0.036 M; pOH = 1.44, pH = 12.56
d. mmol H+ added = 40.0 mL x 0.400 M = 16.0 mmol H+. This is the equivalence
point. Since the H+ will exactly neutralize the OH- from the strong base, all we have
in solution is Ba2+, Cl- and H2O. All are neutral species, so pH = 7.00.
e. mmol H+ added = 80.0 mL x 0.400 M = 32.0 mmol H+
Before
Change
After
H+
→
H2O
OH- +
16.0 mmol
32.00 mmol
-16.00 mmol -16.00 mmol
0
16 mmol
[H+]excess = 16.0 mmol H+/(80.0 mL + 80.0 mL) = 0.100 M; pH = 1.00
72. a. yellow
b. green (Both yellow and blue forms are present)
c. yellow
d. blue
74. a. Ag2CO3(s) ↔ 2Ag+(aq) + CO32-(aq)
b. Ce(IO3)3(s) ↔ Ce3+(aq) + 3IO3- (aq)
c. BaF2(s) ↔ Ba2+(aq) + 2F-(aq)
86. a.
Initial
Equil.
Ag2SO4(s)
↔
s = solubility (mol/L)
Ksp = [Ag+]2[CO32-]
Ksp = [Ce3+][IO3-]3
Ksp = [Ba2+][F-]2
2Ag+(aq)
0
2s
+
SO42-(aq)
0
s
Ksp = 1.2•10-5 = [Ag+]2[SO42-] = (2s)2s = 4s3, s = 1.4•10-2 mol/L
b.
Initial
Equil.
Ag2SO4(s)
↔
s = solubility (mol/L)
2Ag+(aq)
0.10 M
0.10 + 2s
+
SO42-(aq)
0
s
Ksp = 1.2•10-5 = [Ag+]2[SO42-] = (0.10 + 2s)2s ≈ (0.10)2(s), s = 1.2•10-3 mol/L,
Assumption good.
c.
Initial
Equil.
↔
Ag2SO4(s)
s = solubility (mol/L)
2Ag+(aq)
0
2s
+
SO42-(aq)
0.20 M
0.20 + s
Ksp = 1.2•10-5 = [Ag+]2[SO42-] = (2s)2(0.20 + s) ≈ 4s2(0.20), s = 3.9•10-3 mol/L.
Assumption good.
88.
Initial
Equil.
Ce(IO3)3(s)
↔
s = solubility (mol/L)
Ce3+(aq)
0
s
+
3IO3- (aq)
0.20 M
0.20 + 3s
Ksp = [Ce3+][IO3-]3 = s(0.20 + 3s)3
From the problem, s = 4.4•10-8 mol/L; Solving for Ksp:
Ksp = (4.4•10-8)x[0.20 + 3(4.4•10-8)]3 = 3.5•10-10
104.
Ksp = 5.0•10-13
AgBr(s) ↔ Ag+ + BrAg+ + 2S2O32- ↔ Ag(S2O3)23Kf =2.0•1013
________________________________________________
AgBr(s) + 2S2O32- ↔ Ag(S2O3)23- + Br- K = KspxKf =14.5 (Carry extra sig. fig.)
↔
Ag(S2O3)232S2O320.500 M
0
S mol/L AgBr(s) dissolved to reach equilibrium
-s
-2s
+s
0.500 – 2s
s
AgBr(s)
Initial
Change
Equil.
+
+
Br0
+s
s
K = s2/(0.500 – 2s2)2 = 14.5; Taking the square root of both sides:
s/(0.500 – 2s) = 3.81, s = 1.91 – 7.92 s, s = 0.222 mol/L
1.00 L x 0.222 mol AgBr/L x 187.8 g AgBr/mol AgBr = 41.7 g AgBr = 42 g AgBr
126. 50.0 mL x 0.100 M = 5.00 mmol H2SO4; 20.0 ml x 0.100 M = 3.00 mmol HOCl
25.0 mL x 0.200 M = 5.00 mmol NaOH; 10.0 mL x 0.150 M = 1.50 mmol KOH
25.0 mL x 0.100 M = 2.50 mmol Ba(OH)2 = 5.00 mmol OHWe have added 11.50 mmol OH- total.
Let the OH- reacts with the best acid present. This is H2SO4, which is a diprotic acid.
For H2SO4, Ka1 >> 1 and Ka2 = 1.2•10-2. The reaction is:
10.00 mmol OH- + 5.00 mmol H2SO4 → 10.00 mmol H2O + 5.00 mmol SO42OH- still remains, and it reacts with the next best acid, HOCl (Ka= 3.5•10-8). The
remaining 1.50 mmol OH- will convert 1.50 mmol HOCl into 1.50 mmol OCl-,
resulting in a solution containing 1.50 mmol OCl- and (3.00 – 1.50 = ) 1.50 mmol
HOCl. Major species present at this point are: HOCl, OCl-, SO42-, H2O plus cations
that don’t affect the pH. SO42- is an extremely weak base (Kb = 8.3•10-13). Major
equilibrium affecting the pH is therefore:
HOCl ↔ H+ + OCl-.
Since [HOCl] = [OCl-], then:
[H+] = Ka = 3.5•10-8 M, pH = 7.46 Assumptions good.