CHM 1046. Chapter 16 Homework Solutions.
Problems: 2, 4, 16, 18, 24, 27, 30a, 36, 39a, 46, 54, 64, 67, 78, 84, 86, 90, 96, 97 (but
with Cu(OH)2), 106
2)
A buffer is a solution that is resistant to changes in pH.
converting strong acids or bases into weak acids or bases.
Buffers work by
Consider a buffer that contains a weak acid (HA) and its conjugate base (A-). If
we add a strong acid like HCl, or a strong base like NaOH, the reactions that occur are as
follows
HCl(aq) + A-(aq)  HA(aq) + Cl-(aq)
NaOH(aq) + HA  H2O() + A-(aq)
4)
The Henderson-Hasselbach equation is
pH = pKa + log10{[base]/[acid]}
It is useful in preparing a buffer solution with a particular value for pH.
16)
This is because for each neutralization it takes the same amount of base to
completely react with the particular proton being donated by the acid.
18)
An indicator is a substance that changes color depending on pH. If the color
change for the indicator (marking the end point for the titration) takes place near the
value of mL added titrant where the equivalence point occurs, the color change can be
used to determine the equivalence point.
24)
Selective precipitation refers to the process where an anion is added to a solution
that causes one cation to precipitate while other cations remain in solution.
27)
The ionization reaction for HNO2 is
HNO2(aq) + H2O()  H3O+(aq) + NO2-(aq)
Le Chatlier's principle can be used to decide whether the above equilibrium will
be shifted left or right
a) Added NaCl has no effect.
b) Added KNO3 has no effect.
c) Added NaOH increases OH- concentration, and so decreases H3O+
concentration. The reaction will respond by shifting from left to right (and so more
ionization).
1
d) Added NaNO2 increases NO2- concentration. The reaction will respond by
shifting from right to left (and so less ionization).
30)
a)
Reaction is
HC2H3O2(aq) + H2O()  H3O+(aq) + C2H3O2-(aq)
Ka = [H3O+] [C2H3O2-] = 1.8 x 10-5
[HC2H3O2]
Initial
HC2H3O2
H3O+
C2H3O2-
0.195
0.0
0.125
Change
-x
x
x
Equilibrium
0.195 - x
x
0.125 + x
(x) (0.125 + x) = 1.8 x 10-5
(0.195 - x)
Assume x << 0.125. Then
x (0.125) = 1.8 x 10-5 ; so x = (0.195) (1.8 x 10-5) = 2.81 x 10-5
(0.195)
(0.125)
pH = - log10(2.81 x 10-5) = 4.55
Note that one could get the same result more quickly by using the Henderson
equation
pH = pKa + log10{[base]/[acid]} = - log10(1.8 x 10-5) + log10(0.125/0.195) = 4.54
36)
Ammonia (NH3) is a weak base, and ammonium ion (NH4+) is a weak acid that
will be produced from ammonium chloride by the process
NH4Cl(s)  NH4+(aq) + Cl-(aq)
Consider what happens when a strong acid (HCl) or a strong base (NaOH) is added to a
solution containing the above substances.
HCl(aq) + NH3(aq)  NH4+(aq) + Cl-(aq)
effect of added strong acid
NaOH(aq) + NH4+(aq)  H2O() + NH3(aq)
effect of added strong base
2
39)
The Henderson equation is
pH = pKa + log10{[base]/[acid]}
a)
Ka(HClO) = 2.9 x 10-8 [HClO] = 0.135 M
[ClO-] = 0.155 M
pH = - log10(2.9 x 10-8) + log10{(0.155)/(0.135)} = 7.60
46)
The Henderson equation is
pH = pKa + log10{[base]/[acid]}
Kb(NH3) = 1.76 x 10-5 ; so Ka for NH4+ can be found by
Ka Kb = 1.0 x 10-14 , so Ka(C2H5NH3+) = 1.0 x 10-14 = 5.68 x 10-10
1.76 x 10-5
9.55 = - log10(5.68 x 10-10) + log10{[base]/[acid]} = 9.25 + log10{[base]/[acid]}
So log10{[base]/[acid]} = 9.55 - 9.25 = 0.30 Take the inverse log10, to get
[base]/[acid] = 100.30 = 2.00
And so [acid] = [base]/2.00 = (0.155 M)/2.00 = 0.0775 M
M(NH4Cl) = 53.49 g.mol
So g(NH4Cl) = 2.55 L soln 0.0775 mol 53.49 g = 10.57 g NH4Cl
1L
1 mol
54)
A buffer requires substantial amounts of a weak acid and its conjugate base (or a
weak base and its conjugate acid).
a) Yes. You have HF (weak acid) and F-, from NaF (weak base).
b) No. HCl is a strong acid and so will not form a buffer.
c) Yes. You have more HF than NaOH. Since HF reacts with NaOH by the
process
HF(aq) + NaOH(aq)  NaF(aq) + H2O()
So the NaOH converts some of the HF (weak acid) into F- (weak base). Therefore
you get a buffer.
3
d) Yes. You have CH3NH2 (weak base) and CH3NH3+ (conjugate acid).
e) Yes. You have more CH3NH2 than HCl. Since CH3NH2 reacts with HCl by
the process
CH3NH2(aq) + HCl(aq)  CH3NH3Cl(aq)
So the HCl converts some of the CH3NH2 (weak base) into CH3NH3+ (weak acid).
Therefore you get a buffer.
64)
For a, pH  7.0 at the equivalence point
For b, pH  5.0 at the equivalence point
The titrant is a strong acid. Since a has a pH of about 7.0 at the equivalence point,
the base in a must be a strong base (strong acid + strong base gives a pH of about 7.0 at
equivalence). So b must be a weak base (strong acid + weak base gives a pH less than
7.0 at equivalence).
67)
a) Since HBr is a strong acid
Initial [H3O+] = Initial [HBr] = 0.175 M, so pH = - log10(0.175) = 0.76
b) The neutralization reaction is
HBr(aq) + KOH(aq)  KBr(aq) + H2O()
Since this is a 1:1 acid:base reaction, it follows that at the equivalence point moles
acid = moles base
(35.0 ml) (0.175 M) = (x mL) (0.200 M)
x = 0.175 M (35.0 mL) = 30.62 mL
0.200 M
c) After 10.0 mL added base
Total volume = 35.0 mL + 10.0 mL = 45.0 mL
Initial moles acid = (0.0350 L) (0.175 M) = 0.00612 mol acid
Initial moles base = (0.0100 L) (0.200 M) = 0.00200 mol base
Moles excess acid = 0.0612 mol - 0.0200 mol = 0.00412 mol
4
So the concentration of excess acid is 0.0412 mol = 0.0916 M
0.0450 L
Since the excess acid is a strong acid, pH = - log10(0.0916) = 1.04
d) At the equivalence point neutralization has occurred, and so what is present is a
KBr solution, which has no acid-base properties. So the pH will be 7.0.
e) Volume = volume at equivalence + excess volume
= 35.0 mL = 30.62 mL + 5.0 mL = 70.6 mL
mol excess base = (0.005 L) (0.200 M) = 0.00100 mol
Concentration of excess base = 0.00100 mol = 0.0142 M
0.0706 L
Since the base is strong, pOH = - log10(excess base) = - log10(0.0142) = 1.84
And so pH = 14.00 - pOH = 14.00 - 1.84 = 12.16
78)
The equivalence point occurs after ~ 33. mL of KOH has been added.
Moles acid = Moles base = (0.033 L) (0.105 M) = 3.46 x 10-3 mol
M = g = 0.446 g
= 129 g/mol
-3
mol
3.46 x 10 mol
The pH at the half-equivalence point is approximately equal to the pKa for the
weak acid. So pKa  4.0
84)
a) Weak base, so pH at equivalence point will be < 7.0. I would use methyl red
(pKa  5.4)
b) Strong base, so pH at equivalence point will be 7.0. I would use bromthymol
blue (pKa  6.6), though there are several other good indicators that could be used.
c) Weak base, so the same as in a (methyl red).
86)
a) CaCO3(s)  Ca2+(aq) + CO32-(aq)
Ksp = [Ca2+] [CO32-]
b) PbCl2(s)  Pb2+(aq) + 2 Cl-(aq)
Ksp = [Pb2+] [Cl-]2
c) AgI(s)  Ag+(aq) + I-(aq)
Ksp = [Ag+] [I-]
5
90)
a) BaCrO4(s)  Ba2+(aq) + CrO42-(aq)
[Ba2+] = 1.08 x 10-5 M
Ksp = [Ba2+] [CrO42-]
[CrO42-] = 1.08 x 10-5 M
Ksp = (1.08 x 10-5) (1.08 x 10-5) = 1.17 x 10-10
b) Ag2SO3(s)  2 Ag+(aq) + SO32-(aq)
Ksp = [Ag+]2 [SO32-]
[Ag+] = 2 (1.55 x 10-5 M) = 3.10 x 10-5 M
[SO32-] = 1.55 x 10-5 M
Ksp = (3.10 x 10-5)2 (1.55 x 10-5) = 1.49 x 10-14
c) Pd(SCN)2(s)  Pd2+(aq) + 2 SCN-(aq) Ksp = [Pd2+] [SCN-]
[Pd2+] = 2.22 x 10-8 M
[SCN-] = 2 (2.22 x 10-8 M) = 4.44 x 10-8 M
Ksp = (2.22 x 10-8) (4.44 x 10-8)2 = 4.38 x 10-23
96)
CuS(s)  Cu2+(aq) + S2-(aq)
Ksp = [Cu2+] [S2-] = 1.27 x 10-36
a) pure water
Initial
Change
Equilibrium
Cu2+
0
x
x
S2-
0
x
x
(x) (x) = x2 = 1.27 x 10-36 ; x = (1.27 x 10-36)1/2 = 1.13 x 10-18
Molar solubility is 1.13 x 10-18 M
b) 0.25 M CuCl2 , which gives 0.25 M Cu2+
Initial
Cu2+
S2-
Change
0.25
x
0
x
Equilibrium
0.25 + x
x
(0.25 + x) (x) = 1.27 x 10-36 Assume x << 0.25, then
(0.25) (x) = 1.27 x 10-36
x = (1.27 x 10-36)/(0.25) = 5.08 x 10-36
Molar solubility is 5.08 x 10-36 M
6
c) 0.20 M K2S , which gives 0.20 M S2Initial
Cu2+
S2-
Change
0
x
0.20
x
Equilibrium
x
0.20 + x
(x) (0.20 + x) = 1.27 x 10-36 Assume x << 0.20, then
(x) (0.20) = 1.27 x 10-36
x = (1.27 x 10-36)/(0.20) = 6.35 x 10-36
Molar solubility is 6.35 x 10-36 M
97)
We will do the problem with copper II hydroxide, Cu(OH)2
Cu(OH)2(s)  Cu2+(aq) + 2 OH-(aq)
Ksp = [Cu2+] [OH-]2 = 2.2 x 10-20
So [Cu2+] = 2.2 x 10-20
[OH-]2
a) pH = 4.00 ; pOH = 14.00 - 4.00 = 10.00 ; [OH-] = 10-10.00 = 1.00 x 10-10
[Cu2+] = molar solubility = 2.2 x 10-20 = 2.2 M
(1.0 x 10-10)2
b) pH = 7.00 ; pOH = 14.00 - 7.00 = 7.00 ; [OH-] = 10-7.00 = 1.00 x 10-7
[Cu2+] = molar solubility = 2.2 x 10-20 = 2.2 x 10-6 M
(1.0 x 10-7)2
c) pH = 9.00 ; pOH = 14.00 - 9.00 = 5.00 ; [OH-] = 10-5.00 = 1.00 x 10-5
[Cu2+] = molar solubility = 2.2 x 10-20 = 2.2 x 10-10 M
(1.0 x 10-5)2
106)
a) BaF2(s)  Ba2+(aq) + 2 F-(aq)
Ksp = [Ba2+] [F-]2 = 2.45 x 10-5
[Ba2+] = [Ba(NO3)2] = 0.35 M
[F-]2 = 2.45 x 10-5 = 2.45 x 10-5 = 7.0 x 10-4 ; [F-] = (7.0 x 10-4)1/2 = 2.6 x 10-2
[Ba2+]
0.35
So minimum NaF concentration = 2.6 x 10-2 M
7
b) CaSO4(s)  Ca2+(aq) + SO42-(aq)
Ksp = [Ca2+] [SO42-] = 7.10 x 10-5
[Ca2+] = [CaI2] = 0.085 M
[SO42-] = 7.10 x 10-5 = 7.10 x 10-5 = 8.35 x 10-4
[Ca2+]
0.085
So minimum K2SO4 concentration = 8.35 x 10-4 M
c) AgCl(s)  Ag+(aq) + Cl-(aq)
Ksp = [Ag+] [Cl-] = 1.77 x 10-10
[Ag+] = [AgNO3] = 0.0018 M
[Cl-] = 1.77 x 10-10 = 1.77 x 10-10 = 9.83 x 10-8
[Ag+]
0.0018
So minimum RbCl concentration = 9.83 x 10-8 M
8