CHM1046 (Ref #625001)
Chapter 17 and Chapter 18
Exam 4 - Worksheet
Acid-Base Equilibria and Solubility Equilibria:
1. You are asked to prepare a pH = 3.00 buffer starting from 1.50 L of 1.00 M of hydrofluoric acid
(HF) and an excess of sodium fluoride (NaF).
a. What is the pH of the hydrofluoric acid solution prior to adding sodium fluoride?
Weak acid calculation (need to do ICE).
HF (aq)
I
C
E
1.00
-x
1.00 – x
H+ (aq) + F- (aq)
0
+x
x
Ka = 6.8 x 10-4
0
+x
x
Ka = 6.8 x 10-4 =
=
x2 = 6.8 x 10-4
x = 2.61 x 10-2 M = [H+]
pH = -log (2.61 x 10-2)
pH = 1.58
b. How many grams of sodium fluoride should be added to prepare the buffer? Neglect the
small volume change that occurs when the sodium fluoride is added.
Given: pH = 3.00 and [HF] = 1.00 M
Need [F-]
Since this is a buffer we can use the Henderson-Hasselbach equation:
pH = pKa + log
3.00 = 3.167 + log
log
= -0.167
= 10-0.167 = 0.681
[F-] = (0.681) ([HF]) = (0.681) (1.00) = 0.681 M
NaF  Na+ + F-
[NaF] = 0.681 M
Need grams of NaF
(0.681 moles NaF/1 L) (1.50 L) (41.99 g NaF/ mol NaF) = 42.9 g NaF
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2. What is the ratio of HCO3- to H2CO3 in blood of pH 7.4? How is this ratio affected in an
exhausted marathon runner whose blood H is 7.1?
H+ + HCO3-
H2CO3
Ka = 4.3 x 10-7
pH = pKa + log
= 7.4 – 6.37 = 1.03
log
= 101.03 = 10.7
at pH 7.1:
log
= 7.1 – 6.37 = 0.73
= 100.73 = 5.4
3. How many milliliters of 0.0850 M NaOH are required to titrate each of the following solutions to
the equivalence point:
a. 35.0 mL of 0.0850 M CH3COOH
At the equivalence point, the number of moles of CH3COOH equals number of moles of NaOH.
Number of moles of CH3COOH: (35.0 mL) (0.0850 moles/1000 mL) = 2.975 x 10-3 moles CH3COOH.
Number of moles of NaOH has to equal 2.975 x 10-3 moles.
To calculate volume: (volume) x ([NaOH])
(2.975 x 10-3 moles NaOH) (1000 mL/0.0850 moles) = 35.0 mL
b. 40.0 mL of 0.0900 M HF
At the equivalence point, the number of moles of HF equals number of moles of NaOH.
Number of moles of HF: (40.0 mL) (0.0900 moles/1000 mL) = 3.6 x 10-3 moles HF.
Number of moles of NaOH has to equal 3.6 x 10-3 moles.
To calculate volume: (volume) x ([NaOH])
(3.6 x 10-3 moles NaOH) (1000 mL/0.0850 moles) = 42.4 mL
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4. A 35.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution.
Calculate the pH after the following volumes of base have been added:
a. 0 mL
Initial pH, calculation is of weak acid (need to do ICE)
CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
I
C
E
0
0.150
-x
0.150 – x
Ka = 1.8 x 10-5
0
+x
x
+x
x
Ka = 1.8 x 10-5 =
=
x2 = 2.7 x 10-6
x = 1.64 x 10-3 M = [H+]
pH = -log (1.64 x 10-3)
pH = 2.78
b. 17.5 mL
Not yet at equilibrium (number of moles of NaOH less than number of moles of CH3COOH)
# moles CH3COOH: (35.0 mL) (0.150 mol/1000 mL) = 0.00525 moles
# moles of NaOH: (17.5 mL) (0.150 mol/ 1000 mL) = 0.00263 moles
When NaOH is added:
CH3COOH (aq)
Before:
0.00525 mol
Addition:
– 0.00263 mol
After:
OH- (aq) 
+
0
0.00263 mol
0.00262 mol
0
CH3COO- (aq) + H2O (l)
0
+ 0.00263 mol
0.00263 mol
[CH3COOH] = (0.00262 mol/0.0525 L) = 0.050 M
[CH3COO-] = (0.00263 mol/0.0525 L) = 0.050 M
pH = pKa + log
= 4.74 + 0
pH = 4.74
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c. 35.0 mL
CH3COOH (aq)
Before:
0.00525 mol
Addition:
– 0.00525 mol
After:
+
OH- (aq) 
CH3COO- (aq) + H2O (l)
0
0.00525 mol
0 mol
0
0
+ 0.00525 mol
0.00525 mol
[CH3COOH] = (0.00525 mol/0.070 L) = 0.075 M
CH3COO- (aq)
CH3COOH (aq) + OH- (aq)
I
0.075
0
C
-x
+x
+x
E
0.075 – x
x
x
Kb = (Kw/Ka) = 5.56 x 10-10
0
Kb = 5.56 x 10-10 = x2/(0.075 – x) = x2/0.075
x2 = 4.17 x 10-11
x = 6.45 x 10-6 = [OH-]
pOH = 5.19
pH = 8.81
5. For manganese (II) hydroxide, Mn(OH)2 the Ksp = 1.6 x 10-13.
a. Calculate the pH of this solution.
Mn(OH)2 (s)  Mn2+ (aq) + 2 OH- (aq)
Ksp = 1.6 x 10-13
Ksp = 1.6 x 10-13 = [Mn2+] [OH-]2 = (x)(x2)
x = 5.43 x 10-5 M = [Mn(OH)2]
Now need to calculate [OH-]:
(5.43 x 10-5 mol Mn(OH)2 / 1 L) (2 mol OH-/1 mol Mn(OH)2) = 1.09 x 10-4 M = [OH-]\
pOH = 3.96
pH = 10.0
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b. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH.
Initial [OH-] = 0.020 M
Mn(OH)2 (s)  Mn2+ (aq) + 2 OH- (aq)
I
0
C
+x
E
x
0.020
+ 2x
0.020 + 2x
Ksp = 1.6 x 10-13 = (x) (0.020 + 2x)2 = x(0.020)2
[Mn(OH)2] = 4.0 x 10-10 M
6. Will Ag2SO4 precipitate when 100 mL of 0.050 M AgNO3 is mixed with 10 mL of 0.050 M
Na2SO4 solution? Show your work.
2 AgNO3 (aq) + Na2SO4 (aq)  2 NaNO3 (aq) + Ag2SO4 (?)
To determine if Ag2SO4 will precipitate at these conditions, we need to calculate Q:
Ag2SO4 (s)  2Ag+ (aq) + SO42- (aq)
Q = [Ag+]2[SO42-]
To calculate [Ag+]:
(0.100 L) (0.050 mol AgNO3/1L)(1 mol Ag+ / 1 mol AgNO3) = 0.005 moles Ag+
Total volume = 110 mL = 0.110 L
[Ag+] = (0.005 moles/0.110 L) = 0.045 M
To calculate [SO42-]:
(0.010 L) (0.050 mol Na2SO4/1L)(1 mol SO42- / 1 mol Na2SO4) = 5.0 x 10-4 moles SO42Total volume = 110 mL = 0.110 L
[SO42-] = (5.0 x 10-4 moles/0.110 L) = 0.0045 M
Q = [Ag+]2[SO42-] = (0.045)2 (0.0045) = 9.11 x 10-6
Q < Ksp
Precipitate will not form.
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Thermodynamics:
1.
For the following reaction, calculate ∆Suniverse and determine whether the reaction is
spontaneous, nonspontaneous, or at equilibrium:
H2 (g) + I2 (g)  2HI (g)
∆Suniv = ∆Ssys + ∆SSurr
∆SSurr = -(∆Hsys)/T
Values from the Appendix:
∆Hf° (H2) = 0 kJ/mol
∆Hf° (I2) = 62.25 kJ/mol
∆Hf° (HI) = 25.9 kJ/mol
∆Hrxn = 2(25.9 kJ/mol) – [(62.25 kJ/mol) + 0 kJ/mol] = -10.5 kJ/mol = -1.05 x 104 J/mol
∆SSurr = -(∆Hsys)/T = (-1.05 x 104 J/mol)/(273 K) = 38.5 J/mol∙K
Values from the Appendix:
∆S° (H2) = 131.0 J/mol∙K
∆S° (I2) = 260.6 J/mol∙K
∆S° (HI) = 206.3 J/mol∙K
∆Srxn = 2(206.3 J/mol∙K) – [(260.6 J/mol∙K) + 131.0 J/mol∙K] = 21.03 J/mol∙K
∆Suniv = (21.03 J/mol∙K+ (38.5 J/mol∙K) = 59.5 J/mol∙K
process is spontaneous
2. Calculate the standard free-energy change for the following reaction at 298 K, given that
∆H° = 180.7 kJ/mol and ∆S = 24.7 J/mol∙K. Is this reaction spontaneous under these conditions?
N2 (g) + O2 (g)  2 NO (g)
∆G° = ∆H° - T∆S° = (180.7 kJ/mol) – (298 K) (2.47 x 10-2 kJ/mol∙K)
∆G° = 173.3 kJ/mol
∆G° is positive, the reaction is not spontaneous under these conditions.
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3. Calculate ∆G at 298 K for a mixture of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in
the following process:
N2 (g) + 3 H2 (g)  2 NH3 (g)
For this reaction, ∆H° = -92.4 kJ/mol and ∆S° = -198.3 J/mol∙K.
∆G = ∆G° + RT ln Q
We first need to calculate ∆G°
∆G° = ∆H° - T∆S° = (-92.4 kJ/mol) – (298 K)(-0.1983 kJ/mol∙K)
∆G° = -92.4 kJ/mol + 59.1 kJ.mol
∆G° = -33.3 kJ/mol
to calculate ∆G, we need to find Q:
Q=
= 9.3 x 10-3
=
∆G = ∆G° + RT ln Q = (-33.3 kJ/mol) + [(8.314 x 10-3 kJ/mol∙K)(298 K) ( ln 9.3 x 10-3)]
∆G = (-33.3 kJ/mol) + (-11.6 kJ/mol) = - 44.9 kJ/mol
4. The value of Ka for nitrous acid (HNO2) at 25°C is 4.5 x 10-4.
a. Write the chemical equation for the equilibrium.
H+ (aq) + NO2- (aq)
HNO2 (aq)
Ka = 4.5 x 10-4
b. By using the value of Ka, calculate ∆G° for the dissociation of nitrous acid in aqueous
solution.
∆G° = - RT ln K = - (8.314 J/mol∙K) (298 K) (ln 4.5 x 10-4) = + 19093 J/mol = + 19.1 kJ/mol
c. What is the value of ∆G at equilibrium?
At equilibrium ∆G = 0
d. What is the value of ∆G when [H+] = 5.0 x 10-2 M, [NO2-] = 6.0 x 10-4 M, and
[HNO2] = 0.20 M?
Q=
=
= 1.5 x 10-4
∆G = ∆G° + RT ln Q = (+ 19.1 kJ/mol) + [(8.314 x 10-3 kJ/mol∙K)(298 K) ( ln 1.5 x 10-4)]
∆G = (+ 19.1 kJ/mol) + (-21.8 kJ/mol) = -2.7 kJ/mol
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