More EXAM III Practice
CHEM 162-04
Dr. Overway
Instructions: Do not open this test booklet until you are instructed to do so.
Show your work in order to receive partial credit. Keep track of units and use the correct significant
figures. The following information may be useful during the exam.
[A- ]
[BH + ]
)
[B]
𝑝H = 𝑝𝐾𝑎 + log ([HA])
𝑝OH = 𝑝𝐾𝑏 + log (
pH = - log [H3O+]
pOH = - log [OH-]
pH + pOH = 14
Ka ·Kb =Kw
Citric
Nitrous
Hydrofluoric
Formic
Benzoic
Acetic
Carbonic
Hydrogen sulfite ion
Hydrogen sulfide
Hypochlorous
Dihydrogen phosphate ion
Boric
Ammonium ion
Hydrocyanic
Phenol
Hydrogen carbonate ion
Hydrogen peroide
Monohydrogen phosphate ion
Chemical
Pb2+ (aq)
CO32- (aq)
PbCO3 (s)
pKa = - log Ka
Kw=110-14
Acid
M1V1=M2V2
R = 8.3145 J/(K mol)
−∆𝐺
𝐾𝑒𝑞 = 𝑒 𝑅 𝑇
Formula Conjugate Base
H3C6H5O7
H2C6H5O7HNO2
NO2HF
FHCOOH
HCOOC6H5COOH C6H5COOCH3COOH CH3COOH2CO3
HCO3HSO3SO32H2S
HSHClO
ClOH2PO4HPO42H3BO3
H2BO3NH4+
NH3
HCN
CNC6H5OH
C6H5OHCO3CO32H2O2
HO2HPO42PO43-
∆𝐺 = −𝑅𝑇 ln(𝐾𝑒𝑞 )
Ka
7.1  10-4
4.6  10-4
3.5  10-4
1.8  10-4
6.5  10-5
1.8  10-5
4.3  10-7
1.0  10-7
9.1  10-8
3.0  10-8
6.2  10-8
7.3  10-10
5.6  10-10
4.9  10-10
1.3  10-10
5.6  10-11
2.4  10-12
2.2  10-13
Thermodynamic values at 25 C
Hfo (kJ/mol) Sfo (J/(K mol)) Gfo (kJ/mol)
-1.7
10.5
-24.4
-677.1
-56.90
-527.8
-699.1
131.0
-625.5
CHEM 162-04
More Exam III Practice
1. The equilibrium constant for the following reaction:
H2(g) + Br2(g)  2 HBr(g)
6
is 1.9510 at a certain temperature. Find the equilibrium pressure of all of the gases if 5.50 atm of
HBr is introduced into a sealed container at this temperature.
Equilibrium pressures
-3
p(H2) = 3.9410 atm
p(Br2) = 3.9410-3 atm
p(HBr) = 5.49 atm
H2(g) + Br2(g)  2 HBr(g)
I (atm) 0
0
5.5
C
+x
+x
-2x
E
x
x
5.5-2x
Kp = 1.95106 = (5.5-2x)2/(x2)
x = 3.9410-3
assume 2x<< 5.5
0.1%
2. Which of the following solutions will show the properties of a buffer (circle all that apply)?
A) 10 mL of 0.10 M NH4Cl + 50 mL of 0.010 M NaOH
B) 20 moles of NH4Cl + 10 moles of NaOH in 1.0 L of solution.
C) 10 moles of NH4Cl + 20 moles of NaOH in 1.0 L of solution.
strong base left over!
D) 10 mL of 0.10 M CH3COOH + 10 mL of 0.10 M NaCH3COO
E) 10 mL of 0.10 M HCl + 10 mL of 0.10 M NaCl
strong acid left over!
F) 10 mL of 0.10 M CH3COOH + 5 mL of 0.10 M HCl
strong acid left over!
3. (10 pts) In a titration of 75.0 mL of a 0.100 M HCN solution with a 0.200 M NaOH solution, how
many mL of the strong base is required to reach the equivalence (stoichiometric) point?
(75.0 mL) (0.100 M HCN) = (X mL) (0.200 M NaOH)
X mL = 37.5 mL of NaOH
4. What is the equilibrium concentration of [I-] if 12.58 grams of PbI2 (pKsp = 8.10) is added to
500.0 mL of pure water?
PbI2 (s)
I
C
E

Pb2+ (aq) +
0
+x
x
2 I- (aq)
0
+2x
2x
5. Determine the Keq value for the following reaction.
Pb2+ (aq) + CrO42- (aq)  PbCrO4 (s)
Ksp= 10-8.10 = (x)(2x)2
x=1.26×10-3
[I-]=2x=2.51×10-3 M
pKsp = 13.74
This is a reverse Ksp ,so Keq = 1/ Ksp = 1/10-13.74 = 5.50×1013
6. Determine the Gibbs Free Energy (G) and Ksp for the following reaction at 88.7 C.
Pb2+ (aq) + CO42- (aq)  PbCrO4 (s)
The temperature is not 25 C, so I must use G = H - TS
Hrxn = (1)(-699.1) – (1)(-1.7) –(1)(-677.1) = -20.3 kJ/mol
Srxn = (1)(131.0) – (1)(10.5) –(1)(-56.90) = 177.4 J/(K mol)
G = H - TS = (-20.3 kJ/mol) – (273.15 + 88.7 K)[ 177.4 J/(K mol)  1000] = -84.49 kJ/mol
Keq = e-G/RT = exp(+84.49×1000/(8.3145 × 361.85) = 1.57×1012
7. Given the following reaction, how could you get it to do the following:
H2O (l) + CO2 (g) + heat HCO32- (aq) + H+ (aq)
A) Two ways to shift the reaction toward the products.
Increase temperature OR increase pressure OR increase pressure of CO2 OR decrease the
concentrations of HCO32- (aq) or H+ (aq)
B) Shift the reaction toward the reactants.
Increase the concentration of HCO32- (aq) or H+ (aq) OR decrease the pressure of CO2 or
lower the temperature