General Chemistry
Midterm Test
2008. november 12.
a.
We mix two solutions one with pH 2 and a second with pH 4. What should be the volume
ratio if we want to obtain a solution with pH 3 (suppose that the volumes are additive)?
V(pH=2)·10−2 + V(pH=4)·10−4=[V(pH=2)+V(pH=4)]·10−3  V(pH=2)/V(pH=4)=0.1
V(pH=2)/V(pH=4):
b.
0.1
(3p)
We have 100 cm3 pH 10.70 KOH solution and pH 3.1 nitric acid solution. How many
cm3 nitric acid solution is needed to neutralize the KOH solution?
c(KOH)=10−(14−10.7)=5.012·10−4 M (1p) c(HNO3)=10−3.1=7.943·10−4 M
V(HNO3)=100 cm3·(5.012·10−4 M)/(7.943·10−4 M)=63.1 cm3
(1p)
V(HNO3):
(1p)
63.1 cm3
(3p)
c. The Kacid = 1.8·10−5 for the acetic acid. What is the pH and the percent ionization of the
of the 0.5 mol/dm3 acetic acid solution?
[H+](Ks·c0)1/2=3.0·10−3 M  pH=2.52
= [H+]/c0=(3.0·10−3 M)/(0.500 M)=0.006
pH: 2.52
:
(3p)
0.6%
(1p)
d. What is the pH of the of the 0.25 mol/dm3 sodium acetate solution?
Kh=Kw/Ks=10−14/1.8·10−5=5.555·10−10
(1p)
−
1/2
−10
1/2
[OH ](Kh·c0) =(5.555·10 ·0.250) M=1.179·10−5 M
pH:
e. The Henderson-Hasselbalch Equation:
(3p)  pH=9.07 (1p)
(5p)
(5p)
where pKa is acidic ionization exponent, conjugate base concentration is the
concentration of the salt of weak acid with strong base, and acid is the weak acid.
f. The Six Methods of Preparing Buffer Solutions:
(6p)
1. weak acid (HA) + titrate with strong base (NaOH)
2. weak acid (HA) and salt of weak acid (NaA)
3. salt of weak acid (NaA) + titrate with strong acid (HCl)
4. weak base (B) + titrate with strong acid (HCl)
5. weak base (B) and salt of weak base (BHCl)
6. salt of weak base (BHCl) + titrate with strong base (NaOH)
g. Titration of a Weak Acid with a Strong Base (show the curve. and explain)
(9p)
Use the following solubility products:
Ksp(25°C)
1.44·10−10
1.50·10−16
AgCl
AgI
a.
Ksp(25°C)
1.60·10−5
1.35·10−8
PbCl2
PbI2
What is the solubility (S. M=mol/dm3) of the AgCl and the PbI2 in 25°C distilled water?
Ksp(AgCl)=S2  S(AgCl)=[Ksp(AgCl)]1/2=1.20·10−5 M
Ksp(PbI2)=S·(2S)2=4S3  S(PbI2)=[Ksp(PbI2)/4]1/3=1.50·10−3 M
S(AgCl): 1.20·10−5 M
(2p)
S(PbI2): 1.50·10−3 M
(2p)
b. We mix the same volume of saturated AgCl and the PbI2 solutions in water at 25°C. Shall we
observe
precipitate?
Explain
all
the
details
(the
volumes
are
additive).
Shall we observe precipitate? Yes
[Ag ]=6·10−6 M [Cl−]=6·10−6 M [Pb2+]=7.5·10−4 M [I−]=1.5·10−3 M (1p)
[Ag+][I−]=9.0·10−9 >> Ksp(AgI)=1.50·10−16  precipitate is formed (1p)
[Pb2+]·[Cl−]2=2.7·10−14 << Ksp(PbCl2)=1.60·10−5  no precipitate (1p)
If yes. give the formula: AgI
+
c.
(3p)
What is the solubility (S) of the PbI2 in 0.5 mol/dm3 KI solution?
Ksp(PbI2)=S·[I−]2  S=Ksp(PbI2)/[I−]2 = 1.35·10−8/0.5002=5.40·10−8 M
S(PbI2/KI): 5.4 10-8 M
(3p)
d. Effects on the Solubility of PbCl2
add KNO3 to saturated PbCl2 solution
add HCl to saturated PbCl2 solution (draw an approximate solubility diagram versus the
concentration of added salt or acid):
(6p)
General Chemistry
Midterm Test
2008. november 12.
Name:
How much octane (C8H18) have to be burned perfectly to obtain 105 kJ enthalpy? (5p)
How much 0°C ice can be melted by this enthalpy if the efficiency of the enthalpy transfer is
70.0 %-? (5p)
Lmelt(ice)= 35.0 kJ/mol
Hf: C8H18 (g): –269.8 kJ/mol. CO2 (g): –393.5 kJ/mol. H2O(l): –285.8 kJ/mol.
Balance the equation (2p):
1 C8H18 (g) +
12.5 O2 (g) = 8 CO2 (g) +
9 H2O(l)
Explain the reduction and the oxidation. give the half reactions: (5p)
The mass of the octane:
2.09 kg
The mass of the ice: 36.0 kg .
H=[8·(-393.5)+9·(-285.8)] – [1·(-269.8)] kJ/mol= – 5450.4 kJ/mol
n(octane)=105/5450.4 mol=18.3 mol
M=8·12+18·1=114 g/mol
m(octane)=114 g/mol·18.3 mol= 2091.6 g = 2.09 kg
n(ice)=0.700·105/35 mol=2 000 mol  m(ice)= 36.00 kg
Half reactions: (the degree of oxidation of O is zero in O2; it is -2 in CO2 and H2O; the
degree of oxidation of C is +4 in CO2)
O2 + 4e -> 2O2- (reduction)
8 C+4 + 50 e + 18 H+ -> C8H18 (reduction)
Explain the Hess’s law (6p) and give an example (6p):
If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall
process is the sum of the enthalpy changes for the individual steps.
Example:
Hrxn = ∑Hf°products- ∑Hf°reactants
(see above the reaction enthalpy)
The following standard reduction potentials are given:
Sn2+ + 2e  Sn
°= −0.145 V
Sn4+ + 2e  Sn2+
°= +0.150 V
What is the spontaneous chemical reaction that can be derived from these half reactions? (5p)
Sn4+ + 2e → Sn2+
°= +0.150 V (reduction, more positive)
Sn2+ + 2e ← Sn
°= −0.145 V (more negative: oxidation, it goes from right to left)
reduction: Sn4+ + 2e → Sn2+
oxidation: Sn
→ Sn2+ + 2e
total react: Sn + Sn4+ → 2 Sn2+
Ecell0= 0.295 V = 0.15 –(-0.145) V
What is the equilibrium constant of this chemical reaction at 25 C? (5p)
nF Ecell0 = R T ln K → n F Ecell0/RT = ln K
n = 2 (the number of the electrons in the reaction), F=96500 C, R gas constant, T = 298 K
What is the G° of this chemical reaction? (5p) G° = - n F Ecell0
Show the cell diagram (5p)
(anode: oxidation) Sn(s) | Sn2+(aq)|| Sn2+(aq), Sn4+(aq)|Pt(s) (cathode, reduction)
Show Ecell as a Function of Concentration (10p)
ΔG = ΔG° -RT ln Q
-nFEcell = -nFEcell° -RT ln Q
Q is the reaction quotient: Q = [Sn2+]2/[Sn4+]
 
 
2
Ecell
0.0592
Sn 2 
 0.195 
log
V
2
Sn 4 