Answers & Clues to Final Exam Revision Questions:
1.
Arrange the following in order of increasing boiling point:
RbCl, CH3Cl, CH3OH, CH4.
Ans: CH4 < CH3Cl < CH3OH < RbCl
CH4 – non polar: dispersion force only
CH3Cl – polar: dispersion & dipole-dipole forces
CH3OH – polar: dispersion, dipole-dipole & H-bonding
RbCl – ionic salt that contains ionic bonding
2.
Which one of the following substances should exhibit hydrogen bonding in the liquid
state?
SiH4 H2
H2S
CH4
CH3NH2
Ans: CH3NH2
3.
Calculate the molality of a solution containing 14.3 g of NaCl in 42.2 g of water.
Ans: 𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 =
4.
14.3 𝑔
⁄58.5 𝑔/𝑚𝑜𝑙
0.0422 𝑘𝑔
𝑚𝑜𝑙𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
=
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑘𝑔)
= 5.80 𝑚
Calculate the approximate freezing point of a solution made from 21.0 g NaCl and
1.00  102 g of H2O. Kf of water is 1.86C/m.
Ans:
𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 =
21.0 𝑔
⁄58.5 𝑔/𝑚𝑜𝑙
0.100 𝑘𝑔
NaCl  Na+ + Cl–
= 3.5897 𝑚
van’t Hoff factor, i = 2
∆Tf = (i)(Kf)(m) = (2)( 1.86C/m)(3.5897 m) = 13.354 oC
Freezing point of water = 0 oC
Freezing point of solution = 0 oC – 13.354 oC = –13.354 oC
5.
A first-order reaction has a rate constant of 3.00  103 s1. The time required for
the reaction to be 75.0% complete is
Ans:
if 75% complete: [A]0 = 100% and [A]t = 25%
ln [A]t = ln [A]0 – kt
ln 25 = ln 100 – (3.00  103 s1) (t)
t = 462.09 s
6.
When the concentrations of reactant molecules are increased, the rate of reaction
increases. Explain why.
Ans: As the reactant concentration increases, the frequency of molecular collisions
increases.
7.
If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the
reaction vessel when 2NO2(g)
2NO(g) + O2(g) reaches equilibrium is 0.674 atm.
Calculate the equilibrium partial pressure of NO2.
Ans:
2NO2(g)
2NO(g) + O2(g)
I
0.500 atm
0
0
C
–2x
+2x
+x
E
0.500 – 2x
2x
x
At equilibrium, total pressure = 0.674 atm
P(NO2) + P(NO) + P(O2) = 0.674 atm
(0.500 – 2x) + 2x + x = 0.674 atm
x = 0.674 atm – 0.500 atm = 0.174 atm
8.
At 340 K, Kp = 69 for the reaction H2 (g) + I2 (g)
2HI (g). 50.0 g of HI is injected
into an evacuated 5.00-L rigid cylinder at 340 K. What is the total pressure inside the
cylinder when the system comes to equilibrium?
Ans: Calculate initial pressure of HI: 𝑃 =
Given: H2 (g) + I2 (g)
2HI (g)
𝑛𝑅𝑇
𝑉
=
(
50.0
𝑎𝑡𝑚
)(0.0821 𝐿.
.𝐾)(340𝐾)
127.908
𝑚𝑜𝑙
5.00 𝐿
= 2.182 𝑎𝑡𝑚
Kp = 69
HI is injected first, therefore HI now becomes the reactant, the equation has to be inversed:
2HI (g)
H2 (g) + I2 (g)
2HI (g)
𝐾𝑝 =
Kp’ = 1/69 = 0.0145
H2 (g) + I2 (g)
I
2.182 atm
0
0
C
–2x
+x
+x
E
2.182 – 2x
x
x
𝑃𝐻2 ×𝑃𝐼2
𝑥
2.182−2𝑥
2
𝑃𝐻𝐼
=
(𝑥)(𝑥)
(2.182−2𝑥)2
= √0.0145
= 0.0145
x = 0.212
Total pressure = (2.182 – 2(0.212)) + 0.212 + 0.212 = 2.18 atm
9.
Acid strength decreases in the series HI > HSO4 > HF > HCN. Which of the following
anions is the weakest base?
Ans: I
Clue: Use conjugate pair: the weakest base is the anion that comes from the strongest
acid.
10.
Arrange the acids HOCl, HClO3, and HClO2 in order of increasing acid strength.
Ans: HOCl < HClO2 < HClO3
Clue: apply the concept of increasing oxidation number, refer to chlorine – higher
oxidation number means stronger acid.
11.
Which of the following yields a basic solution when dissolved in water?
KNO2 NH4Cl NaCl SO2
Ans: KNO2
Clue: write the hydrolysis equations and find out which ion (H+ or OH–) exist:
KNO2  K+ + NO2–
K+ come from strong base (KOH), it will not hydrolyzed.
NO2– + H2O ⇌ HNO2 + OH– (OH– is present, solution is basic)
12.
Calculate the concentration of oxalate ion (C2O42) in a 0.175 M solution of oxalic acid
(C2H2O4).
[For oxalic acid, Ka1 = 6.5  102, Ka2 = 6.1  10–5.]
Ans:
𝐾𝑎1 =
⇌
H2C2O4
HC2O4–
H+
I
0.175 M
0
0
C
–x
+x
+x
E
0.175 – x
x
x
[𝐻 + ][𝐻𝐶2 𝑂4− ]
[𝐻2 𝐶2 𝑂4 ]
=
𝑥2
0.175−𝑥
= 6.5 × 10−2
*Clue: solve x using quadratic equation
⇌
HC2O4–
𝐾𝑎2 =
+
C2O42–
x = 0.078995 or –0.143995
+
H+
I
0.078995 M
0
0.078995 M
C
–x
+x
+x
E
0.078995 – x
x
0.078995 + x
[𝐻 + ][𝐶2 𝑂42− ]
[𝐻𝐶2 𝑂4− ]
=
(𝑥)(0.078995+𝑥)
0.078995−𝑥
= 6.1 × 10−5
*Clue: solve x using assumption. Check validity!
x = [C2O42–] = 6.1  10–5
13.
Consider the weak bases below and their Kb values:
C6H7O
Kb = 1.3  1010
C2H5NH2
Kb = 5.6  104
C5H5N
Kb = 1.7  109
Arrange the conjugate acids of these weak bases in order of increasing acid strength.
Ans: C2H5NH3+ < C5H5NH+ < C6H7OH
Clue: arrange the weak bases in order of increasing basic strength (higher Kb value means
stronger base): C6H7O < C5H5N < C2H5NH2
Apply this concept: ‘Strong base has weak conjugate acid pair.’ Therefore, the strongest
base has the weakest conjugate acid.
14.
When 2.0  10–2 mole of nicotinic acid (a monoprotic acid) is dissolved in 350 mL of
water, the pH is 3.05. What is the Ka of nicotinic acid?
Ans:
molarity of nicotinic acid = 2.0  10–2 mol/0.350 L = 0.057 M
Assume nicotinic acid = HA
HA
I
0.057 M
C
–x
E
0.057 – x
pH = 3.05
𝐾𝑎 =
15.
⇌
(𝑥)(𝑥)
A–
+
0
0
+x
+x
x
x
; –log [H+] = 3.05
=
0.057−𝑥
H+
(8.913 × 10−4 )2
0.057− (8.913 × 10−4 )
therefore [H+] = 8.913  10–4 M = x
= 1.42 × 10−2
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid
(HCNO) and 0.80 mole of sodium cyanate (NaCNO) in enough water to make 1.0 liter of
solution. [Ka(HCNO) = 2.0  104]
[𝐶𝑁𝑂 − ]
0.80
Ans: 𝑝𝐻 = 𝑝𝐾𝑎 + log [𝐻𝐶𝑁𝑂] = 3.70 + log 0.20 = 4.30
16.
Consider a buffer solution prepared from HOCl and NaOCl. Which is the net ionic
equation for the reaction that occurs when NaOH is added to this buffer?
Ans: OH + HOCl ⇌ H2O + OCl
Clue: OH– from the NaOH added reacts with the acid in the buffer solution
17.
You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and
0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the
addition of 20.0 mL of 1.00 M NaOH solution? [Ka = 1.8  105]
Ans:
Mole CH3COOH =
Mole CH3COO– =
I
C
E
1000
0.20 ×500
1000
1.00 ×20
Mole NaOH =
CH3COOH
0.150 mol
–0.020 mol
0.130 mol
0.30 ×500
OH–
⇌
0.020 mol
–0.020 mol
0
+
Molarity CH3COO– =
= 0.100 𝑚𝑜𝑙
= 0.020 𝑚𝑜𝑙
1000
Molarity CH3COOH =
= 0.150 𝑚𝑜𝑙
0.130 𝑚𝑜𝑙
0.520 𝐿
0.120 𝑚𝑜𝑙
0.520 𝐿
CH3COO–
0.100 mol
+0.020 mol
0.120 mol
+
H2O
= 0.250 𝑀
= 0.231 𝑀
[𝐶𝐻3 𝐶𝑂𝑂− ]
0.231
𝑝𝐻 = 𝑝𝐾𝑎 + log
= 4.74 + log
= 4.71
[𝐶𝐻3 𝐶𝑂𝑂𝐻]
0.250
18.
Calculate the pH at the equivalence point for the titration of 0.20 M HCl with 0.20 M
NH3 (Kb = 1.8  105).
Ans: assume 1 L of each solution:
HCl
+
NH3
⇌
NH4Cl
I
0.20 mol
0.20 mol
0
C
–0.020 mol
–0.020 mol
+0.020 mol
E
0
0
0.20 mol
The salt NH4Cl will undergo hydrolysis:
Molarity of NH4Cl = 0.20 mol / 2 L = 0.10 M
NH4+ +
0.10 M
–x
0.10 – x
I
C
E
𝐾𝑎 =
H2O
𝐾𝑤
1.0 × 10−14
=
𝐾𝑏
1.8 × 10−5
⇌
NH3 + H3O+
0
0
+x
+x
x
x
(𝑥)(𝑥)
=
0.10 − 𝑥
𝑥2
∴ 0.10−𝑥 = 5.56 × 10−10
𝑥 = [𝐻3 𝑂+ ] = 7.45 × 10−6
pH = –log 7.45  10–6 = 5.13
19.
The molar solubility of magnesium carbonate is 1.8  104 mol/L. What is Ksp for this
compound?
Ans:
20.
MgCO3 ⇌ Mg2+ + CO3–2
s
s
Ksp = [Mg2+][ CO3–2] = s2 = (1.8  104)2 = 3.24  10–8
Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb2+ and 0.0050 M
Ag+. (Ksp (PbI2) = 1.4  108; Ksp (AgI) = 8.3  1017). What compound will precipitate
first?
Ans: AgI will precipitate first because AgI has the smallest Ksp.
21.
Calculate the minimum concentration of Mg2+ that must be added to 0.10 M NaF in order
to initiate a precipitate of magnesium fluoride. (For MgF2 , Ksp = 6.9  109.)
Ans:
MgF2 ⇌ Mg2+ + 2F–
s
2s
2+
2
–
Ksp = [Mg ][F ] = 6.9  109
[Mg2+](0.10)2 = 6.9  109
 [Mg2+] = 6.9  107
22.
Will a 0.1 M solution of Na2HPO4 (aq) be acidic, basic or neutral? Given: Ka = 4.8  10–13
and Kb = 1.6  10–7.
Ans:

Na+ is a spectator ion (it comes from strong base, NaOH), so it does not affect the
acidity or basicity of a solution.

There are two possible reactions HPO4– ion can undergo:

23.
#1:
HPO4–2 ⇌ H+ + PO4–3
Ka = 4.8  10–13
#2:
HPO4–2 + H2O ⇌ H2PO4– + OH–
Kb = 1.6  10–7
Kb > Ka, therefore the solution is basic.
What is the missing symbol in this plutonium fission reaction?
239
1
91
1

Pu
+
n
______
+
Sr
+
3
94
0
38
0n
146
Ans: 56 Ba
24.
Radium-226 decays by alpha emission. What is its decay product ?
222
Ans: 86 Rn
25.
Find the nuclear binding energy of potassium-40 (atomic mass = 39.9632591 amu) in
units of joules per nucleon. Data: neutron mass = 1.674928  10–24 g; proton mass =
1.672623  10–24 g; electron mass = 9.109387  10–28 g; NA = 6.0221367  1023 /mol; c =
2.99792458  108 m/s; 1 g = 6.022  1023 amu; 1 J = 1 kg/m2.s2.
Ans:
40
19𝐾
∆m
= [(19  1.672623  10–27 kg) + (21  1.674928  10–27 kg)]
1𝑔
– (39.9632591 amu  6.022 × 1023𝑎𝑚𝑢 ×
→ 19 11𝑝 + 21 10𝑛
1 𝑘𝑔
1000 𝑔
)
= 5.912208818  10–28 kg
∆E = ∆m  c2 = 5.912208818  10–28 kg  (2.99792458  108 m/s)2 = 5.3136  10–11 J
Binding energy per nucleon:
∆E / 40 = 5.3136  10–11 /40 = 1.328  10–12 J per nucleon
26.
A nucleus of carbon-14 lies above the belt of stability. Write an equation for radioactive
decay of carbon-14.
Ans: 146𝐶 →
27.
14
7𝑁
+
0
−1𝛽
A radioactive isotope of Rb-87 decays as follow: (cancel this sentence)
25
11𝑁𝑎
→
25
12𝑀𝑔
+
0
−1𝛽
t½ = 15 hours
Starting with 100 g of Na-25, calculate the quantity of Mg-25 produced after 19 hours.
Ans:
Initial mass of Na-25 = 100 g
At present, assume mass of Mg-25 = x g, therefore mass of Na-25 = (100 – x) g
λ = ln 2 / 15 hours = 0.0462 hours–1
Calculating for Na-25:
ln Nt = ln N0 – λt
ln (100 – x) = ln 100 – (0.0462 hours–1)(19 hours)
ln (100 – x) = 2.7274
(100 – x) = e 2.7274
x = 41.57 g = mass Mg-25
Mass Na-25 left = 100 g – 41.57 g = 58.43 g
28.
What is the monomer of natural rubber? Draw its structure and name it.
Ans: monomer of natural rubber is isoprene.
2-methylbuta-1,3-diene
29.
K2Cr2O7 / H2SO4
Reflux
CH2ClCH2–OH + 2[O]
?
?
+ H2O
Ans: CH2ClCOOH (chloroethanoic acid)
30.
𝑁𝑖
CH2 = CH– CH3 + H2 →
?
Ans: CH3 – CHBr – CH3 (2-bromopropane)
31.
Ans:
32.
(cyclopropene)
Write chemical equations to show the steps involve in the mechanism for the
polymerization of polyethylene using benzoyl peroxide radical as the initiator.
Ans:
Step 1: Initiation
Or can be written as: C6H5COO–OOCC6H5  2(C6H5COO•)
Reaction of radical with monomer (ethylene):
C6H5COO• + CH2 = CH2  C6H5COO– CH2– CH2•
Step 2: Propagation
C6H5COO– CH2– CH2• + CH2 = CH2 
C6H5COO– CH2– CH2– CH2– CH2•
Step 3: Termination
C6H5COO– CH2– CH2– CH2
n
– CH2• + •CH2–CH2– CH2– CH2
 C6H5COO– CH2– CH2– CH2
n
– CH2 –CH2–CH2– CH2– CH2
m
– OOCC6H5
m
– OOCC6H5
From nuclear tutorial:
24.43 A rock contains 270 mol 238U ( t1 / 2 = 4.5  109 years) and 110 mol 206Pb. Assuming
that all the 206Pb comes from the decay of the 238U, estimate the age of the rock.
Ans:
Initial % of 238U = 100%
270𝜇𝑚𝑜𝑙
% of 238U at present = (270+110)𝜇𝑚𝑜𝑙 × 100% = 71.05%
λ = ln 2 / 4.5  109 years = 1.54  10–10 yrs–1
ln Nt = ln N0 – λt
100
71.05
∴𝑡=
= 2.22 × 109 𝑦𝑒𝑎𝑟𝑠
1.54 × 10−10
ln