ANSWERS TO EXERCISES AND REVISION QUESTIONS(iii)
Kc(T)kkf
b
Here,
3.892.4^10- 4
kb
giving,
kb2.4^10- 4
3.89 6.2^10 –^5 mol–^1 dm–^3 s–^1
15.4
K (pNOCl)^2 (0.32)^2
p(T)(pNO) (^2) pCl
2
(0.22) (^2) (0.11)19 atm–^1
15.5
Kc(T)[Sn
4 +(aq)][Fe 2 +(aq)] 2
[Sn^2 +(aq)][Fe^3 +(aq)]^2
The equilibrium concentration of iron(II) ions will be
double that of tin(IV), i.e. [Fe^2 +(aq)]0.080 mol dm–^3.
The following equilibrium concentrations are also known:
[Sn^4 +(aq)]0.040 mol dm–^3 ; [Sn^2 +(aq)]
0.050 mol dm–^3. We also know that Kc1.0 1010.
Substituting into the equilibrium constant expression,
K 0.040(0.080)^2
c0.050[Fe 3 +(aq)] 2
giving
[Fe^3 +(aq)]√(0.040(0.080)
2
0.0501.0 1010 )
7.2 10 –^7 mol dm–^3
15.6No,Kcis so small that virtually no phosgene has
decomposed at this temperature.
15.7We make two observations before tackling this
problem:
(a) We take ‘yield’ to mean the gain in concentration of
methanol at equilibrium.
(b) This reaction involves a reduction in the number of
molecules, going from left to right.
This allows us to make the following conclusions:
(i)Reduces gain in [CH 3 OH] at equilibrium.
(ii)The equilibrium concentration of CH 3 OH increases.
(iii)This decreases the equilibrium concentration of
CH 3 OH.
(iv)No effect.
(v)Equilibrium composition contains more product
(CH 3 OH) at lower temperatures.
Kcis only affected in (v) – Kcincreases with decreasing
temperature for an exothermic reaction.
15.8To see the relationship, substitute the following
equilibrium expressions into the equation
Kc(3)Kc(1) Kc(2):
K [NO(g)]^2 [NO^2 (g)]^2
c(1)[O 2 (g)][N 2 (g)]; Kc(2)[O 2 (g)][NO(g)] 2 ;
K [NO^2 (g)]^2
c(3)[N 2 (g)][O 2 (g)] 2
15.9
(i)Applying the equilibrium law:
Kc(T)[O[O^2 (g)]water
2 (g)]air
or
[O 2 (g)]air[OK^2 (g)]water
c(T)
By the ideal gas equation, the partial pressure of oxygen,
pO 2 , in the air is proportional to its concentration:
pO 2 [O 2 (g)]airRT
so that
pO 2 [O^2 (g)]waterK RT
c(T)
or [O 2 (g)]waterpO^2 RTKc(T)
which may be expressed in the form of Henry’s law (see
Chapter 11):
cKHp
whereppO 2 ,c[O 2 (g)]waterandKHKRTc(T)
(ii)Applying the equilibrium law;
Kc(T)[I[I^2 ]CCl^4
2 ]H 2 O
leads directly to the distribution law. Kc(T)is the
distribution ratio, Kd(T).
Unit 16
Exercises
16A
Kwincreases as the temperature is increased.
16B
(i)[H 3 O+(aq)]0.10 mol dm–^3
[OH–(aq)]1.0 10 –^14 /[H 3 O+(aq)]
1.0 10 –^14 /0.10
1.0 10 –^13 mol dm–^3
pHlog (0.10) 1.0
(ii)[H 3 O+(aq)] 10 – 2.206.3 10 –^3 mol dm–^3
16C
(i)The number of moles of Ca(OH) 2 is
0.10 g
(40 17 17) g mol–^1 0.0014 mol
The concentration of calcium hydroxide is
0.00140.0070 mol dm– 3
0.200
One Ca(OH) 2 provides two hydroxide ions in solution.
Thus, [OH–(aq)]0.014 mol dm–^3.
At room temperature,
1.0 10 –^14 [0.014] [H 3 O+(aq)].
Rearranging;
[H 3 O+(aq)]1.0 10 –^14 /0.014
7.1 10 –^13 mol dm–^3
pHlog [ H 3 O+(aq) ] 12.15
(ii)We answer (b) first:
(b)
[H 3 O+(aq)] 10 – 11.107.9 10 –^12 mol dm–^3
(a)
[OH–(aq)]1.0 10 –^14 /[H 3 O+(aq)]
1.0 10 –^14 / 7.94 10 –^12
1.3 10 –^3 mol dm–^3
16D
pOH 14 pH 14 5.808.20;
[OH–(aq)] 10 – 8.206.3 10 –^9 mol dm–^3
16E
[H 3 O+(aq)] [OH–(aq)] pH pOH
10 –^1411410
10 –^1310 –^11311
10 –^1210 –^21212
10 –^1110 –^31113
10 –^1010 –^41014
10 –^910 –^51915
10 –^810 –^61816
10 –^710 –^71717
10 –^610 –^81618
10 –^510 –^91519
10 –^410 –^101410
10 –^310 –^111311
10 –^210 –^121212
10 –^110 –^131113
Note the pattern: the pH decreases steadily and the pOH
increases steadily. A high pOH value indicates a strongly
acidic solution. A high pH indicates a strongly basic
solution.
16F
(i)
[H 3 O+(aq)] (KaCA)
CA0.020 mol dm–^3
Ka4.9 10 –^10 mol dm–^3 (from Table 16.2)
[H 3 O+(aq)] (4.9 10 –^10 0.020)
3.1 10 –^6 mol dm–^3
The percentage of HCN molecules that are ionised is
calculated as follows:
% ionized [H^3 O
+(aq)]
C ^100
A
3.1^10
– (^6) 100
0.020 0.016%
pHlog[H 3 O+(aq)]log(3.1 10 –^6 )5.5
(ii)Mass of benzoic acid in solution 4000
0.050/1002.0 g.
The formula of benzoic acid is obtained from Table 16.2;
M(C 6 H 5 COOH)122 g mol–^1.
Number of moles of benzoic acid 2.0/1221.6
10 –^2.
Concentration of benzoic acid 1.6 10 –^2 / 1.0 1.6
10 –^2 mol dm–^3.
[H 3 O+(aq) (KaCA)
CA1.6 10 –^2 mol dm–^3 ;
Ka6.5 10 –^5 mol dm–^3 (from Table 16.2)
[H 3 O+(aq)] (6.5 10 –^5 1.60 10 –^2 )
1.0 10 –^3 mol dm–^3
pHlog[H 3 O+(aq)]log(1.0 10 –^3 )3.00
The percentage of benzoic acid molecules ionised is
[1.0 10 –^3 /1.6 10 –^2 ] 100
6 %. Solution of the
equation
K [H^3 O+(aq)]^2
a(T)CA[H 3 O+(aq)]
gives an accurate value of [H 3 O+(aq)] as 9.85
10 –^4 mol dm–^3 , close to the approximate value obtained
above.
16G
(i)CH 3 COOH(aq)H 2 O(l) CH 3 COO–(aq)H 3 O+(aq)
As more hydronium ions are mopped up by OH–(aq) ions
in the neutralisation reaction, more ethanoic acid
molecules ionize. The titration volumes are then
identical with those of a strong acid.
(ii)The members of the first two pairs are strong acids
and strong bases. The CH 3 COOH–NaOH pair contains a
weak acid and most of its molecules must first ionize
before neutralization can take place. The ionization
process is endothermic and reduces the overall amount
of heat given out in the reaction between ethanoic acid
and sodium hydroxide.
16H
(i)Kb 10 – 5.981.1 10 –^6 mol dm–^3
(ii)
Kb(T) [OH
- (aq)] 2
CB
Rearranging;
[OH–(aq)] (Kb(T)CB) (1.8 10 –^5 0.200)
1.9 10 –^3 mol dm–^3
[H 1.0^10 –^14
3 O+(aq)]1.90 10 – 3 5.3^10 –^12 mol dm–^3pH log(5.3 10 –^12 )11.284450230_000118_2 9 _Ans.qxd 3/2/06 2:22 pm Page 445