Page 195 - Basic physical chemistry for the atmospheric sciences

Answers, hints, and solutions to exercise.\· I H I
3
5. 1 3 . Concentration of protons i s 1 . 3 x 1 0 - M . Hint: Proceed 1 1 1
5
t
similar manner o Exercise . 1 2 .
3
5 . 1 4 . [H + (aq)] = [H2P0 - ( aq)] = 8 . 3 x 1 0 - M;
4
1
[HPoi - ( aq)] = 6 . 2 x 1 0 - 8 M ; [POl - (aq)] = 3 . 6 x 0 - 1 8 M
Hint: Assume that the H + (aq) ions derive mainly from the
first stage of dissociation, and that the concentration of
any ion formed in one stage is not significantly affected by
succeeding dissociation .
s
5 . 1 5 . pH = 7 ; fraction hydrolyzed = l . 1 % . Solution: Since
K3(HC2HP2) = Kb(NH 3 ) , the cations and anions hydrolyze
equally. Therefore, the concentration of H 0 + (aq) due to
3
h y drolysis of H .t (aq) will be the same as the concentra­
N
tions of O H - ions from the hydrolysis of C 2 H3 02 - . There­
+
fore, the pH of the solution is 7, and [H 3 0 (aq)] = 1 0 - 7 •
N
For the H 4 + ( aq) hydrolysis,
N
Kh = Kw/Kb( H 3)
= 0 - 1 4 / 1 . 8 X 1 0 - 5 = 5 . 6 X 1 0 - I O
1
fo
The hydrolysis reaction r N H .t is

Therefore,

t
Le , x = [NH3(aq)]. Then,
0.0050 - x = [NH.t (aq)]

Therefore,
10 - 1x
- 5 - 6 x 1 0 - 1 0
-
0 . 00 50 - x
Let s assume that 0.0050 - x = 0 . 0050, then
u
1 0 - 1 X
_ 5 . 6 x 1 0 - I O
-
0.0050
or, x = . 8 x 1 0 - 5 (which j u stifies our assumption that
2
x < < 0.0050). Hence, the quantity of NH4 + ( aq) that is hy­
)
drolyzed is 2 . 8 x 1 0 - 5 M. Since Ka(HC2H 302) = Kb(NH3 ,