Page 113 - Basic physical chemistry for the atmospheric sciences

Acids and bases

Let us now calculate the concentration of H 0 + (aq), H2C03(a ) ,
q
3
HC0 3 ( aq), OH - ( aq) , and Co� - (aq) when C02 from the atmosphere
dissolves in otherwise pure rainwater, given that the solubility of C02
in water is l . O x 1 0 - 5 M at 25°C and 1 atm. Since we have five un­
knowns we need five equations to solve this problem, and so far we
s
have only two equation , namely (5.39) and (5 .40). The other three
equations are provided by the ion-product constant for water
(5. 4 1 )

the material balance relation
l .
[H C03] initial = O X 1 0 - 5M =
2
[H2C03(aq)] + [HC0 3 (aq)] + [Co� - ( aq)] (5 .42)
and the charge balance relation

[H 0 + (aq)] = [HC0 3 (aq)] + 2[Co� - (aq)] + [OH - (aq)] (5.43)
3
where the l . O x 1 0 - 5 M in Eq. (5 .42) follows from the fact that for
every mole of C02 that dissolves in water one mole of H2C0 3 is
formed [see Reaction (5. 1 5 a)] , and the coefficient 2 in Eq. (5.43) allows
for the two units of negative charge on each coj - (aq) ion.
The solution of Eqs. (5.39) through ( . 4 3) is simplified if we make
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some approximations. Since Kai > > K32, the contributions to
[H 0 3 + (aq)] from Reaction (5 . 3 8) is negligible compared to that from
Reaction (5 . 3 7). Also, since the only source of co� - ( aq) is from
Reaction (5. 3 8), [Co� - ( aq)] will be small compared to [H2C03 (aq)]
O
and [HC03(aq)] . Finally, since H - (aq) derives only from the dissoci­
ation of water, and an acid has been added to the water, we can
assume that [H 0 3 + (aq)] > > [OH - (aq)] . Hence, from Eq . (5 .43),
[H30 + (aq)] = [HC03(aq)] , and Eqs . (5 .39) and (5.42) become
0 + (aq)] 2
[H3 1
= 4 . 2 x 1 0 - (5 .39a)
[H2C03(aq)]
and ,
(5.42a)

s
We now have two equations for the two unknowns, which yield :
[H30 + (aq)] = . 8 x 1 0 - 6 M and [H2C0 3 ( aq)] = 8. 1 x 1 0 - 6 M . There­
1
6
3
fore , [HC0 3(aq)J = [H 0 + (aq)] = l . 8 x 1 0 - M. Substituting [Hp +
9
(aq)] = l . 8 x 1 0 - 6 M i n Eq. (5 . 4 1 ) yields [OH-(aq)] = 5 . 6 x 1 0 - M.