Page 104 - Basic physical chemistry for the atmospheric sciences

90 Basic physical chemistry

determine how many moles of C02 would have to be absorbed in I L
of rainwater.
Solution. Since the pH of rainwater is 5.6, the concentration of
H30 + (aq) in natural rainwater is given by
pH = 5 . 6 = - l og[H 0 + (aq)]
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Therefore,

The main source of H 30 + (aq) when C02 dissolves in water is
COz(g) + H20(l) � H2C0 3 (aq) (5. 1 5a)

H2C03 (aq) + H20(l) � HCO)(aq) + H30 + (aq) (5 . 1 5 b)
We see from Reactions (5. 1 5) that for every mole of C02 that is
absorbed in water, one mole of H 30 + (aq) is produced. Therefore, to
produce 0.25 x 1 0 - 5 M of H30 + (aq) , 0.25 x 10 -5 moles of C02 would
have to be absorbed in each liter of rainwater. Since this is about the
solubility of C02 in water at atmospheric pressure, we see that the
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absorption of C02 in rainwater will cause it to have a pH of about . 6.

5. 7 Polyprotic acids
In the above exercise we assumed that Reaction (5 . l 5b) is the only
source of H30 + (aq) when C02 dissolves in water. In fact, following
Reaction (5 . 1 5b) there is another source of H30 + (aq), namely
(5. 1 6)

We see from Reactions (5. 1 5b) and (5. 1 6) that H2C03 (carbonic acid)
contributes two protons to water. Substances that contribute more
than one proton to water are called polyprotic acids. Other polyprotic
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acids are H2C204 (oxalic acid), H3P0 (phosphoric acid), and H 2 S0 3
(sulfurous acid) .
The successive acid-dissociation constants of a polyprotic acid are
often labeled Kai • Kaz• etc. Provided Kai > > Ka2, the second reaction
can be neglected as a source of protons compared to the first reaction.
This is the case for carbonic acid and is the justification for ignoring
Reaction (5 . 1 6) in Exercise . 2 . Note that intuitively we would expect
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the neutral molecule H2C03(aq) to give up a proton more readily than
the negatively charged ion HC0 3 (aq). Of course, if we were inter-