What are the concentrations of $H_2CO_3$ and its conjugate bases in a raindrop with $pH = 5.90$ ?

Question

Calculate the concentrations of carbonic acid, bicarbonate ion ( $HCO_3^-$ ) and carbonate ion ( $CO_3^(2−)$ ) that are in a raindrop that has a pH of 5.90, assuming that the sum of all three species in the raindrop is $1.010^(−5)M$ .

Where,

$K_(a_1) = 4.310^-7$ , and
$K_(a_2) = 5.6*10^-11$

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Answer 1 · 2017-12-08T20:14:10

We must assume,

$[H_2CO_3] + [HCO_3^(-)] + [CO_3^(2-)] = 1.0*10^-5$ , and
$pH = 10^(-5.90) approx 1.26*10^-6M$

Moreover,

$H_2CO_3(aq) rightleftharpoons H^(+)(aq) + HCO_3^(-)(aq)$
$HCO_3^(-)(aq) rightleftharpoons H^(+)(aq) + CO_3^(2-)(aq)$

and,

$K_(a_1) = ([H^(+)][HCO_3^(-)])/([H_2CO_3])$
$K_(a_2) = ([H^(+)][CO_3^(2-)])/([HCO_3^(-)])$

So, we will relate these equilibrium expressions,

$[H_2CO_3] = ([H^(+)][HCO_3^(-)])/(K_(a_1)$

$[CO_3^(2-)] = (K_(a_2)[HCO_3^(-)])/([H^(+)])$

and our assumption,

$([H^(+)][HCO_3^(-)])/(K_(a_1)) + [HCO_3^(-)] + (K_(a_2)[HCO_3^(-)])/([H^(+)]) = 1.0*10^-5$

Now, to simplify a bit,

$[H^+]^2[HCO_3^(-)] + K_(a_1)[H^+][HCO_3^-] + K_(a_1)K_(a_2)[HCO_3^(-)] = 1.0*10^-5 * K_(a_1) [H^+]=>$

$[HCO_3^(-)]([H^+]^2 + K_(a_1)[H^+] + K_(a_1)K_(a_2))= 1.0*10^-5 * K_(a_1) [H^+]$

$therefore [HCO_3^(-)] approx 2.5*10^-6M$

From here, it's a lot easier! We'll use the former equilibrium expressions,

$[H_2CO_3] = ((1.3*10^-6)(2.5*10^-6))/(4.3*10^-7) approx 7.5*10^-6$ , and

$[CO_3^(2-)] = ((5.6*10^-11)(2.5*10^-6))/(1.3*10^-6) approx 1.1*10^-10$