Question

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

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 `1.0*10^(−5)M`.

Where,

`K_(a_1) = 4.3*10^-7`, and
`K_(a_2) = 5.6*10^-11`

Answer 1

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`