Question

What is the pH of a 0.325 M solution of H2SO4? Ka2 = 1.20×10–2?

Answer 1

Let's consider that sulfuric acid is very strong, and losing its first proton is very easy. Thus, let's assume we are left worrying about `HSO_4^-` losing its proton in the equilibrium,

`HSO_4^(-)(aq) rightleftharpoons SO_4^(2-)(aq) + H^(+)(aq)`

where `K_("a"_2) = ([SO_4^(2-)][H^+])/([HSO_4^(-)]) = 1.2*10^-2`

Let's "calculate" `[H^+]` for the first proton,

`[H^+] = [H_2SO_4] = 0.325M`

and work this in to our second proton dissociation calculation,

`K_("a"_2) = (x(0.325+x))/(0.325-x) = 1.2*10^-2`

`therefore x = 0.0112`

Hence, the pH of our solution is approximately,

`"pH" = -log(0.325+x) approx 0.473`

which is reasonable because sulfuric acid is a very strong acid.