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

1 Answer
Apr 19, 2018

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.