Question

Calculate the pH of a 2 M solution of H2SO4.

Answer 1

$H_2SO_4->H^++HSO_4^-$

2 0 0

2-2=0 2 2

$HSO_4^-->H^++SO_4^{2-}$

2 2 0

2-x 2+x x

$Ka_2=1.1×10^{-2}$

$\frac{x(2+x)}{2-x}=0.011$

$x^2+2.011x-0.022=0$

$x=[H^+]=0.011$

$pH=-log[H^+]=-log(0.011)=1.96$

$pH=1.96$