determine the ph of 0.015 M solution of h2so4. the dissociation occurs in two steps ka1 is extremely large ka2 is 1.2*10^-2

Question

aaronq · Answer

Since the first Ka is extremely large, we assume that the dissociation of the first proton goes to completion, meaning $[H^+]=[H_2SO_4]$

after we use the equilibrium expression for the second proton, 

$HSO_4^-\rightleftharpoons  H^++SO_4^{2-}$

$K_{a2}=\dfrac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

we also know that $[HSO_4^-]$ is equal to initial concentration of $H_2SO_4$

aaronq · Answer

Add up the concentrations of $H^+$ then use them to find the final pH