Answer
pH of HCl solution is 1.40
pH of $H_{2}SO_{4}$ solution is 1.31
Work Step by Step
-- HCl is a strong acid, [$H^+$]=[HCl]=0.040M
pH=-log[$H^+$]=-log(0.040)$\approx$1.40
-- $H_{2}SO_{4}$ dissociates to give its two protons in two steps.
First step: $H_{2}SO_{4}$(aq)--> $H^+$(aq) + $HSO_{4}^{-}$(aq)
In this very first step, $H_{2}SO_{4}$ completely dissociates, so the concentration of $H^+$ formed [$H^+$]=[$H_{2}SO_{4}$]=0.040M, and in the meantime the concentration of $HSO_{4}^{-}$ formed [$HSO_{4}^{-}$]=0.040M.
Second step: $HSO_{4}^{-}$(aq) $H^+$(aq) +$SO_4^{2-}$(aq)
In this step, $HSO_{4}^{-}$ partially dissociates, so we need calculate the final [$H^+$] using K$_a$($HSO_{4}^{-}$).
The final concentration of $HSO_{4}^{-}$= (0.040-x)M
The final concentration of $H^+$= (0.040+x)M
The final concentration of $SO_4^{2-}$=x M
K$_a$($HSO_{4}^{-}$)=$\frac{[H^+][SO_4^{2-}]}{[HSO_{4}^{-}]}$$\frac{(0.040+x)(x)}{0.040-x}$=1.3$\times$10$^{-2}$.
The quadratic equation we form is x$^{2}$+0.053x-5.2$\times$10$^{-4}$=0; using the quadratic formula we find x=8.46$\times$10$^{-3}$, and we cancel the other negative solution.
The final concentration of $H^+$= (0.040+x)M=(0.040+8.46$\times$10$^{-3}$)M$\approx$ 0.048M
So the pH of $H_{2}SO_{4}$ solution =-log[$H^+$]=-log(0.048)$\approx$1.31
