Question

The $pH$ of ${10}^{-3}M{H}_{2}S{O}_4$ is :

• A. 3
• B. 11.7
• C. 2.69
• D. None of the above

Answer 1

The correct option is C 2.69

Given, $pH={10}^{-3}$
After ionisation, $H_{2}S{O}_4\left(aq\right)\rightleftharpoons 2H^{+}\left(g\right)+S{O}_4^{2-}\left(g\right)$
$\left[H^{+}\right]$ from $H_{2}S{O}_4$ = $2\times {10}^{-3}M$
we know that $pH=-lo{g}_{10}\left[H^{+}\right]$ $pH=-lo{g}_{10}(2\times {10}^{-3}=3-lo{g}_{10}2=3-1.414=2.6989M)$
hence, $pH$ of ${10}^{-3}M{H}_{2}S{O}_4$ is 2.69