Question

Calculate the $pH$ of the resulting solution formed by mixing the following solutions.
$10mL$ of $0.1M{H}_{2}S{O}_4$ + $10mL$ of $0.01MKOH$.

Answer 1

$m$ moles of $H^{+}=10\times 0.1\times 2=2$

$m$ moles of $O{H}^{-}=10\times 0.01\times 0.1=0.1$

$\therefore m$ moles of $H^{+}$ left= $2-0.1=1.9$

$\left[H^{+}\right]=\frac{1.9}{10+10}=0.095$

$\therefore pH=-\log \left[H^{+}\right]=-\log \left(0.095\right)=1.022$