Question

What is the $pH$ of a solution containing $0.01moles{H}_2{SO}_4$ and $0.01moles$ $KOH$ in $2litre$ water.

Answer 1

Total mass of $H^{+}=0.01\times 2=0.02$

Total moles of $O{H}^{-}=0.01\times 1=0.01$

$\therefore$ Moles of $H^{+}$ left= $0.01$

$\left[H^{+}\right]=\frac{{10}^{-2}}{2}=5\times {10}^{-3}M$

$\therefore pH=-\log \left[H^{+}\right]=-\log (5\times {10}^{-3})=2.3010$