Question

$0.01$ mole $NaOH$ is added to $10$ litres of water. The $pH$ of the solution is_____.

Answer 1

Answer: 11

$[OH]=\frac{0.01}{10}=10^{-3}M$
$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{10^{-14}}{10^{-3}}=10^{-11}M$
$\therefore pH=-{\log }_{10}\left[H^{+}\right]=-{\log }_{10}\left(10^{-11}\right)=11$