Question

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

Answer 1

$\left[\right.OH\left]\right.=\frac{0 . 01}{10}=10^{- 3}M$
$\left[H^{+}\right]=\frac{K_{w}}{\left[OH^{-}\right]}=\frac{10^{- 14}}{10^{- 3}}=10^{- 11}M$
$\therefore \quad pH =-\log _{10}\left[ H ^{+}\right]=-\log _{10}\left(10^{-11}\right)=11$