Question

${10}^{-2}$ mole of $NaOH$ was added to 10 litres of water then calculate the change in $pH$ value.

Answer 1

The solution of $NaOH$ on mixing with water with becomes basic. Therefore, its' $pH$ will be more than $7$. This is because,

$pH=-\log \left[OH\right]=-\log \left[0.01/10\right]=-\log {10}^{-3};pH=3$

Therefore $pH=14-pOH=14-3=11$