Question

Calculate the $pH$ of an $NaOH$ solution, the concentration of which is $0.1g/L$. Assume the dissociation of $NaOH$ to be complete.

Answer 1

density of $NaOH=0.1g$/litre

Molarity of $NaOH=\frac{0.1}{40}$ mole / litre $=2.5\times {10}^{-5}M$

If $NaOH$ is dissociate completely, then

$\left[O{H}^{-}\right]=2.5\times {10}^{-3}M$

$pOH=-\log \left[O{H}^{-}\right]=-\log (2.5\times {10}^{-3})=3-\log \left(2.5\right)$

$\therefore pH=14-pOH=11+\log \left(2.5\right)=11.40$