Question

Calculate $pH$ of $0.00424M$ aqueous solution of $KOH$.

Answer 1

$pOH=-\log \left[O{H}^{-}\right]$

Here, $\left[O{H}^{-}\right]=0.00424$

$=-\log [424\times {10}^{-5}]$

$=5-\log \left(424\right)=3-\log \left(4.24\right)$

$\Rightarrow pH=14-pOH=14-(3-\log (4.24\left)\right)$

$=11+\log \left(4.24\right)$