The $P^H$  of a sample of KOH solution is 12.3979. The weight of solid KOH of 70% pure required to prepare 2.5 lit of this solution is

$P^H=12.3979;P^{OH}=1.6021$

$\left[O{H}^{-}\right]=AL\left(-1.6021\right)(Al=Anti\log )$

               $=AL\left(-2+0.3979\right)$

               $=0.025N$

$W_{KOH}=N\times GEW\times V$

            $=0.025\times 56\times 2.5$

            $=3.5gr$      

$\text{Required}\text{wt}=\frac{3.5\times 100}{70}=5gr$