Question

$pH$ of ${10}^{-8}$ M $Ba(OH{)}_2$ solution will be:

• A. $6.96$
• B. $6.31$
• C. $7.04$
• D. $7.08$

Answer 1

The correct option is B $6.31$

$Ba(OH{)}_2⟶B{a}^{2+}+2O{H}^{-}$

$\Rightarrow \left[O{H}^{-}\right]={10}^{-8}\times 2=2\times {10}^{-8}$

$\therefore p^{OH}=-\log (2\times {10}^{-8})=7.69$

$\therefore p^H=14-p^{OH}=14-7.69=6.31$