Question

What is the PH of ${Ba\left(OH\right)}_2$ if normality is ${10}^{-4}$:

• A. 4
• B. 10
• C. 7
• D. 9

Answer 1

The correct option is B 10

$Ba(OH{)}_2$ is a strong base so it will completely dissociate in solution:
$Ba\left(OH{)}_2\right(s)\to B{a}^{2+}+2O{H}^{-}$
each $Ba(OH{)}_2$ dissociates to make 2$O{H}^{-}$
Molarity = Normality/n-factor
$M=\frac{{10}^{-4}}{2}=0.00005$
$pOH=-lo{g}_{1}0\left[O{H}^{-}\right]=-lo{g}_{1}0\left[0.00005\right]=4.301$
As $pH+pOH=14$
$pH=14-4.301=9.69$