The pH of 0.005 M $Ba(OH{)}_2$ is ________.

It is given to find out the pH of 0.005 M $\mathrm{Ba}(\mathrm{OH}{)}_2$.

 pH + pOH = 14

 ${\mathrm{BaOH}}_2\rightleftharpoons {\mathrm{Ba}}^{2+}+2{\mathrm{OH}}^{-}$

$\left[{\mathrm{OH}}^{-}\right]=2\times 0.005$ = 0.01 M

pH + pOH = 14

pH = 14 – ($-\log \left[{\mathrm{OH}}^{-}\right]$) = 14 – (-log [0.01]) = 14 – 1.70 = 12.

Hence, the answer is 12.