Question

Assuming that $Ba(OH{)}_2$ is completely ionised in aqueous solution under the given conditions the concentration of $H_3{O}^{+}$ ions in 0.005 M acqueous solution of $Ba{\left(OH\right)}_{2}at298K$ is ______ $\times {10}^{-12}mol{L}^{-1}$. (Nearest integer)

Answer 1

Assuming $Ba(OH{)}_2$ is completely ionised,
$Ba{\left(OH\right)}_2\to B{a}^{+2}+\underset{\downarrow }{2O{H}^{-}}$
$2\times 0.005=0.01={10}^{-2}$

We know
$\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$
$\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

$\left[H_3{O}^{+}\right]=\frac{{10}^{-14}}{{10}^{-2}}={10}^{-12}$

Hence, answer is 1.