Question

The $pH$ of an aqueous solution of $Ba(OH{)}_2$ is $10$. If the $K_{sp}$ of $Ba(OH{)}_2$ is $1\times {10}^{-9}$, then the conc. of $B{a}^{+2}$ ions is the solution in mol $L^{-1}$ is:

• A. $1\times {10}^{-2}$
• B. $1\times {10}^{-4}$
• C. $1\times {10}^{-1}$
• D. $5\times {10}^{-5}$

Answer 1

The correct option is D $5\times {10}^{-5}$

$Ba(OH{)}_2\leftrightharpoons B{a}^{2+}+2O{H}^{-}$

Eqb. $1-x$ $x$ $2x$

$Ksp\left(Ba\right(OH{)}_2)=[B{a}^{2+}\left]\right[O{H}^{-}{]}^2$

$pH\left[Ba\right(OH{)}_2]=10$

$⟹pOH=4$ $\left[O{H}^{-}\right]={10}^{-4}$

$2[OH{]}^{-}⟹2x={10}^{-4}$

$\left[B{a}^{2+}\right]⟹x=\frac{{10}^{-4}}{2}=5\times {10}^{-5}$

$B{a}^{2+}$ concentration is $5\times {10}^{-5}$ $mol/L$