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 concentration of $B{a}^{2+}$ ions in the solution in mol $L^{-1}$ is:

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

Answer 1

The correct option is C $1\times {10}^{-1}$

As $pH$ is $10$, so $pOH=14-10=4$

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

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

$K_{sp}=\left[B{a}^{2+}\right][O{H}^{-}{]}^2$

As given, ${10}^{-9}=\left[B{a}^{2+}\right]{10}^{-8}$

$\left[B{a}^{2+}\right]={10}^{-1}$