Question

An aqueous solution contains an unknown concentration of ${\mathrm{Ba}}^{2+}$ . When $50\mathrm{mL}$ of a $1\mathrm{M}$ solution of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is added, ${\mathrm{BaSO}}_4$ just begins to precipitate. The final volume is $500\mathrm{mL}$. The solubility product of ${\mathrm{BaSO}}_4$ is $1\times {10}^{-10}$ . What is the original concentration of ${\mathrm{Ba}}^{2+}$ ?

• A. $2\times {10}^{-9}\mathrm{M}$
• B. $1.1\times {10}^{-9}\mathrm{M}$
• C. $1.0\times {10}^{-9}\mathrm{M}$
• D. $5\times {10}^{-9}\mathrm{M}$

Answer 1

The correct option is B

$1.1\times {10}^{-9}\mathrm{M}$

Explanation for the correct option:

The concentration of ${\mathrm{SO}}_4^{2-}$in ${\mathrm{BaSO}}_4$ Solution,

$M_1{V}_1=M_2{V}_2$

$M_2=\frac{M_1{V}_1}{V_2}$

$M_2=\frac{1\times 50}{500}$=$\frac{1}{10}$.

For precipitation,

$\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]=K_{sp}$

$$\begin{gathered} \left[B{a}^{2+}\right]=\frac{K_{sp}}{0.1} \\ \left[B{a}^{2+}\right]=\frac{{10}^{-10}}{0.1} \\ \left[B{a}^{2+}\right]={10}^{-9} \end{gathered}$$ in $500\mathrm{mL}$ solution

Now for the calculation of $\left[B{a}^{2+}\right]$ in the original solution,

$M_1\times 450={10}^{-9}\times 500$

$M_1=1.11\times {10}^{-9}$

Hence the concentration of ${\mathrm{Ba}}^{2+}$ in the original solution $=$$1.11\times {10}^{-9}M$

So, option B is correct.