Question

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

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

Answer 1

Answer: (A)

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

$BaS{O}_4\rightleftharpoons B{a}^{+2}+S{O}_4^{-2}$Initially $S{O}_4^{-2}$ is

Present in 50ml of 1M solution. Later it is present in 500ml solution having molarity $M_2$.

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

$1\times 50=M_2\times 500$

$M_2=\frac{1}{10}M$

$M_2$ is the concentration of $S{O}_4^{-2}$. So, $\left[S{O}_4^{-2}\right]=\frac{1}{10}$

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

$\left[B{a}^{+2}\right]=\frac{{10}^{-10}}{{10}^{-1}}={10}^{-9}$

This is the final concentration of $B{a}^{+2}$ in 500ml solution.

Suppose initial concentration of $B{a}^{+2}$ in 450ml solution was $x$. Applying $M_1{V}_1=M_2{V}_2$

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

$x=\frac{{10}^{-9}\times 500}{450}=1.1\times {10}^{-9}M$

So, the correct option is $A$