Question

An aqueous solution contains an unknown concentration of ${Ba}^{2+}$. When $50ml$ of a $1M$ solution of ${Na}_2{SO}_4$ is added, ${BaSO}_4$ just begins to precipitate. The final volume is $500ml$. The solubility product of ${BaSO}_4$ 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

The correct option is A $1.1\times {10}^{-9}M$

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

$1\times 50=M_2\times 500$

$\therefore M_2=0.1M$ (Concentration of ${SO}_4^{2-}$ in ${Ba}^{+2}$ solution)

For precipitation,

$I.P+K_{sp}$

Now, $K_{sp}=\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]$

$\therefore {10}^{-10}=\left[{Ba}^{+2}\right]\times 0.1$

$\therefore \left[{Ba}^{2+}\right]={10}^{-9}M$ in $500ml$ solution

In $400ml$, $M_1\times 450={10}^{-9}\times 500$

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