Question

An aqueous solution contains an unknown concentration of $B{a}^{2+}$. When 50 mL of 1 M solution of $N{a}_{2}S{O}_4$ is added, $BaS{O}_4$ just begins to precipitate. The final volume is 500 mL. The solubility product of $BaS{O}_4$ is $1\times {10}^{-10}$. What is the original concentration of $B{a}^{2+}$?

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

Answer 1

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

$B{a}^{2+}+N{a}_{2}S{O}_4\to BaS{O}_4+2N{a}^{+}$
Given,
the solubility product of $BaS{O}_4$ is $1\times {10}^{10}$

$1\times {10}^{-10}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$....(1)
Let the concentation of $S{O}_4^{2-}$ in 500 mL solution be M. Then,

$500\times M=50\times 1$
$\Rightarrow M=0.1$

$\left(1\right)\Rightarrow \left[B{a}^{2+}\right]=\frac{1\times {10}^{-10}}{0.1}={10}^{-9}$ (in 500 mL).

Now to calculate the concentration of $\left[B{a}^{2+}\right]$ in the original solution, i.e., in 450 mL,
$450\times M=500\times {10}^{-9}$
$\Rightarrow M=1.1\times {10}^{-9}$