Question

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

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

Answer 1

Answer: (C)

Final concentration of $\left[S{O}_4^{--}\right]=\frac{[50\times 1]}{[500]}=0.1M$

$K_{sp}$ of $BaS{O}_4,$

$\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]=1\times 10^{-10}$

$\left[B{a}^{2+}\right][0.1]=\frac{10^{-10}}{0.1}=10^{-9}M$

Concentration of $B{a}^{2+}$ in final solution $=10^{-9}M$

Concentration of $B{a}^{2+}$ in the original solution.

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

$M_1(500-50)=10^{-9}(500)$

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