Question

An aqueous solution contains an unknown concentration of $Ba^{2+}$. When $50 \,mL$ of a $1\, M$ solution of $Na_2SO_4$ is added, $BaSO_4$ just begins to precipitate. The final volume is $500 \,mL$. The solubility product of $BaSO_4$ is $1 \times 10^{-10}$. What is the original concentration of $Ba^{2+}$ ?

• 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 $\left[ SO _{4}^{--}\right]=\frac{[50 \times 1]}{[500]}=0.1\, M$

$K _{ sp }$ of $BaSO _{4},$

${\left[ Ba ^{2+}\right]\left[ SO _{4}^{2-}\right]=1 \times 10^{-10}} $

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

Concentration of $Ba ^{2+}$ in final solution $=10^{-9}\, M$

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

$M _{1} \,V _{1}= M _{2}\, V _{2}$

$M _{1}(500-50)=10^{-9}(500)$

$M _{1}=1.11 \times 10^{-9} \,M$