Question

A saturated solution of $BaSO _{4}$ at $25^{\circ} C$ is $4 \times 10^{-5} M$. The solubility of $BaSO _{4}$ in $0.1 \,M\, Na _{2} SO _{4}$ at this temperature will be

• A. $1.6 \times 10^{-9} M$
• B. $1.6 \times 10^{-8} M$
• C. $4 \times 10^{-6} M$
• D. $4 \times 10^{-4} M$

Answer 1

Answer: (D)

$BaSO_4 \rightleftharpoons \underset{s}{Ba^{2+}} + \underset{s + 0.1}{SO_4^{2-}}$
$Na_2SO_4 \rightarrow \underset{0.2}{2Na^+} + \underset{0.1}{SO_4^{2-}}$
$K_{sp} = [Ba^{+2}][SO_4^{-2}]$
$\therefore 4 \times 10^{-5} = s \times (s + 0.1)$
or $ 4 \times 10^{-5} \approx s\times 0.1$
$\therefore s = 4 \times 10^{-4}$
as $s < < < 0.1$
$\therefore s + 0.1 \approx 0.1$