Question

What is the minimum concentration of $S{O}_4^{2-}$ required to precipitate $BaS{O}_4$ in a solution contining $1.0\times {10}^{-4}$ mole of $B{a}^{2+}?$ ($K_{sp}$ for $BaS{O}_4=4\times {10}^{-10}$)

• A. $4\times {10}^{-10}$ M
• B. $2\times {10}^{-7}M$
• C. $4\times {10}^{-6}M$
• D. $2\times {10}^{-3}M$

Answer 1

The correct option is C $4\times {10}^{-6}M$

$K_{ip}=\text{ionic product}$
$K_{sp}=\text{solubility product}$
$K_{ip}>K_{sp}$
$BaS{O}_4\left(s\right)\rightleftharpoons \left[B{a}^{2+}\right(aq\left)\right]\left[S{O}_4^{2-}\right(aq\left)\right]K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]\left[S{O}_4^{-2}\right]=\frac{K_{sp}}{\left[B{a}^{2+}\right]}=\frac{4\times {10}^{-10}}{1\times {10}^{-4}}\therefore \left[S{O}_4^{2-}\right]=4\times {10}^{-6}M$