Question

What is the minimum concentration of $S{O}_4^{2-}$ required to precipitate $BaS{O}_4$ in a solution containing $1.0\times {10}^{-4}$ 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 $4\times {10}^{-6}M$

$BaS{O}_4\rightleftharpoons B{a}^{2+}+S{O}_4^{2-}$

$K_{sp}=S^2\Rightarrow S=\sqrt{K_{sp}};K_{sp}=\left[B{a}^{2+}\right]\times \left[S{O}_4^{2-}\right]$

$4\times {10}^{-10}=[1\times {10}^{-4}]\times \left[S{O}_4^{2-}\right]$

$\left[S{O}_4^{2-}\right]=\frac{4\times {10}^{-10}}{1\times {10}^{-4}}=4\times {10}^{-6}$