Question

What is the minimum concentration of $S{O}_4^{2-}$ required to precipitate $BaS{O}_4$ in a solution containing $1\times {10}^{-4}$ mole of $B{a}^{2+}(K_{sp}$ for $BaS{O}_4=4\times {10}^{-10})$?

• A. $4\times {10}^{-10}$M
• B. $2\times {10}^{-10}$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

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

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

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

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