Question

Solid $Ba{\left({NO}_3\right)}_2$ is gradually dissolved in a $1\times {10}^{-4}M$ ${Na}_2{CO}_3$ solution. At what concentration of ${Ba}^{2+}$ will precipitate begin to form? ($K_{sp}$ for $Ba{CO}_3=5.1\times {10}^{-9}$)

• A. $4.1\times {10}^{-5}M$
• B. $5.1\times {10}^{-5}M$
• C. $8.1\times {10}^{-8}M$
• D. $8.1\times {10}^{-7}M$

Answer 1

The correct option is B $5.1\times {10}^{-5}M$

$K_{sp}=\left[B{a}^{2+}\right]\left[C{O}_3^{2-}\right]=5.1\times {10}^{-9}$

${10}^{-4}MN{a}_{2}C{O}_3$ means ${10}^{-4}M$ of $C{O}_3^{2-}$ are present.

So $K_{sp}=\left[B{a}^{2+}\right]\left[{10}^{-4}\right]=5.1\times {10}^{-9}$

$\left[B{a}^{2+}\right]=5.1\times {10}^{-5}$ so precipitation will begin at $\left[B{a}^{2+}\right]=5.1\times {10}^{-5}$.