Question

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

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

Answer 1

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

Alc to the problem

$\begin{array}{cc}{Na}_2{CO}_3⟶& 2{Na}^{+}+{CO}_3^{2-}\\ \left[{cO}_3^{2-}\right]& =1\times {10}^{-4}M\end{array}$

For, ${BaCO}_3\rightleftharpoons \left[{Ba}^{2+}\right]+\left[{CO}_3^{2-}\right]$

${KG}_p=\left[{Ba}^{2+}\right]\cdot \left[{CO}_3^{2-}\right]$

$\begin{array}{cc}\Rightarrow 5.1\times {10}^{-9}& =\left[B{a}^{2+}\right]\cdot [1\times {10}^{-4}]\\ & [\begin{array}{c}\left[B{a}^{2+}\right]=5.1\times {10}^{-9+4}\\ \left[B{a}^{2+}\right]=5.1\times {10}^{-5}M\end{array}\end{array}$

$\text{(c) is correct option}$