Question

Solid $Ba{\left({NO}_3\right)}_2$ is gradually dissolved in a $1.0\times {10}^{-4}M{Na}_2{CO}_3$ solution. At which concentration of ${Ba}^{2+}$, precipitate of ${BaCO}_3$ begins to form?$(K_{sp}$ for ${BaCO}_3=5.1\times {10}^{-9})$

• (A) $5.1\times {10}^{-5}M$
• (B) $7.1\times {10}^{-5}M$
• (C) $4.1\times {10}^{-5}M$
• (D) $8.1\times {10}^{-7}M$

Answer 1

The Correct Option is A

Explanation:
Given that: Concentration of ${Na}_2{CO}_3=1.0\times {10}^{-4}M$$\therefore \left[{CO}_3^{-}\right]=1.0\times {10}^{-4}M$i.e., $s=1.0\times {10}^{-4}M$At equilibrium:$\left[{Ba}^{++}\right]\left[{CO}_3^{--}\right]=K_{sp}$ of ${BaCO}_3$$\left[{Ba}^{++}\right]=\frac{K_{sp}}{\left[{CO}_3^{--}\right]}$$=\frac{5.1\times {10}^{-9}}{1.0\times {10}^{-4}}$$=5.1\times {10}^{-5}M$Hence, the correct option is (A).