Question

Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-4}$ M $Na_2CO_3$ 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. $7.1 \times 10^{-8}\,M$
• C. $4.1 \times 10^{-5}\, M$
• D. $8.1 \times 10^{-7}\,M$

Answer 1

Answer: (A)

Given $\quad Na_{2}CO_{3} = 1.0 \times 10^{-4}M$
$\therefore \left[CO_{3} ^{--}\right] = 1.0 \times 10^{-4} M$
$i.e.\quad s=1.0\times10^{-4}M$
At equilibrium
$\left[Ba^{++}\right] \left[CO_{3}^{--}\right] = K_{sp}$ of $BaCO_{3}$
$\left[Ba^{++}\right] = \frac{K_{sp}}{ \left[CO_{3}^{--}\right]} = \frac{5.1\times10^{-9}}{1.0\times10^{-4}}$
$= 5.1\times 10^{-5} M$