Solid $\mathrm{Ba}\left(\mathrm{NO}_{3}ight)_{2}$ is gradually dissolved $\mathrm{in}$ a $1.0 \times 10^{-4}…

Solid $\mathrm{Ba}\left(\mathrm{NO}_{3}ight)_{2}$ is gradually dissolved $\mathrm{in}$ a $1.0 \times 10^{-4} \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ solution. At which
concentration of $\mathrm{Ba}^{2+}$, precipitate of $\mathrm{BaCO}_{3}$ begins to form $?\left(K_{s p}ight.$ for $\left.\mathrm{BaCO}_{3}=5.1 \times 10^{-9}ight)$
  1. $5.1 \times 10^{-5} \mathrm{M}$
  2. $7.1 \times 10^{-8} \mathrm{M}$
  3. $4.1 \times 10^{-5} \mathrm{M}$
  4. $8.1 \times 10^{-7} \mathrm{M}$

Solution

Given $\quad \mathrm{Na}_{2} \mathrm{CO}_{3}=1.0 \times 10^{-4} \mathrm{M}$
$\therefore\left[\mathrm{CO}_{3}^{2-}ight]=1.0 \times 10^{-4} \mathrm{M}$
i.e. $\mathrm{s}=1.0 \times 10^{-4} \mathrm{M}$
At equilibrium $\begin{aligned}\left[\mathrm{Ba}^{2+}ight] &\left[\mathrm{CO}_{3}^{2-}ight]=\mathrm{K}_{\mathrm{sp}} \text { of } \mathrm{BaCO}_{3} \\\left[\mathrm{Ba}^{2+}ight] &=\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{CO}_{3}^{2-}ight]}=\frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} \\ &=5.1 \times 10^{-5} \mathrm{M} \end{aligned}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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