# The Ksp of manganese(II) carbonate, MnCO_3, is 2.42 * 10^-11. What is the solubility of this compound in g/L?

Dec 10, 2015

$s = 5.65 \times {10}^{- 5} \frac{g}{L}$

#### Explanation:

Manganese(II) carbonate dissociate in water according to the following equation:

$M n C {O}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s M {n}^{2 +} \left(a q\right) + C {O}_{3}^{2 -} \left(a q\right) \text{ " } {K}_{s p} = 2.42 \times {10}^{- 11}$

The expression of ${K}_{s p}$ is given by: ${K}_{s p} = \left[M {n}^{2 +}\right] \left[C {O}_{3}^{2 -}\right]$

Let us use ICE table to solve this question:
$\text{ " " " " " } M n C {O}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s M {n}^{2 +} \left(a q\right) + C {O}_{3}^{2 -} \left(a q\right)$
$\text{Initial" " " " "C_M" " " " " " " "0 M" " " " " "0 M}$
$\text{Change" " " " -sM " " " " " " " +s M" " " "" +s M}$
$\text{Equilibrium" " " " " " " " " " " " "s M" " " " " } s M$

Therefore, the equilibrium concentrations are:
$\left[M {n}^{2 +}\right] = s M$
$\left[C {O}_{3}^{2 -}\right] = s M$

Therefore, ${K}_{s p} = s \times s = {s}^{2} \implies s = \sqrt{{K}_{s p}} = \sqrt{2.42 \times {10}^{- 11}}$

$\implies s = 4.92 \times {10}^{- 6} M$

Therefore, the solubility of Manganese(II) carbonate is $s = 4.92 \times {10}^{- 6} M$

To convert it to $\frac{g}{L}$ using the molar mass of manganese(II) carbonate $114.9469 \frac{g}{\text{mol}}$:

$s = \frac{4.92 \times {10}^{- 6} \cancel{m o l}}{1 L} \times \frac{114.9469 g}{1 \cancel{m o l}} = 5.65 \times {10}^{- 5} \frac{g}{L}$

Here is a video that will be of great help on solubility equilibria:
Solubility Equilibria | Solubility Product.