Question

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

Answer 1

`s=5.65xx10^(-5)g/L`

Explanation:

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

`MnCO_3(s)rightleftharpoonsMn^(2+)(aq)+CO_3^(2-)(aq)" " "K_(sp)=2.42xx10^(-11)`

The expression of `K_(sp)` is given by: `K_(sp)=[Mn^(2+)][CO_3^(2-)]`

Let us use ICE table to solve this question:
`" " " " " " "MnCO_3(s)rightleftharpoonsMn^(2+)(aq)+CO_3^(2-)(aq)`
`"Initial" " " " "C_M" " " " " " " "0 M" " " " " "0 M"`
`"Change" " " " -sM " " " " " " " +s M" " " "" +s M"`
`"Equilibrium" " " " " " " " " " " " "s M" " " " " " sM`

Therefore, the equilibrium concentrations are:
`[Mn^(2+)]=sM`
`[CO_3^(2-)]=sM`

Therefore, `K_(sp)=sxxs=s^2=>s=sqrt(K_(sp))=sqrt(2.42xx10^(-11))`

`=>s=4.92xx10^(-6)M`

Therefore, the solubility of Manganese(II) carbonate is `s=4.92xx10^(-6)M`

To convert it to `g/L` using the molar mass of manganese(II) carbonate `114.9469g/"mol"`:

`s=(4.92xx10^(-6)cancel(mol))/(1L)xx(114.9469g)/(1cancel(mol))=5.65xx10^(-5)g/L`

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