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

1 Answer
Dec 10, 2015

Answer:

#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.