Question

The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.

Answer 1

Answer :  The solubility of this compound in g/L is $565.414\times 10^{-6}g/L$.

Solution : Given,

$K_{sp}=2.42\times 10^{-11}$

Molar mass of $MnCO_3$ = 114.945g/mole

The balanced equilibrium reaction is,

                      $MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3$

At equilibrium                         s       s

The expression for solubility constant is,

$K_{sp}=[Mn^{2+}][CO^{2-}_3]$

Now put the given values in this expression, we get

$2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L$

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

$s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L$

Therefore, the solubility of this compound in g/L is $565.414\times 10^{-6}g/L$.