Respuesta :
Answer : The solubility of this compound in g/L is [tex]565.414\times 10^{-6}g/L[/tex].
Solution : Given,
[tex]K_{sp}=2.42\times 10^{-11}[/tex]
Molar mass of [tex]MnCO_3[/tex] = 114.945g/mole
The balanced equilibrium reaction is,
[tex]MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3[/tex]
At equilibrium s s
The expression for solubility constant is,
[tex]K_{sp}=[Mn^{2+}][CO^{2-}_3][/tex]
Now put the given values in this expression, we get
[tex]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[/tex]
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.
[tex]s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L[/tex]
Therefore, the solubility of this compound in g/L is [tex]565.414\times 10^{-6}g/L[/tex].
