Chapter 17, Problem 6STP

Interpretation Introduction

Interpretation ;

The molar solubility of ${\text{MnCO}}_{\text{3}}$ is to be calculated.

Concept introduction:

Molar solubility is the number of moles of a substance that can be dissolved per liter of solution after the solution has achieved saturation. It is measured in mol/L.

Answer to Problem 6STP

Correct answer: Option A − he molar solubility is $\text{4}{\text{.73×10}}^{\text{-6}}$.

Explanation of Solution

Reason for correct option:

Given information :

Equilibrium concentration of manganese and carbonate ion is $\text{1}{\text{.00×10}}^{\text{-2}}$ mol/L and $\text{2}{\text{.24×10}}^{\text{-9}}$ mol/L respectively.

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

${\text{MnCO}}_{\text{3}}{}_{}\underset{}{\rightleftharpoons }{\text{Mn}}^{\text{2+}}{\text{+CO}}_{\text{3}}{}^{\text{2-}}$

I $\text{= 0}\text{.0100 0}\text{.0000 }$ (initial concentration)

C $\text{= +x +x }$ (change in concentration or change in amount from initial state to equilibrium)

E $\text{= 1}{\text{.00×10}}^{\text{-2}}\text{ 2}{\text{.24×10}}^{\text{-9}}$ (equilibrium concentration)

The solubility product is given as

$\text{KSP = }\left[{\text{Mn}}^{\text{2+}}\right]\text{ × }\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]$

$\begin{array}{l}\text{= ( 1}{\text{.00×10}}^{\text{-2)}}\text{ × ( 2}{\text{.24×10}}^{\text{-9)}}\hfill \\ \text{ = 2}{\text{.24 ×10}}^{\text{-11}}\hfill \end{array}$

Now molar solubility is given as

$\begin{array}{l}{\text{K}}_{\text{sp}}\text{=x×x}\\ {\text{K}}_{\text{sp}}{\text{=x}}^{\text{2}}\\ \sqrt{{\text{K}}_{\text{sp}}}\text{=x}\\ \sqrt{\text{2}{\text{.24×10}}^{\text{-11}}}\text{=x}\\ \text{or}\text{x=}\sqrt{\text{22}{\text{.4×10}}^{\text{-12}}}\\ \text{x=4}{\text{.73×10}}^{\text{-6}}\end{array}$

The molar solubility is

$\text{4}{\text{.73×10}}^{\text{-6}}\text{M}$

Conclusion

Reasons for incorrect options:

B. This one is not a calculated value

C. This value is not correct

D. This value is not correct