Chapter 17.3, Problem 31SSC

Interpretation Introduction

Interpretation:

The solubility of magnesium carbonate is to be determined when solubility product is given at a particular temperature.

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

The solubility product is given as

${\text{K}}_{\text{SP}}\text{= }\left[{\text{Mg}}^{\text{+2}}\right]\text{×}\left[{\text{CO}}_{\text{3}}{}^{\text{-2}}\right]$

Answer to Problem 31SSC

The solubility of magnesium carbonate is $\text{5}{\text{.1×10}}^{\text{-5}}\text{mol/L}$.

Explanation of Solution

Given information : The solubility product of magnesium carbonate is $\text{2}{\text{.6×10}}^{\text{-9}}$.

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

${\text{MgCO}}_{\text{3}}\underset{}{\rightleftharpoons }{\text{Mg}}^{\text{2+}}{\text{+CO}}_{\text{3}}{}^{\text{-2}}$

$\text{I = 0 0 }$ (Initial concentration)

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

$\text{ E= x x }$ (Equilibrium concentration)

The solubility product is given as

${\text{K}}_{\text{sp}}\text{= }\left[{\text{Mg}}^{\text{2+}}\right]\text{×}\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]$

$\text{= x×x}$

${\text{= x}}^{\text{2}}$

$\sqrt{\text{Ksp}}{\text{ = x}}^{}$

Or, $\text{x = }\sqrt{\text{2}{\text{.6 × 10}}^{\text{-9}}}$

$\text{= 5}{\text{.1×10}}^{\text{-5}}\text{mol/L}$

Conclusion

The solubility of lead chromate is $\text{5}{\text{.1×10}}^{\text{-5}}\text{mol/L}$.