Question

The Ksp of CaSO₄ is 4.93× 10^-5 M2. Calculate the solubility (in g/L) of CaSO₄(s) in 0.350 M Na₂SO₄(aq) at 25 ℃.

Answer 1

Concepts and reason

Solubility product a substance is the mathematical product of its dissolved ion concentrations raised to the power of their stoichiometric coefficients.

First write the expression of solubility product for the substance ${\rm{CaS}}{{\rm{O}}_4}$ . Calculate the solubility of ${\rm{CaS}}{{\rm{O}}_4}$ by using the concentration of common ion present in ${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$ and ${\rm{CaS}}{{\rm{O}}_4}$ .

Fundamentals

‎The solubility product is a heterogeneous equilibrium constant, and it depends on temperature. The larger the solubility product of a substance, the higher is its solubility.

Expression of solubility product for ${\rm{CaS}}{{\rm{O}}_4}$ is as follows:

$\begin{array}{l}\\{\rm{CaS}}{{\rm{O}}_{\rm{4}}} \to {\rm{ C}}{{\rm{a}}^{2 + }}{\rm{ + S}}{{\rm{O}}_4}^{2 - }\\\\S{\rm{ }}S{\rm{ }}S\\\\{\rm{Ksp }} = \left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]{\rm{ }}\left[ {{\rm{S}}{{\rm{O}}_4}^{2 - }} \right]\\\\{\rm{Ksp }} = \left( S \right)\left( S \right) = {\rm{ }}{S^2}\\\end{array}$

${\rm{Ksp }} = \left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]{\rm{ }}\left[ {{\rm{S}}{{\rm{O}}_4}^{2 - }} \right]$

Concentration of ${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$ is 0.350 M.

So,

$\left[ {{\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}} \right] = \left[ {{\rm{S}}{{\rm{O}}_4}^{2 - }} \right] = 0.350{\rm{ M}}$

Substitute these values.

$\begin{array}{l}\\{\rm{4}}{\rm{.93}} \times {\rm{1}}{{\rm{0}}^{ - 5}}{\rm{ }} = \left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]{\rm{ }} \times \left( {0.350} \right)\\\\\left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]{\rm{ }}\, = 1.41 \times \times {\rm{1}}{{\rm{0}}^{ - 4}}{\rm{ M}}\\\end{array}$

Solubility of is equal to solubility of ${\rm{C}}{{\rm{a}}^{2 + }}$ .

Thus, solubility of ${\rm{CaS}}{{\rm{O}}_4}$ in ${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$ is $1.41 \times \times {\rm{1}}{{\rm{0}}^{ - 4}}{\rm{ M}}$ .

$\begin{array}{c}\\{\rm{Solubility of CaS}}{{\rm{O}}_4}{\rm{in g/mL = }}1.41 \times \times {\rm{1}}{{\rm{0}}^{ - 4}}{\rm{ }}\left( {\frac{{{\rm{mol}}}}{{\rm{L}}}} \right) \times 136\left( {\frac{{\rm{g}}}{{{\rm{mol}}}}} \right)\\\\ = 0.019{\rm{ g/L}}\\\end{array}$

Ans:

Solubility of ${\rm{CaS}}{{\rm{O}}_4}$ in ${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$ is $0.019{\rm{ g/L}}$ .