Question

The Ksp of CaSO4 is 4.93× 10–5. Calculate the solubility (in g/L) of CaSO4(s) in 0.250 M Na2SO4(aq) at 25 °C.

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 \(\mathrm{CaSO}_{4}\). Calculate the solubility of \(\mathrm{CaSO}_{4}\) by using the concentration of common ion present in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{CaSO}_{4}\).

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 \(\mathrm{CaSO}_{4}\) is as follows:

\(\mathrm{CaSO}_{4} \rightarrow \mathrm{Ca}^{2+}+\mathrm{SO}_{4}^{2-}\)

\(S S S\)

\(\mathrm{Ksp}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_{4}^{2-}\right]\)

\(\mathrm{Ksp}=(S)(S)=S^{2}\)

Solubility of a substance is represented by \(\mathrm{S}\). In that expression solubility of \(\mathrm{Ca}^{2+}\) ion and \(\mathrm{SO}_{4}^{2-}\) ion is equal to \(S\). The final expression will be \(S^{2}\).

 

\(\mathrm{Ksp}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_{4}^{2-}\right]\)

Concentration of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is \(0.350 \mathrm{M}\) So,

$$ \left[\mathrm{Na}_{2} \mathrm{SO}_{4}\right]=\left[\mathrm{SO}_{4}^{2-}\right]=0.350 \mathrm{M} $$

Substitute these values.

$$ \begin{array}{l} 4.93 \times 10^{-5}=\left[\mathrm{Ca}^{2+}\right] \times(0.350) \\ {\left[\mathrm{Ca}^{2+}\right]=1.41 \times \times 10^{-4} \mathrm{M}} \end{array} $$

Solubility of is equal to solubility of \(\mathrm{Ca}^{2+}\). Thus, solubility of \(\mathrm{CaSO}_{4}\) in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is \(1.41 \times \times 10^{-4} \mathrm{M}\).

$$ \begin{array}{l} \text { Solubilityof } \mathrm{CaSO}_{4} \text { ing } / \mathrm{mL}=1.41 \times \times 10^{-4}\left(\frac{\mathrm{mol}}{\mathrm{L}}\right) \times 136\left(\frac{\mathrm{g}}{\mathrm{mol}}\right) \\ \quad=0.019 \mathrm{~g} / \mathrm{L} \end{array} $$

Solubility of \(\mathrm{CaSO}_{4}\) in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is \(0.019 \mathrm{~g} / \mathrm{L}\).

Solubility of \(\mathrm{CaSO}_{4}\) and solubility of \(\mathrm{Ca}^{2+}\) is equal to \(S\). Solubility product for \(\mathrm{CaSO}_{4}\) Ksp is \(4.93 \times 10^{-5}\). Molar mass of \(\mathrm{CaSO}_{4}\) is \(136 \mathrm{~g} / \mathrm{mol}\). Convert the solubility of \(\mathrm{CaSO}_{4}\) into \(\mathrm{g} / \mathrm{L}\) by using its molar mass.