Question

The solubility of $\text{PbS}{{\text{O}}_{\text{4}}}$ in water is $0.038$ $\text{g }{{\text{L}}^{\text{-1}}}$ at $\text{25}{{\text{ }}^{\text{0}}}\text{C}$ . Calculate its solubility product at the same temperature.

Answer 1

The solubility of a substance is defined as the ability of the substance to get dissolved in a particular substance. The temperature of the solution is an important parameter that determines the solubility of the solute. We shall calculate the molarity of the compound and use it to calculate the solubility product.

Complete step by step solution:
Let us first determine the number of moles of the solute (lead sulphate) that dissolves in the particular solvent (water).
The molecular weight of lead sulphate = 303 $\text{g mo}{{\text{l}}^{\text{-1}}}$
Therefore the molarity of the solution, which is the number of moles of the solute dissolved per litre of the solution is = $\dfrac{0.038}{303}=1.254\times {{10}^{-4}}$ $\text{mol }{{\text{L}}^{\text{-1}}}$
At equilibrium, the concentration of the species after dissociation are as follows:
$\text{PbS}{{\text{O}}_{\text{4}}}\rightleftharpoons \text{P}{{\text{b}}^{\text{2+}}}\text{ + S}{{\text{O}}_{\text{4}}}^{2-}$
Let the solubility of lead sulphate in the solution be equal to “S”. As the number of the lead and the sulphate ions released are equal, therefore, the solubility constant,
${{\text{K}}_{\text{s}}}=\left[ \text{P}{{\text{b}}^{\text{2+}}} \right]\text{ }\left[ \text{S}{{\text{O}}_{\text{4}}}^{2-} \right]=\text{S }\times\text{ S = }{{\text{S}}^{\text{2}}}$
The solubility of the lead sulphate is as found above: $1.254\times {{10}^{-4}}$ $\text{mol }{{\text{L}}^{\text{-1}}}$
Therefore putting the value of the solubility in the above equation is,
${{\text{K}}_{\text{s}}}=\text{ }{{\text{S}}^{\text{2}}}={{\left( 1.254\times {{10}^{-4}} \right)}^{2}}=1.573\times {{10}^{-8}}$
So the Solubility product of Lead Sulphate at the same temperature is $1.573\times {{10}^{-8}}$ .

Note:
The rule that often works in case of solubility is that like polar solvents dissolve ionic or polar covalent compounds and nonpolar solvents dissolve nonpolar solutes or homo-atomic solutes. For most of the solutions, with the rise in temperature, the solubility increases but only for some solutes the solubility of the solutes decreases with the increase in the temperature. Other factors that affect the solubility of the solute in a solution are pressure and the concentration of the solution.