Question

A very dilute saturated solution of a sparingly soluble salt $A_3$$B_4$ has a vapour pressure of 20 mm of Hg at temperature T, while pure water exerts a pressure of 20.0126 mm Hg at the same temperature. Calculate the solubility product constant of $A_3$$B_4$ at the same temperature.

• A. $2.7\times {10}^{-13}$
• B. $4.45\times {10}^{-7}$
• C. $6.8\times {10}^{-12}$
• D. $13.6\times {10}^{-7}$

Answer 1

The correct option is B $4.45\times {10}^{-7}$

For very dilute solution, the relative lowering in the vapour pressure is equal to the mole fraction of solute

$\frac{p^0-p}{p^0}=\frac{n_B}{n_A}$

For 1 L of very dilute solution, the number of moles of solution is equal to $\frac{1000}{18}=55.56$

$\frac{20.0126-20}{20}=\frac{n_B}{55.56}{n}_B=0.035$

For 1 L of very dilute solution, the number of moles of solute is equal to its molarity (or solubility S). Thus $S=0.035$

The solubility product constant $K_{sp}=\left[A^{4+}{]}^3\right[B^{3+}{]}^4=\left(3S{)}^3\right(4S{)}^4=6912S^7$

$K_{sp}=6912\times {0.035}^7=4.45\times {10}^{-7}$