The solubility of LiNa(AlF) is 0.0744 g per 100 ml at 298 K. Calculate the solubility product of the — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Molar mass $= 3(7) + 3(23) + 2[27 + 6(19)] = 21 + 69 + 2(141) = 372 \text{ g/mol}$. Solubility $S = 0.0744 \text{ g/100 ml} = 0.744 \text{ g/L} = 0.744 / 372 = 2 \times 10^{-3} \text{ M}$. The salt dissociates into 3 $\text{Li}^+$, 3 $\text{Na}^+$, and 2 $\text{AlF}_6^{3-}$ ions. $K_{sp} = (3S)^3(3S)^3(2S)^2 = 27S^3 \cdot 27S^3 \cdot 4S^2 = 2916 S^8$. Substituting $S = 2 \times 10^{-3} \text{ M}$, $K_{sp} = 2916 \times (2 \times 10^{-3})^8 = 2916 \times 256 \times 10^{-24} = 746496 \times 10^{-24} \approx 7.46 \times 10^{-19}$.