The conductivity of saturated solution of sparingly soluble salt, , is . The limiting equivalent con — Electrochemistry Chemistry Question
Question
The conductivity of saturated solution of sparingly soluble salt, $Ba_3(PO_4)_2$, is $1.2 \times 10^{-5} \text{ ohm}^{-1} \text{ cm}^{-1}$. The limiting equivalent conductances of $BaCl_2, K_3PO_4 \text{ and } KCl$ are 160, 140 and 100 $\text{ ohm}^{-1}\text{cm}^2\text{eq}^{-1}$, respectively. The $K_{sp}$ of $Ba_3(PO_4)_2$ (in the order of $10^{-25}$) is
💡 Solution & Explanation
Equivalent conductance of $Ba_3(PO_4)_2$ is $\Lambda^\circ_{eq} = \Lambda^\circ_{eq}(BaCl_2) + \Lambda^\circ_{eq}(K_3PO_4) - \Lambda^\circ_{eq}(KCl) = 160 + 140 - 100 = 200 \text{ ohm}^{-1} \text{ cm}^2 \text{ eq}^{-1}$. Total cationic charge $n=6$. Molar conductance $\Lambda^\circ_m = \Lambda^\circ_{eq} \times 6 = 1200 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1}$. Solubility $S = (\kappa \times 1000) / \Lambda^\circ_m = (1.2 \times 10^{-5} \times 1000) / 1200 = 10^{-5}$ M. For $Ba_3(PO_4)_2 \rightarrow 3Ba^{2+} + 2PO_4^{3-}$, $K_{sp} = (3S)^3 (2S)^2 = 108 S^5 = 108 \times (10^{-5})^5 = 108 \times 10^{-25}$. The integer part is 108. Therefore, correct answer is 0108.