[Four-digit Integer] The conductivity of saturated solution of sparingly soluble salt, Ba3(PO4)2, is — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Step 1: Find equivalent conductance at infinite dilution.} Using Kohlrausch's law (on equivalent conductance basis): \[ \lambda^\circ_{eq}(\text{Ba}_3(\text{PO}_4)_2) = \lambda^\circ_{eq}(\text{BaCl}_2) + \lambda^\circ_{eq}(\text{K}_3\text{PO}_4) - \lambda^\circ_{eq}(\text{KCl}) \] \[ = 160 + 140 - 100 = 200\ \text{ohm}^{-1}\text{cm}^2\text{eq}^{-1} \] \textbf{Step 2: Find equivalent concentration.} \[ C_{eq} = \frac{\kappa}{\lambda^\circ_{eq}} = \frac{1.2 \times 10^{-5}}{200} = 6.0 \times 10^{-8}\ \text{eq cm}^{-3} = 6.0 \times 10^{-5}\ \text{eq L}^{-1} \] \textbf{Step 3: Find molar solubility S.} For Ba\textsubscript{3}(PO\textsubscript{4})\textsubscript{2}, the $n$-factor = 6 (3 Ba$^{2+}$ × 2 + 2 PO$_4^{3-}$ × 3 = 12, but 12/2 = 6 per formula unit): \[ S = \frac{C_{eq}}{6} = \frac{6.0 \times 10^{-5}}{6} = 1.0 \times 10^{-5}\ \text{mol L}^{-1} \] \textbf{Step 4: Calculate Ksp.} \[ \text{Ba}_3(\text{PO}_4)_2 \rightarrow 3\text{Ba}^{2+} + 2\text{PO}_4^{3-} \] \[ [\text{Ba}^{2+}] = 3S = 3 \times 10^{-5}\ \text{M}; \quad [\text{PO}_4^{3-}] = 2S = 2 \times 10^{-5}\ \text{M} \] \[ K_{sp} = [\text{Ba}^{2+}]^3[\text{PO}_4^{3-}]^2 = (3 \times 10^{-5})^3 \times (2 \times 10^{-5})^2 \] \[ = 27 \times 10^{-15} \times 4 \times 10^{-10} = 108 \times 10^{-25} \] In units of $10^{-25}$: answer $= \boxed{108}$ → four-digit: \textbf{0108}