Conductivity of a saturated solution of a sparingly soluble salt $ A B $ at $ 298 K $ is $ 1.85 \times 10^{-5} \mathrm{Sm}^{-1} $. Solubility product of the salt AB at $ 298 \mathrm{~K} $ is Given $ \Lambda{ }_{\mathrm{m}}^{\circ}(A B)=140 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} $

Question

Conductivity of a saturated solution of a sparingly soluble salt $ A B $ at $ 298 K $ is $ 1.85 \times 10^{-5} \mathrm{Sm}^{-1} $. Solubility product of the salt AB at $ 298 \mathrm{~K} $ is Given $ \Lambda{ }_{\mathrm{m}}^{\circ}(A B)=140 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} $, $ 1.32 \times 10^{-12} $, $ 1.74 \times 10^{-12} $, $ 5.7 \times 10^{-12} $, $ 7.5 \times 10^{-12} $

MARKS App · Accepted Answer

Given, $ \Lambda{ }_{\mathrm{m}}^{\circ}=140 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1} $ $ K=1.85 \times 10^{-5} \mathrm{Sm}^{-1} $ Now, molar conductivity of the saturated salt is, $$ \begin{array}{l} \left.\Lambda_{m}{ }^{\circ}=\frac{K}{1000 \times C}=\frac{1.85 \times 10^{-5}}{1000 \times S} \text { [Concentration }(\mathrm{C})=\text { Solubility }(S)\right] \ 140 \times 10^{-4}=\frac{1.85 \times 10^{-5}}{1000 \times S} \ S=\frac{1.85 \times 10^{-5}}{1000 \times 140 \times 10^{-4}} \ S=1.32 \times 10^{-6} \end{array} $$ For the saturated salt $$ \begin{array}{l} A B \rightarrow A^{+}+B^{-} \ K_{s p}=\left[A^{+}\right]\left[B^{-}\right]=S^{2} \ K_{s p}\left[A^{+}\right]\left[B^{-}\right]=S^{2} \ K_{s p}=S^{2} \ =\left(1.32 \times 10^{-6}\right)^{2} \ =1.74 \times 10^{-12} \end{array} $$

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