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 \( \Lambda{ }_{\mathrm{m}}^{\circ}(A B)=140 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \)

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
\[
\begin{array}{l}
A B \rightarrow A^{+}+B^{-} \\
K_{s p}=\left[A^{+}\right]\left[B^{-}\right]=S^{2} \\
K_{s p}\left[A^{+}\right]\left[B^{-}\right]=S^{2} \\
K_{s p}=S^{2} \\
=\left(1.32 \times 10^{-6}\right)^{2} \\
=1.74 \times 10^{-12}
\end{array}
\]