Question

At 298 K, the conductivity of a saturated solution of AgCl in water is $2.6\times {10}^{-6}Sc{m}^{-1}$. Its solubility product at 298 K (given : ${\lambda }^{\infty }\left(A{g}^{+}\right)=63.0Sc{m}^{2}mo{l}^{-1},{\lambda }^{\infty }\left(C{l}^{-}\right)=67.0Sc{m}^{2}mo{l}^{-1})$

• A. $2.0\times {10}^{-5}{M}^2$
• B. $4.0\times {10}^{10}{M}^2$
• C. $4.0\times {10}^{-16}{M}^2$
• D. $2\times {10}^{-8}{M}^2$

Answer 1

The correct option is B

$4.0\times {10}^{10}{M}^2$

$SolubilityS=\frac{1000k}{{\lambda }_{AgCl}^{\circ }}=\frac{1000\times 2.6\times {10}^{-6}}{{\lambda }_A^{\circ }g+{\lambda }_{C{l}^{-}}^{\circ }}=\frac{2.6\times {10}^{-3}}{63+67}=2\times {10}^{-5}mol{L}^{-1};K_{sp}=S^2$