Question

The specific conductivity of a saturated solution of $AgCl$ is $3.40\times {10}^{-6}{\text{ohm}}^{-1}{\text{cm}}^{-1}$ at ${25}^{\circ }C$. If ${\lambda }_{Ag+}=62.3{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$ and ${\lambda }_{C{l}^{-}}=67.7{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$, the solubility of $AgCl$ at ${25}^{\circ }C$ is:
[1 mark]

• A. $2.6\times {10}^{-5}\text{M}$
• B. $4.5\times {10}^{-3}\text{M}$
• C. $3.6\times {10}^{-5}\text{M}$
• D. $3.6\times {10}^{-3}\text{M}$

Answer 1

The correct option is A $2.6\times {10}^{-5}\text{M}$

${\lambda }_{A{g}^{+}}=62.3{\text{S cm}}^2{\text{mol}}^{-1},{\lambda }_{C{l}^{-}}=67.7{\text{S cm}}^2{\text{mol}}^{-1},{\kappa }_{AgCl}=3.4\times {10}^{-6}{\text{S cm}}^{-1}{\lambda }_{AgCl}={\lambda }_{A{g}^{+}}+{\lambda }_{C{l}^{-}}{\lambda }_{AgCl}^{\infty }=62.3+67.7=130{\text{S cm}}^2{\text{mol}}^{-1}\text{By using formula}S=\frac{\kappa \times 1000}{\lambda }S=\frac{3.4\times {10}^{-6}\times 1000}{130}=2.6\times {10}^{-5}\text{M}$