Question

The conductivity of saturated solution of $AgCl$ is found to be $1.86\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$ and that of water is $6\times {10}^{-8}oh{m}^{-1}c{m}^{-1}$. If ${\lambda }^0$ $AgCl$ is 138 $oh{m}^{-1}c{m}^{2}e{q}^{-1}$ the solubility product of $AgCl$ is:

• A. $1.3\times {10}^{-5}$
• B. $1.69\times {10}^{-10}$
• C. $2\times {10}^{-10}$
• D. $2.7\times {10}^{-10}$

Answer 1

Answer: (B)

$1.69\times {10}^{-10}$

Conductivity of distilled water= $6\times {10}^{-8}oh{m}^{-1}c{m}^{-1}$

Conductivity of $AgCl$ solution= $1.86\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$

Thus corrected conductivity of $AgCl=K_{AgCl}-K_{H_{2}O}$

=$(1.86\times {10}^{-6}-6\times {10}^{-8})oh{m}^{-1}c{m}^{-1}$

$R_{AgCl}=1.8\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$

as concentration= $\frac{conductivity}{{\lambda }^o}$

$\Rightarrow C=\frac{1.8\times {10}^{-6}oh{m}^{-1}c{m}^{-1}}{138oh{m}^{-1}c{m}^{2}e{q}^{-1}}$

$\Rightarrow C=1.3\times {10}^{-8}c{m}^{-3}eq$

For $AgCl$ $1eq$= $1$ moles

and $1c{m}^3={10}^{-3}d{m}^3$

or $1c{m}^{-3}={10}^{3}d{m}^{-3}$

$\Rightarrow C=1.3\times {20}^{-5}$ mole $d{m}^{-3}$

$\Rightarrow AgCl\rightleftharpoons [A{g}^{+}{]}_{aq}+C{l}_{aq}^{-}$

$\Rightarrow K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=C^2$

$\Rightarrow K_{sp}=(1.3\times {10}^{-5}{)}^{2}mo{l}^{2}d{m}^{-6}$

$\Rightarrow K_{sp}=1.69\times {10}^{-10}mo{l}^{2}d{m}^{-6}$