Question

Specific conductance of saturated solution of $AgCl$ is $2.68\times {10}^{-4}S{m}^{-1}$ and that of water is $0.86\times {10}^{-4}S{m}^{-1}$ at $298K$. Calculate the solubility product in $mol/{dm}^3$ unit if ${\lambda }_m^o$ for $H{NO}_3,Ag{NO}_3$ and $HCl$ are $421\times {10}^{-4}S{m}^2$ ${mol}^{-1}$,$144\times {10}^{-4}S{m}^2$ ${mol}^{-1}$,$426\times {10}^{-4}S{m}^2$ ${mol}^{-1}$ respectively:

• A. $1.74\times {10}^{-10}$
• B. $1.32\times {10}^{-5}$
• C. $1.32\times {10}^{-2}$
• D. $1.74\times {10}^{-4}$

Answer 1

The correct option is D $1.74\times {10}^{-4}$

Here answer is option D

$K_{sol}=K_{Agcl}+K_{water}$

$K_{Agcl}=K_{sol}-K_{water}$

$=(2.68-0.86)\times {10}^{-4}S{m}^{-1}$

$A_g{NO}_3+HCl\to Agcl+{HNO}_3$

By Kohlrauchs law

${\lambda }_m^{0}AgCl={\lambda }_m^0{AgNO}_3+{\lambda }_m^{0}HCl-{\lambda }_m^0{HNO}_3$

$=(144+426-421)\times {10}^{-4}S{m}^2{mol}^{-1}$

$=149{\times 10}^{-4}S{m}^2{mol}^{-1}$

${\lambda }_m=\frac{K}{C}$

$C=\frac{K}{{\lambda }_m^0}$

$=\frac{1.82\times {10}^{-4}}{149{\times 10}^{-4}}$

$=1.22\times {10}^{-2}mol{m}^{-3}$

$=1.22\times {10}^{-5}mold{m}^{-3}$

solubility of $AgCl$ =($1.22\times {10}^{-5}$)($143.5$g/mol$)

=$1.75\times {10}^{-4}$