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# The EMF of cell $Ag,AgCl$ in 0.1M $KCl\parallel sat\;NH_4NO_3\parallel 0.1M\;AgNO_3,Ag$ is $0.45\;volt$ at $25^oC$.Calculate solubility product & solubility of $AgCl(0.1M\; KCl$ is $85\%$dissociated & $0.1M$ $AgNO_3$ is $82\%$ dissociated)

$\begin{array}{1 1}(a)\;1.8\times 10^{-10},1.7\times 10^{-3}\\(b)\;1.7\times 10^{-10},1.8\times 10^{-3}\\(c)\;1.9\times 10^{-3},1.6\times 10^{-9}\\(d)\;1.2\times 10^{-9},1.6\times 10^{-3}\end{array}$

Since at $25^o$C 0.1M $AgNO_3$ is 82% dissociated hence $Ag^+$ ion concentration $(C_2)$ on $AgNO_3$ side =$\large\frac{0.1\times 82}{100}$$=0.082mol\;dm^{-3} Let C_1 be concentration of Ag^+ ions on AgCl.Then assuming activity coefficient=1,the EMF of cell is E=\large\frac{0.059}{\eta}$$\log\large\frac{C_1}{C_2}$
$E=0.45v$
$\eta=1$
$C_2=0.082mol\;dm^{-3}$
$0.45v=0.0591v\log 0.082mol\;dm^{-3}$
$C_1=2.008\times 10^{-9}mol\;dm^{-3}$
Since at $25^o$C KCl is 85% dissociated