Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
0 votes

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)

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

Can you answer this question?

1 Answer

0 votes
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}$$=0.082mol\;dm^{-3}$
Let $C_1$ be concentration of $Ag^+$ ions on AgCl.Then assuming activity coefficient=1,the EMF of cell is
$0.45v=0.0591v\log 0.082mol\;dm^{-3}$
$C_1=2.008\times 10^{-9}mol\;dm^{-3}$
Since at $25^o$C KCl is 85% dissociated
$[Cl^-]=\large\frac{0.1\times 85}{100}=$$0.085mol\;dm^{-3}$
$K_{sp}$ of $AgCl=[Ag^+][Cl^-]$
$\Rightarrow 2.008\times 10^{-9}\times 0.085$
$\Rightarrow 1.7068\times 10^{-10}$
Solubility of AgCl =$\sqrt{solubility\;product\;of\;AgCl}$
$\Rightarrow \sqrt{1.7068\times 10^{-10}}$
$\Rightarrow 1.3\times 10^{-5}$
$\Rightarrow 1.3\times 10^{-5}mol\;dm^{-3}\times 143.5g\;mol^{-1}$
$\Rightarrow 1.875\times 10^{-3}g\;dm^{-3}$
Hence (b) is the correct answer.
answered Dec 12, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App