Question

Given that for the reaction, $Ag{\left({NH}_3\right)}_2^{+}\rightleftharpoons {Ag}^{+}+2{NH}_3,K_c=6.2\times {10}^{-8}$ and $K_{sp}$ of $AgCl=1.8\times {10}^{-10}$ at $298K$. The concentration of the complex ion in $1.0M$ aqueous ammonia is:

• A. $0.0539M$
• B. $0.0589M$
• C. $0.0639M$
• D. None of these

Answer 1

Answer: (A)

$0.0539M$

$Ag{\left({NH}_3\right)}_2^{+}\left(aq\right)\rightleftharpoons {Ag}^{+}\left(aq\right)+2{NH}_3\left(aq\right)$

$K_c=\frac{\left[{Ag}^{+}\right]{\left[{NH}_3\right]}^2}{\left[Ag{\left({NH}_3\right)}_2^{+}\right]}=6.2\times {10}^{-8}....\left(i\right)$

$AgCl\left(s\right)\rightleftharpoons {Ag}^{+}\left(aq\right)+{Cl}^{-}\left(aq\right);K_{sp}=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]....\left(ii\right)$

By Eqs. $\left(i\right)$ and $\left(ii\right)$ $\frac{K_c}{K_{sp}}=\frac{{\left[{NH}_3\right]}^2}{\left[Ag{\left({NH}_3\right)}_2^{+}\right]\times \left[{Cl}^{-}\right]}.......\left(iii\right)$

Given, $\left[{NH}_3\right]=1M$;

Also $\left[Ag{\left({NH}_3\right)}_2^{+}\right]=\left[{Cl}^{-}\right]=a$ because ${Ag}^{+}$ obtained from $AgCl$ passes in $\left[Ag{\left({NH}_3\right)}_2^{+}\right]$ state.

Thus, $\frac{K_c}{K_{sp}}=\frac{1}{a\times a}or{a}^2=\frac{K_{sp}}{K_c}$

or $a^2=\frac{1.8\times {10}^{-10}}{6.2\times {10}^{-8}}=0.29\times {10}^{-2}$

$\therefore$ $a=0.539\times {10}^{-1}$

$a=0.0539M$

or $\left[Ag{\left({NH}_3\right)}_2^{+}\right]=0.0539M$