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$\begin{aligned} & \text { If } \mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+} ; K_1=1.7 \times 10^7 \\ & \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{AgCl} ; K_2=5.4 \times 10^9\end{aligned}$
Then, for
$\mathrm{AgCl}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}+\mathrm{Cl}^{-}$
equilibrium constant will be
Options:
Then, for
$\mathrm{AgCl}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}+\mathrm{Cl}^{-}$
equilibrium constant will be
Solution:
1696 Upvotes
Verified Answer
The correct answer is:
$0.31 \times 10^{-2}$
For the given reaction,
$\begin{aligned} \mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+} ; K_1=1.7 \times 10^7 \\ \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{AgCl} ; K_2=5.4 \times 10^9\end{aligned}$
Hence for
$\begin{aligned} & \mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & K_2^{\prime}=\frac{1}{K_2}=\frac{1}{5.4 \times 10^9}\end{aligned}$
Now for the reaction,
$\mathrm{AgCl}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}+\mathrm{Cl}^{-}$
$\begin{aligned} K=K_1 \times K_2^{\prime} & =1.7 \times 10^7 \times \frac{1}{5.4 \times 10^9} \\ & =0.31 \times 10^{-2}\end{aligned}$
$\begin{aligned} \mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+} ; K_1=1.7 \times 10^7 \\ \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{AgCl} ; K_2=5.4 \times 10^9\end{aligned}$
Hence for
$\begin{aligned} & \mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & K_2^{\prime}=\frac{1}{K_2}=\frac{1}{5.4 \times 10^9}\end{aligned}$
Now for the reaction,
$\mathrm{AgCl}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}+\mathrm{Cl}^{-}$
$\begin{aligned} K=K_1 \times K_2^{\prime} & =1.7 \times 10^7 \times \frac{1}{5.4 \times 10^9} \\ & =0.31 \times 10^{-2}\end{aligned}$
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