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Observe the following equations
$\begin{aligned}
& \mathrm{NH}_3+\mathrm{Ag}^{+} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}, \quad K_1=1.6 \times 10^3 \\
& {\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+},} \\
& K_2=6.8 \times 10^3
\end{aligned}$
The equilibrium constant for the following reaction, $\mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$is
Options:
$\begin{aligned}
& \mathrm{NH}_3+\mathrm{Ag}^{+} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}, \quad K_1=1.6 \times 10^3 \\
& {\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+},} \\
& K_2=6.8 \times 10^3
\end{aligned}$
The equilibrium constant for the following reaction, $\mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$is
Solution:
2686 Upvotes
Verified Answer
The correct answer is:
$1.088 \times 10^7$
When silver ion reacts with $\mathrm{NH}_3$, then diammine silver (I) ion is formed.
$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$,
The equilibrium constant for the following reaction,
Here,
$\begin{aligned}
& K_3=K_1 \times K_2 \\
& \therefore \quad K_3=1.6 \times 10^3 \times 6.8 \times 10^3 \\
& =1.088 \times 10^7
\end{aligned}$
$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$,
The equilibrium constant for the following reaction,
Here,
$\begin{aligned}
& K_3=K_1 \times K_2 \\
& \therefore \quad K_3=1.6 \times 10^3 \times 6.8 \times 10^3 \\
& =1.088 \times 10^7
\end{aligned}$
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