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JEE Advanced · Chemistry · 7. Chemical Equilibrium

\(\mathrm{Ag}^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; k_1=3.5 \times 10^{-3}\) \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} ; k_2\) \(=1.7 \times 10^{-3}\) then the formation constant of \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}\)is

  1. A \(6.08 \times 10^{-6}\)
  2. B \(6.08 \times 10^6\)
  3. C \(6.08 \times 10^{-9}\)
  4. D None of these
Verified Solution

Answer & Solution

Correct Answer

(A) \(6.08 \times 10^{-6}\)

Step-by-step Solution

Detailed explanation

(i) \(\mathrm{Ag}^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; k_1\) \(=3.5 \times 10^{-3}\)
(ii) \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+},\) \(k_2=1.7 \times 10^{-3}\)
on the basis of above reaction
\(\begin{aligned}
& k_1=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_3\right]} \\
& k_2=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}}{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}\left[\mathrm{NH}_3\right]}
\end{aligned}\)
For the formation of \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}\)
\(\mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}\)
Formation constant \((K)=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_3\right]^2}\)
From eq. (i) and (ii)
\(K=k_1 \times k_2=3.5 \times 10^{-3} \times 1.7 \times 10^{-3}\) \(=5.95 \times 10^{-6} \approx 6.08 \times 10^{-6}\)
From JEE Advanced
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