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Question: Answered & Verified by Expert
$\mathrm{Ag}^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; k_1=3.5 \times 10^{-3}$ $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} ; k_2=1.7 \times 10^{-3}$ then the formation constant of $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$is
ChemistryChemical EquilibriumJEE AdvancedJEE Advanced 2006
Options:
  • A
    $6.08 \times 10^{-6}$
  • B
    $6.08 \times 10^6$
  • C
    $6.08 \times 10^{-9}$
  • D
    None of these
Solution:
2741 Upvotes Verified Answer
The correct answer is:
$6.08 \times 10^{-6}$
(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)
$$
K=k_1 \times k_2=3.5 \times 10^{-3} \times 1.7 \times 10^{-3}=5.95 \times 10^{-6} \approx 6.08 \times 10^{-6}
$$

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