JEE PYQ: Chemical Equilibrium Question 6
Question 6 - 2021 (24 Feb Shift 1)
The stepwise formation of $[\text{Cu}(\text{NH}_3)_4]^{2+}$ is given below:
$\text{Cu}^{2+} + \text{NH}_3 \xrightleftharpoons{K_1} [\text{Cu}(\text{NH}_3)]^{2+}$
$[\text{Cu}(\text{NH}_3)]^{2+} + \text{NH}_3 \xrightleftharpoons{K_2} [\text{Cu}(\text{NH}_3)_2]^{2+}$
$[\text{Cu}(\text{NH}_3)_2]^{2+} + \text{NH}_3 \xrightleftharpoons{K_3} [\text{Cu}(\text{NH}_3)_3]^{2+}$
$[\text{Cu}(\text{NH}_3)_3]^{2+} + \text{NH}_3 \xrightleftharpoons{K_4} [\text{Cu}(\text{NH}_3)_4]^{2+}$
The value of stability constants $K_1$, $K_2$, $K_3$ and $K_4$ are $10^4$, $1.58 \times 10^3$, $5 \times 10^2$ and $10^2$ respectively. The overall equilibrium constant for dissociation of $[\text{Cu}(\text{NH}_3)_4]^{2+}$ is $x \times 10^{-12}$. The value of $x$ is ____. (Rounded off to the nearest integer)
Show Answer
Answer: 1
Solution
$K = \frac{1}{K_1 K_2 K_3 K_4} = 1.26 \times 10^{-12}$, so $x = 1$.