Calculate the overall complex dissociation equilibrium constant for the…

Calculate the overall complex dissociation equilibrium constant for the $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ ions, given that stability constant $\left(\beta_4\right)$ for this complex is $2.1 \times 10^{13}$.
  1. $8.27 \times 10^{-13}$
  2. $4.76 \times 10^{-14}$
  3. $2.39 \times 10^{-7}$
  4. $1.83 \times 10^{14}$

Solution

Dissociation constant is the reciprocal of the stability constant $(\beta=1 / K)$. Overall complex dissociation equilibrium constant, $\begin{aligned} K & =\frac{1}{\beta_4} \\ & =\frac{1}{2.1 \times 10^{13}}=4.76 \times 10^{-14}\end{aligned}$

Asked in: NEET 2022 (Phase 2)

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