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