Calculate the overall complex dissociation equilibrium constant for the $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{t}^{2+}$ ion, given that $\beta_{4}$ for this complex is $2.1 \times 10^{18}$.

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Calculate the overall complex dissociation equilibrium constant for the $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{t}^{2+}$ ion, given that $\beta_{4}$ for this complex is $2.1 \times 10^{18}$ .

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Calculate the overall complex dissociation equilibrium constant for the $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{t}^{2+}$ ion, given that $\beta_{4}$ for this complex is $2.1 \times 10^{18}$ .

Ans: We are given the overall stability constant $\left(\beta_{s}\right)=2.1 \times 10^{13}$ .
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant. This is given below:
Overall dissociation constant $=\frac{1}{\beta_{a}}=\frac{1}{2.1 \times 10^{17}}=4.7 \times 10^{-14}$

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