Question
Calculate the overall complex dissociation equilibrium constant for the $\mathrm{Cu}\left(\mathrm{NH}_3\right)_4{ }^{2+}$ ion, given that $\beta_4$ for this complex is $2.1 \times 10^{13}$.
✓
Answer
$\beta_{4} =2.1\times10^{13}$
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, $\beta_{4}$
$\therefore\frac{1}{\beta_{4}}=\frac{1}{2.1\times10^{13}}$
$=4.7\times10^{-14}$
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, $\beta_{4}$
$\therefore\frac{1}{\beta_{4}}=\frac{1}{2.1\times10^{13}}$
$=4.7\times10^{-14}$
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