Question

Calculate the overall complex dissociation equilibrium constant for the $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ ions , given that stability constant $\left({\beta }_4\right)$ for this complex is $2.1\times {10}^{13}$:

• A. $8.27\times {10}^{-13}$
• B. $4.76\times {10}^{-14}$
• C. $2.39\times {10}^{-7}$
• D. $1.83\times {10}^{14}$

Answer 1

The correct option is B $4.76\times {10}^{-14}$

Dissociation constant is the reciprocal of the stability constant $(k=1/\beta )$.
Overall complex dissociation equilibrium constant, $k=\frac{1}{{\beta }_4}$

$k=\frac{1}{2.1\times {10}^{13}}$

$k=4.76\times {10}^{-14}$