Question

Calculate the overall complex dissociation equilibrium constant for the $Cu(N{H}_3{)}_4^{2+}$ ion, given that ${\beta }_4$ for this complex is $2.1\times {10}^3$.

Answer 1

Dissociation constant is the reciprocal of the stability constant ${\beta }_4$.

$\therefore$ Overall complex dissociation constant $=\frac{1}{{\beta }_4}=\frac{1}{2.1\times {10}^3}=4.7\times {10}^{-4}$