Calculate the overall complex dissociation equilibrium constant for the $\text{Cu(N}{{\text{H}}_{\text{3}}}\text{)}_{\text{4}}^{\text{2+}}$ ion, given that ${{\text{ }\!\!\beta\!\!\text{ }}_{\text{4}}}$ for this complex is $\text{2}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{13}}}$.
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Ans: We are given the overall stability constant (${{\text{ }\!\!\beta\!\!\text{ }}_{\text{4}}}$) = $\text{2}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{13}}}$.
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant. This is given below:\[\text{Overall dissociation constant = }\frac{\text{1}}{{{\text{ }\!\!\beta\!\!\text{ }}_{\text{4}}}}\text{ = }\frac{\text{1}}{\text{2}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{13}}}}\text{ = 4}\text{.7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-14}}}\]
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