Question

The overall complex dissociation equilibrium constant for the complex $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ ion will be ? (${\beta }_4$ for this complex 2.1 $\times {10}^{13})$

• A. $4.7\times {10}^{-14}$
• B. $2.1\times {10}^{13}$
• C. $11.9\times {10}^{-2}$
• D. $2.1\times {10}^{-13}$

Answer 1

Answer: (A)

$4.7\times {10}^{-14}$

Instability constant $=\frac{1}{{\beta }_4}=\frac{1}{2.1\times {10}^{13}}=4.76\times {10}^{-14}$.

$\left[Cu\right(N{H}_3{)}_4{]}^{2+}\to \left[Cu\right(N{H}_3{)}_3{]}^{2+}+N{H}_3$

Instability constant $=\frac{\left[Cu\right(N{H}_3{)}_3{]}^{2+}\left[N{H}_3\right]}{\left[Cu\right(N{H}_3{)}_4{]}^{2+}}$

$\left[Cu\right(N{H}_3{)}_3{]}^{2+}+N{H}_3\to \left[Cu\right(N{H}_3{)}_4{]}^{2+}$

Stability constant $={\beta }_4=\frac{\left[Cu\right(N{H}_3{)}_4{]}^{2+}}{\left[Cu\right(N{H}_3{)}_3{]}^{2+}\left[N{H}_3\right]}$