Question

Find ${\mathit{\alpha }}_{\mathbf{Z}{\mathbf{n}}^{\mathbf{2}\mathbf{+}}}$if free, unprotonated [NH₃] = 0.02 M.

Short answer

${\alpha }_{{\mathrm{Zn}}^{2+}}$=0.00673 if free, unprotonated [NH₃] = 0.02 M.

Step-by-step solution

Introduction

Zn²⁺ and NH₃ together form the following complexes

$$\begin{gathered} Zn\left(N{H}_3{)}^{2+} \\ Zn\right(N{H}_3{)}_2^{2+} \\ Zn\left(N{H}_3{)}_3^{2+} \\ Zn\right(N{H}_3{)}_4^{2+} \end{gathered}$$

The concentration of free, unprotonated NH₃ is 0.02 M

Equations need to use

${\alpha }_{Z{n}^{2+}}=\frac{1}{1+{\beta }_{1}L+{\beta }_2{L}^2+{\beta }_3{L}^3+{\beta }_4{L}^4}$

From appendix -I formation constants of the complexes are obtained below

$$\begin{gathered} Zn(N{H}_3{)}^{2+}\to {\beta }_1={10}^{2.8} \\ Zn(N{H}_3{)}_2^{2+}\to {\beta }_2={10}^{4.43} \\ Zn(N{H}_3{)}_3^{2+}\to {\beta }_3={10}^{6.74} \\ Zn(N{H}_3{)}_4^{2+}\to {\beta }_4={10}^{8.7} \end{gathered}$$

L=0.02M

Determine fraction of zinc

Plugging the respective values in the equation given

$$\begin{gathered} {\alpha }_{Z{n}^{2+}}=\frac{1}{1+{10}^{2.8}\times 0.02+{10}^{4.3}\times 0.{02}^2+{10}^{6.74}\times 0.{02}^3+{10}^{8.7}\times 0.{02}^4} \\ =0.00673 \end{gathered}$$

${\alpha }_{Z{n}^{2+}}=$0.00673 if free, unprotonated [NH₃] = 0.02 M.