Question

Calculate the activity coefficient of ${\mathbf{Zn}}^{\mathbf{2}\mathbf{+}}$when $\mathbf{\mu }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{083}\mathbf{M}$by using (a) Equation $\mathbf{8}\mathbf{-}\mathbf{6}$; linear interpolation in Table $\mathbf{8}\mathbf{-}\mathbf{1}$.

Short answer

Thus the activity coefficient of ${\mathrm{Zn}}^{2+}$when $\mathrm{\mu }=0.083\mathrm{M}$is

(a) 0.422 M

(b) 0.432 M

Step-by-step solution

Calculation the Activity Coefficient using Debye Huckle Equation.

The task is to calculate the activity coefficient of ${\mathrm{Zn}}^{2+}$when $\mathrm{\mu }=0.083\mathrm{M}$.

In the task (a) we need to find using the equation $8-6$.

Debye Huckle equation role="math" localid="1654853836629" $\log \gamma =\frac{-0.51\times {\mathrm{z}}^2\times \sqrt{\mathrm{\mu }}}{1+\mathrm{\alpha }\sqrt{{\displaystyle \raisebox{1ex}{$\mathrm{\mu }$}\!\left/ \!\raisebox{-1ex}{$305$}\right.}}}$

role="math" localid="1654853927016" $$\begin{gathered} \log \gamma =\frac{-0.51\times {\mathrm{z}}^2\times \sqrt{0.083}}{1+600\sqrt{{\displaystyle \raisebox{1ex}{$0.083$}\!\left/ \!\raisebox{-1ex}{$305$}\right.}}} \\ \log \gamma =-0.375 \\ \mathrm{\gamma }={10}^{-0.375}=0.422\hspace{0.17em}\mathrm{M} \end{gathered}$$

Calculating the Activity Coefficient using Interpolation.

In the task (b) , we need to find the activity coefficient using the equation for interpolation.

$\gamma =\left(\frac{0.083-0.05}{0.7-0.05}\times \left(0.405-0.485\right)\right)+0.485=0.432\mathrm{M}$

Thus the activity coefficient of ${\mathrm{Zn}}^{2+}$when $\mathrm{\mu }=0.083\mathrm{M}$is

(a) 0.422 M

(b) 0.432 M