Question

Suppose that the electroosmotic mobility of a solution is $\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{61}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\left(V\times s\right)$) How many plates are required to separate sulfate from bromide with a resolution of 2.0? Use Table 15-1 for mobilities and Equation 26-14 for resolution.

Short answer

Plates that are required to separate sulfate from bromide with a resolution of 2.0 is$\mathrm{N}=2\times 5^{10}\mathrm{plates}$

Step-by-step solution

Plates that are required to separate sulfate from bromide with a resolution of 2.0

From Table 15-1 we will use the following:

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{ep}}\left({\mathrm{SO}}_4^{2-}\right)=-8.27\times {10}^{-8}\mathrm{m}/\mathrm{Vs} \\ {\mathrm{\mu }}_{\mathrm{app}}={\mathrm{\mu }}_{\mathrm{eo}}={\mathrm{\mu }}_{\mathrm{ep}} \\ {\mathrm{\mu }}_{\mathrm{app}}=\left(16.1\times {10}^{-8}\right)+{\left(-8.27\times {10}^{-8}\right)}^{\mathrm{a}} \\ {\mathrm{\mu }}_{\mathrm{app}}=7.83\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs} \\ {\mathrm{\mu }}_{\mathrm{ep}}\left({\mathrm{Br}}^{-}\right)=-8.13\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs} \\ {\mathrm{\mu }}_{\mathrm{app}}={\mathrm{\mu }}_{\mathrm{eo}}={\mathrm{\mu }}_{\mathrm{ep}} \\ {\mathrm{\mu }}_{\mathrm{app}}=\left(16.1\times {10}^{-8}\right)+\left(-8.13\times {10}^{-8}\right) \\ {\mathrm{\mu }}_{\mathrm{app}}=7.97\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs} \end{gathered}$$

Step 2:Calculate the average

$\mathrm{Next}\mathrm{we}\mathrm{will}\mathrm{calculate}\mathrm{the}\mathrm{average}\mathrm{\mu }\mathrm{and}∆{\mathrm{\mu }}_{\mathrm{ep}}$

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{av}}=\frac{\left(7.83\times {10}^{-8}\right)+\left(7.97\times {10}^{-8}\right)}{2} \\ {\mathrm{\mu }}_{\mathrm{av}}=7.90\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs} \\ ∆{\mathrm{\mu }}_{\mathrm{ep}}=\left(8.27\times {10}^{-8}\right)-\left(8.13\times {10}^{-8}\right) \\ ∆{\mathrm{\mu }}_{\mathrm{ep}}=0.14\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs} \end{gathered}$$

Step 3:Equation 26-14 for resolution

Then we will use the Equation 26-14 for resolution:

$$\begin{gathered} \mathrm{N}={\left(4\times \left(\mathrm{Resolution}\right)\frac{{\mathrm{\mu }}_{\mathrm{av}}}{∆{\mathrm{\mu }}_{\mathrm{ep}}}\right)}^2 \\ \mathrm{N}={\left(4\times \left(2.0\right)\frac{7.90}{0.14}\right)}^2 \\ \mathrm{N}=2\times {10}^5\mathrm{plates} \end{gathered}$$