Question

A certain inorganic cation has electrophoretic mobility of $5.27\times {10}^{-4}{\mathrm{cm}}^2{\mathrm{s}}^{-1}{\mathrm{V}}^{-1}$. This same ion has a diffusion coefficient of $9.5\times {10}^{-6}{\mathrm{cm}}^2{\mathrm{s}}^{-1}$.If this ion is separated from other cations by CZE with a

data-custom-editor="chemistry" $50.0\mathrm{cm}$capillary, what is the expected plate count N at applied voltages of

(a) data-custom-editor="chemistry" $10.0\mathrm{kV}$?

(b) data-custom-editor="chemistry" $20.0\mathrm{kV}$?

(c) data-custom-editor="chemistry" $30.0\mathrm{kV}$?

Short answer

(a) The expected plate count is $2.8\times {10}^5$.

(b) The expected plate count is $5.5\times {10}^5$.

(c) The expected plate count is $8.3\times {10}^5$.

Step-by-step solution

Part (a) Step 1. Given information

The diffusion coefficient is $9.5\times {10}^{-6}{\mathrm{cm}}^2{\mathrm{s}}^{-1}$.

The electrophoretic mobility is $5.27\times {10}^{-4}{\mathrm{cm}}^2{\mathrm{s}}^{-1}{\mathrm{V}}^{-1}$.

The applied voltage islocalid="1645523631391" $10\mathrm{kV}$.

Part (a) Step 2. Calculate plate count

The expression for plate count is as follows:

$N=\frac{{\mu }_{e}V}{2D}.......\left(1\right)$

Here, The plate count is $N$, the diffusion coefficient is $D$, the electrophoretic mobility is ${\mu }_e$, the applied voltage is $V$.

Substitute the values in equation (1) as follows:

localid="1645523852805" $$\begin{gathered} N=\frac{5.27\times {10}^{-4}{\mathrm{cm}}^2{\mathrm{s}}^{-1}{\mathrm{V}}^{-1}\left(10\mathrm{kV}\times {\displaystyle \frac{1000\mathrm{V}}{1\mathrm{kV}}}\right)}{2\times 9.5\times {10}^{-6}{\mathrm{cm}}^2{\mathrm{s}}^{-1}} \\ =2.8\times {10}^5 \end{gathered}$$

Part (b) Step 1. Given information

The applied voltage is$20.0\mathrm{kV}$.

Part (b) Step 2. Calculate plate count 

Substitute the values in equation (1) as follows:

$$\begin{gathered} N=\frac{5.27\times {10}^{-4}{\mathrm{cm}}^2{\mathrm{s}}^{-1}{\mathrm{V}}^{-1}\left(20\mathrm{kV}\times {\displaystyle \frac{1000\mathrm{V}}{1\mathrm{kV}}}\right)}{2\times 9.5\times {10}^{-6}{\mathrm{cm}}^2{\mathrm{s}}^{-1}} \\ =5.5\times {10}^5 \end{gathered}$$

Part (c) Step 1. Given information 

The applied voltage is$30.0\mathrm{kV}$.

Part (c) Step 2. Calculate plate count 

Substitute the values in equation (1) as follows:

$$\begin{gathered} N=\frac{5.27\times {10}^{-4}{\mathrm{cm}}^2{\mathrm{s}}^{-1}{\mathrm{V}}^{-1}\left(30\mathrm{kV}\times {\displaystyle \frac{1000\mathrm{V}}{1\mathrm{kV}}}\right)}{2\times 9.5\times {10}^{-6}{\mathrm{cm}}^2{\mathrm{s}}^{-1}} \\ =8.3\times {10}^5 \end{gathered}$$