Figure 26-23 shows the separation of substituted benzoates. There is a peak of unknown identity at 86.0 seconds. (a) Is the unknown a cation, neutral, or an anion? (b) Find the apparent mobility and electrophoretic mobility of the unknown peak.

The electrophoretic flow of the compound depends on its charge and in contrast to its magnitude and viscosity of the medium.

(a)

It is given that normal order of capillary zone electrophoresis elution is:

(1) Cations (highest mobility)

(2) Neutrals (unseparated)

(3) Anions (highest mobility)

The following is $\mathrm{t}=86\mathrm{s}=1\min +26$s, so it can only be a cation because it is the first to be eluted)

(b)

Calculate the apparent mobility:$\left({\mu }_{app}\right)$

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{app}}=\frac{{\mathrm{u}}_{\mathrm{net}}}{\mathrm{E}} \\ {\mathrm{\mu }}_{\mathrm{app}}=\frac{0.400\mathrm{m}/86\mathrm{s}}{5\times {10}^4\mathrm{V}/\mathrm{m}} \\ {\mathrm{\mu }}_{\mathrm{app}}=9.3\times {10}^8{\mathrm{m}}^2/\mathrm{vs} \end{gathered}$$

The electroosmotic velocity $\left({\mathrm{u}}_{\mathrm{eo}}\right)$:

$$\begin{gathered} {\mathrm{u}}_{\mathrm{eo}}=\frac{\mathrm{distance}\mathrm{to}\mathrm{detector}}{\mathrm{migration}\mathrm{time}} \\ {\mathrm{u}}_{\mathrm{eo}}=\frac{0.400\mathrm{m}}{188\mathrm{s}} \\ {\mathrm{u}}_{\mathrm{eo}}=0.00213\mathrm{m}/\mathrm{s} \end{gathered}$$

And electroosmotic mobility

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{eo}}=\frac{{\mathrm{u}}_{\mathrm{eo}}}{\mathrm{E}} \\ {\mathrm{\mu }}_{\mathrm{eo}}=\frac{0.00213\mathrm{m}/\mathrm{s}}{5\times {10}^4\mathrm{V}/\mathrm{m}} \\ {\mathrm{\mu }}_{\mathrm{eo}}=4.26\times {10}^8{\mathrm{m}}^2/\mathrm{Vs} \end{gathered}$$

The electrophoretic mobility $\left({\mathrm{\mu }}_{\mathrm{ep}}\right)$:

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{app}}={\mathrm{\mu }}_{\mathrm{ep}}+{\mathrm{\mu }}_{\mathrm{eo}} \\ {\mathrm{\mu }}_{\mathrm{ep}}={\mathrm{\mu }}_{\mathrm{app}}-{\mathrm{\mu }}_{\mathrm{eo}} \\ {\mathrm{\mu }}_{\mathrm{ep}}=\left(9.3\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}\right)-\left(4.26\times {10}^8{\mathrm{m}}^2/\mathrm{Vs}\right) \\ {\mathrm{\mu }}_{\mathrm{ep}}=5.04\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs} \end{gathered}$$