91Ó°ÊÓ

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:$\left({\mathrm{μ}}_{app}\right)$

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

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

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

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

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