Question

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.

Short answer

The electrophoretic mobility of the unknown peak is${\mathrm{\mu }}_{\mathrm{ep}}=5.04\times {10}^8{\mathrm{m}}^2/\mathrm{Vs}$

Step-by-step solution

Define electrophoretic mobility:

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

Determine whether the unknown is a cation, neutral or an anion:

(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)

Find the apparent mobility and electrophoretic mobility of the unknown peak:

(b)

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

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

Calculation for the electroosmotic velocity:

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

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

 Calculation for the electrophoretic mobility:

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

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