Question

(a) A particular solution in a particular capillary has an electroosmotic mobility of $\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\mathbf{(}\mathbf{V}\mathbf{\times }\mathbf{s}\mathbf{)}\mathbf{atpH}\mathbf{2}$and 8.1 ${\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\mathbf{(}\mathbf{V}\mathbf{\times }\mathbf{s}\mathbf{)}$. How long will it take a neutral solute to travel 52 cm from the injector to the detector if 27 kV is applied across the 62-cm-long capillary tube at pH 2? At pH 12?(b) An analyte anion has an electrophoretic mobility of $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\mathbf{(}\mathbf{V}\mathbf{\times }\mathbf{s}\mathbf{)}$How long will it take to reach the detector at pH 2?At pH 12?

Short answer

a)A neutral solute to travel at pH=2 and pH=12 is$t=9.2\times {10}^{2}s$and$\mathrm{t}=1.47\times {10}^2\mathrm{s}$.

b)An analyte anion to reach the detector at pH=2 and pH=12 is ${\mathrm{\mu }}_{\mathrm{app}}=-0.3\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}$and$\mathrm{t}=1.84\times {10}^2\mathrm{s}$

Step-by-step solution

A neutral solute to travel at pH=2 and pH=12

a)At pH=2 use the following requations

$$\begin{gathered} \mathrm{u}={\mathrm{\mu }}_{\mathrm{eo}}\mathrm{E} \\ \mathrm{u}=\left(1.3\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}\right)\times \left(\frac{27\times {10}^3\mathrm{V}}{0.62\mathrm{m}}\right) \\ \mathrm{u}=5.66\times {10}^{-4}\mathrm{m}/\mathrm{s} \\ \mathrm{t}=\mathrm{s}/\mathrm{u} \\ \mathrm{t}=0.52\mathrm{m}/\left(5.66\times {10}^{-4}\mathrm{m}/\mathrm{s}\right) \\ \mathrm{t}=9.2\times {10}^2\mathrm{s} \end{gathered}$$

b)At pH=12 use the following requations

$$\begin{gathered} \mathrm{u}={\mathrm{\mu }}_{\mathrm{eo}}\mathrm{E} \\ \mathrm{u}=\left(1.3\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}\right)\times \left(\frac{27\times {10}^3\mathrm{V}}{0.62\mathrm{m}}\right) \\ \mathrm{u}=3.53\times {10}^{-3}\mathrm{m}/\mathrm{s} \\ \mathrm{t}=\mathrm{s}.\mathrm{u} \\ \mathrm{t}=0.52\mathrm{m}/\left(3.53\times {10}^{-3}\mathrm{m}/\mathrm{s}\right) \\ \mathrm{t}=1.47\times {10}^2\mathrm{s} \end{gathered}$$

An analyte anion to reach the detector at pH=2 and pH=12

a)At pH=2 use the following requations

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{app}}={\mathrm{\mu }}_{\mathrm{ap}}+{\mathrm{\mu }}_{\mathrm{eo}} \\ {\mathrm{\mu }}_{\mathrm{app}}=\left(-1.6\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}\right)+\left(1.3\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}\right) \\ {\mathrm{\mu }}_{\mathrm{app}}=-0.3\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs} \end{gathered}$$

Anion wouldn’t migrate towards the detector.

b)At pH=12 use the following requations

$$\begin{gathered} \mathrm{u}={\mathrm{\mu }}_{app}\mathrm{E} \\ \mathrm{u}=\left(6.5\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}\right)\times \left(\frac{27\times {10}^3\mathrm{V}}{0.62\mathrm{m}}\right) \\ \mathrm{u}=2.83\times {10}^{-3}\mathrm{m}/\mathrm{s} \\ \mathrm{t}=\mathrm{s}/\mathrm{u} \\ \mathrm{t}=0.52\mathrm{m}/\left(2.83\times {10}^{-3}\mathrm{m}/\mathrm{s}\right) \\ \mathrm{t}=1.84\times {10}^2\mathrm{s} \end{gathered}$$