Question

The migration time for ${\mathbf{Cl}}^{\mathbf{-}}$in a capillary zone electrophoresis experiment is 17.12 min and the migration time for ${\mathbf{l}}^{\mathbf{-}}$is 17.78 min. From mobilities in Table 15-1, predict the migration time of Br(The observed value is 19.6 min.)

Short answer

Expressions for ${\mathrm{Cl}}^{-}$and ${\mathrm{l}}^{-}\mathrm{is}{\mathrm{\mu }}_{\mathrm{app}}=\frac{\mathrm{L}}{\mathrm{tE}}$

The migration time is calculated as x=20.5min

Step-by-step solution

Step 1:Expressions forCl-andl-using the following equations

We need to write the expressions for ${\mathrm{Cl}}^{-}$and ${\mathrm{l}}^{-}$using the following equations

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{L}}{{\mathrm{\mu }}_{\mathrm{net}}}=\frac{\mathrm{L}}{{\mathrm{\mu }}_{\mathrm{app}}\mathrm{E}} \\ {\mathrm{\mu }}_{\mathrm{app}}=\frac{\mathrm{L}}{\mathrm{tE}} \end{gathered}$$

Where t is migration time,L is length,E is electric field and is speed.

$$\begin{gathered} \mathrm{for}{\mathrm{Cl}}^{-}{\mathrm{\mu }}_{\mathrm{app}}=\frac{\mathrm{L}}{17.2\mathrm{E}} \\ {\mathrm{l}}^{-}{\mathrm{\mu }}_{\mathrm{app}}=\frac{\mathrm{L}}{17.78\mathrm{E}} \end{gathered}$$

Step 2:Difference in mobilities

Next we write the difference in mobilities

$∆{\mathrm{\mu }}_{\mathrm{app}}\left(\mathrm{l}-\mathrm{Cl}\right)=\frac{\mathrm{L}}{17.2\mathrm{E}}-\frac{\mathrm{L}}{17.78\mathrm{E}}$$\mathrm{from}\mathrm{Table}15-\mathrm{l}\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that}\left(\mathrm{l}-\mathrm{Cl}\right)=0.05\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}$

Step 3:Difference in mobilities for  and 

$\mathrm{Then}\mathrm{we}\mathrm{will}\mathrm{write}\mathrm{the}\mathrm{difference}\mathrm{in}\mathrm{mobilities}\mathrm{for}{\mathrm{Br}}^{-}\mathrm{and}{\mathrm{Cl}}^{-}:$

$∆{\mathrm{\mu }}_{\mathrm{app}}\left(\mathrm{Br}-\mathrm{Cl}\right)=\frac{\mathrm{L}}{17.2\mathrm{E}}-\frac{\mathrm{L}}{\mathrm{xE}}$

$\mathrm{from}\mathrm{Table}15-\mathrm{l}\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that}\left(\mathrm{Br}-\mathrm{Cl}\right)=0.22\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}$

Step 4:Calculate the migration time

Using step 2 and step 3 calculating the migration time

$$\begin{gathered} \frac{∆{\mathrm{\mu }}_{\mathrm{app}}\left(\mathrm{Br}-\mathrm{Cl}\right)}{∆\mathrm{\mu }\left(\mathrm{l}-\mathrm{Cl}\right)}=\frac{0.22}{0.05} \\ 4.4=\frac{{\displaystyle \frac{\mathrm{L}}{17.2\mathrm{E}}}-{\displaystyle \frac{\mathrm{L}}{\mathrm{xE}}}}{{\displaystyle \frac{\mathrm{L}}{17.2\mathrm{E}}}-{\displaystyle \frac{\mathrm{L}}{17.78\mathrm{E}}}} \\ \mathrm{x}=20.5\min \end{gathered}$$