How long will the cation take?

Question

The electroosmotic velocity of a solution in a particular capillary is 0.736 mm/s. A 21.9 kV potential is applied across the 57.6 cm long capillary. How long will it take a cation with an electrophoretic mobility of 4.58×10−4 cm2/(V⋅s) to travel the 46.7 cm from the injector to the detector?

Answer 1

Short answer: about 189 s, or 3.1 minutes.

The whole problem turns on one idea: in capillary electrophoresis a cation is carried along by two things at once. The bulk electroosmotic flow (EOF) sweeps the entire solution toward the detector, and on top of that the ion does its own electrophoretic migration. For a cation those two point the same way, so the velocities add.

The trap that ruins this problem: the two lengths are not interchangeable. The 57.6 cm is the total capillary, and that is what the 21.9 kV is dropped across, so it sets the field. The 46.7 cm is only how far the ion has to travel to reach the detector. Both numbers appear, doing different jobs, and swapping them gives an answer that looks perfectly reasonable.

Step 1 - the electric field. Voltage divided by the length it is applied across:

E = V / L(total)
E = 21900 V / 57.6 cm
E = 380.2 V/cm

Step 2 - the ion's own migration velocity. Mobility times field. The units cancel to a velocity, which is itself the check that you used the field and not the raw voltage:

v(ep) = mobility x E
v(ep) = (4.58e-4 cm^2/(V*s)) x (380.2 V/cm)
v(ep) = 0.1741 cm/s

Step 3 - add the electroosmotic flow. Convert it first, since it was handed to you in mm/s while everything else is in cm:

v(eo) = 0.736 mm/s = 0.0736 cm/s
v(app) = v(ep) + v(eo)
v(app) = 0.1741 + 0.0736
v(app) = 0.2477 cm/s

Step 4 - time to reach the detector. Now, and only now, use the 46.7 cm:

t = L(detector) / v(app)
t = 46.7 cm / 0.2477 cm/s
t = 188.5 s = 189 s (3 sig figs)
t = 3.14 minutes

A cross-check that costs one line and proves both lengths belong. Combine the mobilities first, then use the standard CE migration-time formula:

mobility(eo) = v(eo)/E = 0.0736/380.2 = 1.94e-4 cm^2/(V*s)
mobility(app) = 4.58e-4 + 1.94e-4 = 6.52e-4 cm^2/(V*s)
t = L(detector) x L(total) / (mobility(app) x V)
t = (46.7)(57.6) / ((6.52e-4)(21900))
t = 2690 / 14.28 = 188.5 s

Same answer by a completely independent route. Notice the numerator carries both lengths multiplied together - that is the algebraic reason a CE problem always hands you two lengths, and it is worth memorizing in exactly that form.

Three errors to watch for:

1. Using 46.7 cm to get the field. That gives E = 468.9 V/cm, v(app) = 0.2884 cm/s and t = 162 s. Nothing about 162 s looks wrong, which is precisely what makes this one dangerous.

2. Forgetting the EOF and using only the ion's own migration. That gives 46.7/0.1741 = 268 s, too slow by 42 percent.

3. Leaving the EOF in mm/s. Adding 0.736 to 0.1741 gives 0.910 in mixed units and an answer of 51 s.

Why the EOF points at the detector in the first place, since most textbooks state this without explaining it: bare fused silica carries surface silanol groups that deprotonate above roughly pH 3, leaving the wall negatively charged. The solution immediately next to that wall is therefore cation-rich, and when the field is applied that cationic layer drags the entire bulk solution toward the cathode. Because the driving force sits at the wall rather than in a pressure drop, the flow profile is flat rather than parabolic - which is the real reason CE gives much narrower peaks than HPLC.

One consequence worth carrying into the next problem: when the EOF is strong enough it outruns even the anions, which are migrating against it. So cations arrive first, neutrals arrive together at exactly v(eo) because they have no mobility of their own, and anions arrive last - all in a single run. That is the feature that makes CE worth the trouble.