a lil help orgo question..

[egots]

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mixture of alanine(pI=6) and aspartic acid(pI=3) subted to electrophoresis at pH=3...what would occur

i know aspartic acid has no movement but i thought that alanine would migrate to the anode but the answer in kaplan says cathode...i thought regardless of protonation the relationship is

pH (positive goes to cathode) < pI < pH (negative goes to anode)

 

[marylanddds]

ur right.

 

[dat_student]

...
pH (positive goes to cathode) < pI < pH (negative goes to anode)

In a battery or other source of direct current, the cathode is the positive terminal.

In electrolysis, cathode is the negative electrode in the electrolytic cell.

 

[slayerdeus]

No you are wrong. If the pI of aniline is 6 it will still have to be protonated at pH 3. If aniline is still protonated then that means it will still have a positive charge, thus it will go towards the cathode.

 

[slayerdeus]

No you are wrong. If the pI of aniline is 6 it will still have to be protonated at pH 3. If aniline is still protonated then that means it will still have a positive charge, thus it will go towards the cathode.

Don't forget that pI is the pH where the molecule has no net charge. Thus putting the compound with a certain pI in a solution that has lower pH than its pI would add protons and positive charge making it more positive. Conversely putting the compound with a certain pI in a solution that has a high pH than its pI would removed protons and negative charge making it more negative.

 

[egots]

I know what you mean but Kaplan threw me off with the pH < pI think negative and ph> pI think positive…but in reality if the solutions pH is less then pI it will protonate those molecues in the solution and make it positive…and when it is greater it will deprotonate

For solutions:
pH < pI … travel to cathode b/c of protonation
pH > pI … travel to anode b/c of deprotonation