Question about acids at low pH and pKa

[Little Etoile]

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Gchem book passage , 54 #D:



“At low ph, COOH and NH3+ are protonated.” But if it is low pH, it is acidic and there is a high concentration of H+ --> so shouldn’t carboxylic acid be donating its proton too? This always gets me really confused. Is it because carboxylic acids are weak acids, so they get protonated at low pH rather than donating? And/or is it that there are too many protons in solution already? I don't know why this always trips me up.



Also, in regards to pKa, if an acid's pKa is 3.5, does that mean that below (e.g. pH 2.2) that, it will remain protonated, but above it, it donates its proton? ..

As always, thanks in advance!

 

[sleepy425]

Gchem book passage , 54 #D:



“At low ph, COOH and NH3+ are protonated.” But if it is low pH, it is acidic and there is a high concentration of H+ --> so shouldn’t carboxylic acid be donating its proton too? This always gets me really confused. Is it because carboxylic acids are weak acids, so they get protonated at low pH rather than donating? And/or is it that there are too many protons in solution already? I don't know why this always trips me up.



Also, in regards to pKa, if an acid's pKa is 3.5, does that mean that below (e.g. pH 2.2) that, it will remain protonated, but above it, it donates its proton?

As always, thanks in advance!

Actually, your second question is the key. At pH below the pKa, the group will be protonated. At pH above the pKa, the group will be deprotonated.

Here's why:

HA<---->(H+) + (A-)

the double headed arrow is an equilibrium arrow, not a resonance arrow, i just couldn't draw it that way.

so the equilibrium expression, Ka (same as Keq for an acid) is:
Ka=[H+][A-]/[HA]

Ok, so pKa is just another way of writing Ka, right, because it's just the negative log of Ka.

As pH drops, H+ concentration increases. By mass action (LeChatlier's principle), that means you're increasing the concentration of a product so you drive the equilibrium towards formation of HA, or the protonated state. As pH increases, H+ concentration drops. By mass action, that means you're decreasing the concentration of the product so you drive equilibrium towards production of A-.

So at low pH, you have more of the protonated state because the proton concentration is higher in solution. At high pH, you have less of the protonated state because proton concentration is low in solution.

So that's why the carboxyl group is protonated. The pH is below the pKa. Incidentally, when the pH is at the pKa, you have 50% protonated, 50% deprotonated. One pH unit lower than pKa means 90% protonated, 10% deprotonated. Two pH units lower than pKa means 99% protonated, 1% deprotonated. One pH unit above the pKa means 10% protonated, 90% deprotonated. Two pH units above pKa means 1% protonated, 99% deprotonated. This comes straight from the Henderson-Hasselbach equation.

So if you have an amino acid with pKa1=3 and pKa2=10:

below pH 3, most of the species in solution are fully protoanted (both amino group and carboxyl group are protonated).

between pH 3 and pH 10, most of the species are zwitterionic (carboxyl group is deprotonated because the pH is above 3 which is the pKa for the carboxyl group, and amino group is protonated because the pH is below 10 which is the pKa for the amino group)

above pH 10, most of the species are fully deprotonated (pH is above both pKas)

that's all there is to it

 

[Farcus]

ok sleepy here is my reiteration of what you typed tell me if I am right.

Since the OP's problem was in low pH, meaning high concentration H+, by le chatlier the overall reaction would be moving to the reverse reaction of
HA<---->(H+) + (A-). Thus the compounds in the low pH would be protonated.

So by the same concept, if we put those compound in high pH, low H concentration, those compounds such as carboxylic acid would easily deprotonate and give off lots of H+ to make even the equilibrium that always goes into effect "Le Chatilier" I know I spelled it wrong 🙂.

 

[sleepy425]

ok sleepy here is my reiteration of what you typed tell me if I am right.

Since the OP's problem was in low pH, meaning high concentration H+, by le chatlier the overall reaction would be moving to the reverse reaction of
HA<---->(H+) + (A-). Thus the compounds in the low pH would be protonated.

So by the same concept, if we put those compound in high pH, low H concentration, those compounds such as carboxylic acid would easily deprotonate and give off lots of H+ to make even the equilibrium that always goes into effect "Le Chatilier" I know I spelled it wrong 🙂.

Sounds good to me!