1
$\begingroup$

Layman here. I was looking at vitamin C serums in skincare and wanted to know more about it. It has been awhile since taking chemistry in college so my understanding is rusty.

I want to better understand this diagram (Source: U of Iowa) that talks about the different forms vitamin C:

enter image description here

All I remember of $\mathrm{p}K_\mathrm{a}$s was that high $\mathrm{p}K_\mathrm{a}$s means the acid is very strong and dissociates all the way while lower $\mathrm{p}K_\mathrm{a}$ means a weaker acid. So I'm a little lost on how to interpret this diagram in terms of the $\mathrm{p}K$ values, ie the $\ce{OH}$ group is deprotonated from $\ce{AsCH2}$ to $\ce{AsCH-}$ but I'm not sure how the $\mathrm{p}K$ value fits in all of this? Layman me wants to say at $\mathrm{pH}=4.1$ is when deprotonation happens, but I have a feeling that isn't right.

$\endgroup$
  • 1
    $\begingroup$ That's backwards. Lower $\mathrm{pK}_{a}$ values correspond to stronger acids. $\endgroup$ – Zhe Dec 7 '16 at 23:26
  • 1
    $\begingroup$ $\mathrm{pH} = 4.1$ would imply that there are equal concentrations of the neutral and singly-protonated species in solution via the Henderson-Hasselbach equation. $\endgroup$ – Zhe Dec 7 '16 at 23:29
  • 1
    $\begingroup$ @Zhe, oops my mistake, you're right. I got pka and Ka mixed up. Okay so when Ph=pka in this case when pH= 4.1 there are equal amounts of AscH2 and AscH-, correct? And would it be correct to say that at ph below 4.1, the form is mostly AsCH2 and at ph's above 4.1, the form steadily increases to be more and more of AsCH- ? $\endgroup$ – Arcuvia Dec 7 '16 at 23:39
  • $\begingroup$ That's correct. Take a look at the equation in ringo's answer if you're still confused. $\endgroup$ – Zhe Dec 7 '16 at 23:44
  • $\begingroup$ There are three more, chemically partially relevant $\mathrm{p}K_\mathrm{a}$ values — although the corresponding anions do not exist in water. $\endgroup$ – Jan Dec 8 '16 at 22:03
2
$\begingroup$

You actually have it backwards. A low $\mathrm{p}K_\mathrm{a}$ is a strongly dissociating acid, and a high $\mathrm{p}K_\mathrm{a}$ is a weakly dissociating acid. It would do you well to remember the origin of $\mathrm{p}K_\mathrm{a}$ \eqref{pKa}, as well as the Henderson–Hasselbalch equation \eqref{HHeq}:

$$K_\mathrm{a}=\ce{\frac{[A-][H3O+]}{[HA]}} \Rightarrow \mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})= -\log \left( \ce{\frac{[A-][H3O+]}{[HA]}}\right)\tag1\label{pKa}$$

$${\mathrm {pH}}={\mathrm {p}}K_{{\mathrm {a}}}+\log _{{10}}\left({\frac {[{\mathrm {A}}^{-}]}{[{\mathrm {HA}}]}}\right)\tag2\label{HHeq}$$

From this we can surmise a physical meaning behind what the $\mathrm{p}K_\mathrm{a}$ of an acid is—it is the $\mathrm{pH}$ of the solution when $\ce{[A-]=[HA]}$. This means that for $\mathrm{pH}>\mathrm{p}K_\mathrm{a}$, $\ce{[A-]>[HA]}$. Since ascorbic acid is a diprotic acid, it has two $\mathrm{p}K_\mathrm{a}$ values, one for each deprotonation.

$\endgroup$
  • $\begingroup$ okay this helps a lot. I would like some clarification on this relationship: pH>pKa, [A−]>[HA]. Not sure if I have this right, so if I dissolve AsCH2 in a solution where the ph is higher than 4.1, let's say 6.0 for example. The solvent solution (let's say slightly acidic water) at ph= 6.0. pH>pKa , [A−]>[HA] says there there would be more of AsCH- than AsCH2. Is it b/c there would be more OH- ions in the water which then deprotonates ASCH2 leaving more ASCH- ? $\endgroup$ – Arcuvia Dec 7 '16 at 23:58
  • $\begingroup$ The $\mathrm{p}K_\mathrm{a}$ of an acid is a constant and does not change. This means that you can think of $\mathrm{pH}$ as a variable that responds to changes in $\ce{[HA]}$. When the $\mathrm{pH}$ is higher than $\mathrm{p}K_\mathrm{a}$, the other term in the Henderson–Hasselbalch equation has to be positive. Logarithm terms are only positive when the argument is >1; therefore, $\ce{[A−]>[HA]}$. $\endgroup$ – ringo Dec 8 '16 at 0:04
  • $\begingroup$ I would say that it is because there aren't enough $\ce{H3O+}$, so there will be a shift in equilibria to balance this. Think of the reaction in terms of $\ce{AscH2 + H2O <=> AscH- + H3O+}$ $\endgroup$ – ringo Dec 8 '16 at 0:14
  • $\begingroup$ I get what you are saying from a math standpoint when you were mentioning how pka is a constant and a log value has to be > 1 so A- > HA. But you mentioned how pH is a variable that responds to changes in HA. But I was thinking how the pH of a solution can also be the driver of what form Vit C can take on. I have to break this comment into two, read below: $\endgroup$ – Arcuvia Dec 8 '16 at 0:28
  • $\begingroup$ Here's how I was thinking about it. I take a Vit C powder in the AscH2 form and I dissolve it in alkaline water of pH=13. Water dissociates freely into hydroxide and H+, there is more OH- in this case b/c the water is more basic. I know OH- is a very strong base so it then deprotonates AscH- so that there is more AscH2-. The HH equation you showed me says when ph=pka is when we have equal [ ] of both. But our alkaline water is greater than the pka2 value, so I can intuitively just guess that there is greater than 50% AscH2- in the solution. I mean this logic is correct also or no? $\endgroup$ – Arcuvia Dec 8 '16 at 0:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.