- Acid-base definitions
- Organic acid-base mechanisms
- Ka and acid strength
- Ka and pKa review
- Using a pKa table
- Using pKa values to predict the position of equilibrium
- Stabilization of a conjugate base: electronegativity
- Acid strength, anion size, and bond energy
- Stabilization of a conjugate base: resonance
- Stabilization of a conjugate base: induction
- Stabilization of a conjugate base: hybridization
- Stabilization of a conjugate base: solvation
Using pKa values to predict the position of equilibrium
How to use pKa values to predict the position of equilibrium for organic acid-base reactions.
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- At1:34, it was said that a specific proton in the molecule had a pKa value. Is there any meaning to refer a single proton as having a pKa value versus the entire molecule? In general chemistry, I only had teachers refer to an entire molecule as having a pKa value.(11 votes)
- Your Gen Chem teacher probably only used examples that had one acidic proton and could only be involved in one acid base reaction with the other reactants in the example. If you look at a molecule like Sulfuric acid, H2SO4, a molecule with more than one acidic hydrogen, it's going to have two pKa's, one for each proton. In these cases, you can avoid confusion if you talk about the pKa of the proton instead of the pKa of the molecule. It's not wrong to talk about the first and second pKa's of H2SO4 , but having the ability to talk about the pKa of a specific Hydrogen can make communication more clear. In other words, the chemistry doesn't change between the two sets of terminology, but referring to the pKa of a specific proton can be of value in clarifying communication.(16 votes)
- Where is the equation: pKeq = pKa(acid-left) - pKa(acid-right) derived? Is this in another video? Why is this true?(8 votes)
- start with the equilibrium equation
Keq = [leftA-] [HrightA] / ( [HleftA] [rightA-] )
rearrange and multiply top and bottom by
Keq = ([leftA-][H+]/[HleftA]) * ([HrightA]/([rightA-][H+]))
this is now the same as
Keq = Ka(left) / Ka(right)
take negative logs of both sides
-log( Keq ) = -log( Ka(left) / Ka(right) )
-log( Keq ) = -log( Ka(left) ) - -log( Ka(right) )
pK = -logK
pKeq = pKa(left) - pKa(right)(8 votes)
- In the previous video ('Using pKa table'), acetic acid was marked as a weak acid relative to hydrochloric acid - "relative" being the operative term.
For the above example, is acetic acid considered to be a weak/moderate/strong acid relative to the presence of sodium hydroxide? I suppose what I'm trying to ask is whether acid strength changes relative to the presence of the base and its strength. For example, does acetic acid's pKa value change (from 5) to something like 10 in the presence of a weak base because the base will not readily deprotonate acetic acid?(1 vote)
- The pKa of the acid isn't dependent on the pKb or strength of the base. The equilibrium constant, Keq, which roughly measures how how readily the reactants will become the products, will.
Think of it like two children who want the same toy. Suzy has the toy. Billy wants to take it. How badly Billy wants the toy won't effect how much Suzy wants the toy, but it will effect who gets the toy in the end. (Example assumes that Suzy's desire of the toy doesn't change when she finds out Billy also wants it.)(6 votes)
- acetic acid basically is considered weak acid, then why is the k(eq) much greater than 1 ,even though acetic acid is a weak acid?(2 votes)
- K_a is much *less''than 1.
K_a = 0.000 0175, and this number is much less than 1.(2 votes)
- I'm terrible with math so this may be an incredibly dumb question but oh well:
in the first example to re-arrange the equation the - from -log was taken from the right to the left (makes sense) and resulted in 10^-.....
In the second equation the right side of the equation is positive - and yet we still end up with a 10^ -.....
can anybody explain why this is so? (probably a simple answer) :/(2 votes)
- Actually, in the second equation, the right side isn't positive. Since it's still a pKeq, it's still going to be -log(Keq). He just doesn't show that step. Rearrange the problem like you did with the first, and you get the same result as he did. I don't think that was a dumb question. :P(2 votes)
- At4:29, what does the crosshair-like symbol below the reversible double-headed arrow symbol indicate? It isn't present in the previous example. Thanks!(2 votes)
- What do we do when there are more than one Pkas in the acids, for example if i have a COOH at both ends of my molecule. COOH.........COOH with pka 1= 3 and pka2=4.3(2 votes)
- This is the pattern I'm noticing so far(can someone confirm, or give me an example when this is not the case?):
When determining the direction of equilibrium, and say I don't know the pKa values between the acid and its conjugate base, I would compare the two and the one with more resonance structures will be the one that's a better acid. Since more resonance = more carbons for the electrons to localize to, making the oxygen that those electrons belonged to less negative, which lessens the oxygen's need for the proton to stick around. Whereas in the acid that doesn't have as many resonance structures (or any), the electrons will stay localized to the oxygen, making it less likely to give up that proton. , and thus, a worse acid.(1 vote)
- At1:49we see that the other sides pKa is about 16. How do we know that? Is there a formula or is that a given number?(1 vote)
- pKa is the negative log of the equilibrium constant Ka, so -log(Ka). So you'd need to calculate the Ka and you can do that by measuring the concentrations of your reactants and products at equilibrium and plug them into the equilibrium expression for the deprotonation of water. Hope that helps.(1 vote)
- Could we also use pKb values to get the pKeq?(1 vote)
- [Voiceover] We've already seen the mechanism for this acid based reaction in the video on "Organic Acid Based Mechanisms." But I wanna run through it really quickly again. So on the left we have acetic acid which is gonna function as our acid, and then we have hydroxide over here, which will be our base. Our base is going to take the acidic proton on acetic acid, leaving these electrons behind on the oxygen, to give us the acetate anion here. And if we protonate OH minus, then we would form H2O. So those would be our two products. Let's think about the reverse reaction. So the acetate anion would function as a base, and take this proton on water, leaving these electrons behind. That would give us back acetic acid and hydroxide. So for the reverse reaction, the acetate anion here is functioning as a base, and water is functioning as an acid. So it's donating a proton. What if our goal was to find the equilibrium constant for the forward reaction? So what is the the equilibrium constant for the forward reaction? Which would be, the stuff on the left would be the reactants and the stuff on the right would, of course, be the products. So we could figure that out just using pKa values. So if we know the pKa values for the two acids in our reaction, we can figure out the equilibrium constant for that reaction. So we need to know the pKa of the acid on the left. So we already know that acetic acid is the acid on the left side here, and acetic acid has a pKa, this proton right here has a pKa of approximately five. On the right what's functioning as an acid? That's of course water. So what is the pKa of this proton right here on water? That's approximately 16. So let's plug that in to our equation. So the pKeq for the forward reaction is equal to the pKa of the acid on the left, which would be approximately five, minus the pKa of the acid on the right, which is approximately 16. So five minus 16 gives us a pKeq equal to negative 11. So how do we go from the pKeq to the Keq? Well we can do that because we know from general chemistry pKeq is equal to the negative log of Keq. So to solve for Keq first we need to put the negative sign on the left, so we have negative pKeq is equal to the log of Keq. And how do I get rid of that log? I have to take 10 to both sides. If I take 10 to both sides that gets rid of our log, so we know that Keq is equal to 10 to the negative pKeq. So we need to take our answer here for pKeq and we just need to plug that in to our equation here. So we get that the Keq, let me go ahead and put that right here, Keq is equal to 10 to the negative. So that negative sign that I just wrote here is this negative sign, 10 to the negative pKeq, which was negative 11, so 10 to the negative negative 11. Which of course is 10 to the eleventh. So the equilibrium constant for the forward reaction is equal to 10 to the eleventh. And we know what that means from general chemistry. We know that when K is much greater than one like this, at equilibrium we have way more products than we do reactants. So the equilibrium lies to the right, the equilibrium lies to the right, and we have a large amount of our products compared to our reactants. If you wanted to do that a faster way, if you just wanted to figure out which direction the equilibrium lies, look at your pKa values, let's go back up here, and you can see on the left it's five and on the right it's 16. Now we figured out that the equilibrium lies to the right, so therefore the equilibrium lies to the side that has the acid with the higher pKa value. So the equilibrium favors the weaker acid. So that's the short way of figuring out the position of equilibrium using pKa values. Here's another organic acid based mechanism that we've seen before. Acetone on the left functions as a base and takes a proton from H3O plus, which is hydronium, leaving these electrons behind on the oxygen. So hydronium functions as an acid. If you protonate acetone you would get this, so this must be the conjugate acid to acetone. And if you take away an H plus from hydronium you are left with water, so water must be the conjugate base to H3O plus. So for the reverse reaction if water functions as a base, water's going to take this proton leaving these electrons behind on the oxygen, giving us back acetone and forming hydronium, H3O plus. To use our pKa values to predict the position of equilibrium we need to find the pKa for the acid on the left and from that we subtract the pKa for the acid on the right. The acid on the left is hydronium and hydronium has a pKA of approximately negative two. The acid on the right, is right here and the pKa for this proton is approximately negative three. So to find the pKeq for the forward reaction right, for the forward reaction, meaning the stuff on the left is the reactants, and the stuff on the right is the products. The pKeq is equal to the pKa of the acid on the left, which would be negative two, minus the pKa of the acid on the right, which is negative three. So negative two minus negative three is equal to plus one. So the pKeq for the forward reaction is plus one. We know how to find the Keq because from the previous example we saw that the Keq is equal to 10 to the negative pKeq. So we plug in our pKeq, which is one, into here. And so we get the Keq for the forward reaction is equal to 10 to the negative first. So K is less than one, and we know what it means when K is less than one. That means that equilibrium, we have more reactants than products. So the equilibrium lies to the left. And we have more reactants than we do products at equilibrium. So we could also do this using the shorter way right? To figure out the position of equilibrium we could look at our pKa values and say, "Alright on the left we have negative two, on the right we have negative three." And we know that the equilibrium, the equilibrium favors the acid with the higher pKa value, favors the formation of the acid with the higher pKa value. And it's a little bit tricky cause we have two negative values for our pKa. But negative two is closer to zero than negative three, so negative two is the higher pKa. And our equilibrium favors the formation of the acid with the higher pKa. Our equilibrium favors the formation of the weaker acid. So the equilibrium lies to the left. So again that's the fast way of figuring out the position of equilibrium using your pKa values.