pH and pKa relationship for buffers (video)

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pH and pKa relationship for buffers

How the Henderson-Hasselbalch equation can be used to look at the ratio of conjugate acid and base using relationship between buffer pH and pKa.

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• cant we have buffers with weak bases??
  if yes, then what would be its general equation..(8 votes)
  • Chemical equation: B + H₂O ⇌ BH⁺ + OH⁻.
    Henderson-Hasselbalch equation: pOH = pKb + log([BH⁺]/[B})(18 votes)
• at, correct me if I’m wrong, but isn’t the concentration of water 55.56 M, not 1? 1:32(2 votes)
  • We consider H2O(l) as 1 because it is not going to be relevant in finding our pH or pOH. To find pH and pOH we are only worried about aqueous solutions, meaning we throw them into water. So if we are throwing them into water, the Hydrogen Ions begin to mix making more H2O(l) molecules. This is why we excluded H2O (L) in the ICE tables as well.(15 votes)
• Why are all these eqaution dealing with weak acids, why doesn’t it work for strong acids?! Thanks!(1 vote)
  • We can construct an acid dissociation expression for strong acids and calculate their Ka and pKa values. These values indicate that the products of a strong acid reaction are heavily favored compared to the reactants. So very large numbers for Ka and negative values for pKa for strong acids. This tells us that the reaction is essentially irreversible, or it only goes in one direction; towards the products with very little reactants reformed in the reverse direction.
    The Henderson-Hasselbalch equation works best with weak acids because since weak acids dissociate only partially, there are comparable amounts of the weak acid and its conjugate base. You can use the Henderson-Hasselbalch equation on strong acids but we’ll have a situation where [A-] >> [HA]. However it’s unnecessary to calculate the pH of strong acid solution using the Henderson-Hasselbalch equation since we can do so using stoichiometry of H+ and assume that strong acids dissociate ~100%. So it’s possible to use the Henderson-Hasselbalch equation, but it creates unneeded work when simpler methods are possible.
    Hope that helps.(4 votes)
• what happens to the buffer ratio if the pH is raised?(2 votes)
  • The Buffer ratio increases with an increase in pH.(1 vote)
• What do you call a reaction where the pH of two substances is brought closer to 7? *(1 vote)
• At, shouldn’t the pH and pKa be negative when converted into the exponent for 10 since pH=-log[H+]? 3:55(1 vote)
  • No, as the 10 is added due to the log of [A-]/[HA] which is positive. The log in pH is left untouched. If you alter that the equation, by removing the p as well, then it gets kind of complex to look at but it can be done. Then what you said would be right.(1 vote)
• At, she mentions the half equivalence point. Is this the same thing as the end point during a tritration? 5:46(0 votes)
• Calculate the composition of buffer solution containing CH3COOH
  and CH3COOK having a total concentration of 0.2 and pH equal to 4.4(ka=1.8×10-5)(0 votes)
• If exactly half the amount of a weak acid in solution is ionized at some pH, say 5, how do you find the pKa of the weak acid if you don’t know anything else?(0 votes)
  • The pKa is equal to the pH in that scenario(1 vote)

Video transcript

– [Voiceover] We’re gonna talk about the relationship between pH and pK_a and buffers. And specifically, we’re going to be talking mostly about this in terms of the Henderson-Hasselbalch equation. But before we go to the Henderson-Hasselbalch equation, which I am gonna assume you’ve seen before, and if not we have some other videos introducing it and also deriving it, let’s do a quick review of what exactly is a buffer. So a buffer is something that contains an aqueous solution. It’s something that contains both a weak acid, which generically we write as HA, and it also contains in equilibrium the conjugate base of our acid, so A minus. This is, as written, it’s the acid dissociation on reaction for HA and since it’s an equilibrium we can write an expression called K_a, which is just the equilibrium constant for this equation. And it just has a special name, ’cause it happens to be for the dissociation of an acid. K_a is just equal to H_three O plus times A minus all over the concentration of HA, and we don’t include water because it’s a pure liquid, so we assume that the concentration is always one. Based on this expression for K_a, we can, and do, in the separate video, derive the Henderson-Hasselbalch equation. So the Henderson-Hasselbalch equation just says that the pH is equal to the pK_a plus the log of A minus over HA, where HA is our weak acid and A minus is its conjugate base. And as you can see up here, an acid and its conjugate base are just related by the fact that the acid has an extra H. We can rearrange the Henderson-Hasselbalch equation to get a lot of different kinds of information. One kind of problem you see a lot is for some buffer, you know, they might ask you, oh, what’s the pH? And then that means you probably know the pK_a and you know the concentrations of A minus and HA. The other thing that you can use the Henderson-Hasselbalch equation to tell you is the relationship between A minus and HA which is something you might wanna know. A lot of times, you just wanna know, you know, what’s in your solution, depending on what you wanna do to your solution, if you wanna add things to it, maybe you wanna add some acid, you wanna add some base. You wanna know what’s going on. The Henderson-Hasselbalch equation gives you a really quick and easy way of doing that. So what we’re gonna do, is we’re gonna rearrange this equation to solve for this ratio that we might be interested in. So we’re gonna subtract pK_a from both sides and that gives us a log of A minus over HA. And I don’t know about you, but I actually find, well, (laughs) I find logs not super-intuitive sometimes. So I’m actually going to get rid of the log by raising both sides to the 10th power. So that gives us 10 to the pH minus pK_a is equal to A minus over HA. So what does this tell us? It may not look like it tells us a whole lot more, but actually, it tells us a lot. It tells us about the relative relationship and size between A minus and HA concentration. And it’s saying that these two things are related to the relative size of pH and pK_a. So if we look at this, we can derive a couple relationships. We’re saying that pH versus pK_a, this relationship, can tell us about A minus over HA, which is a ratio, but that in turn we can relate to just the relative size of HA versus A. So let’s go ahead and look at all the possible scenarios for these three things. We’re going to start with the simplest possible scenario, which is that pH is equal to pK_a. When pH is equal to pK_a, we’re raising 10 to the zeroth power. So anything to the zeroth power is equal to one. Which tells us that this ratio is equal to one. And if A minus concentration over HA concentration is equal to one, that means that they have the same concentration. I forgot a minus sign there. This is a really helpful thing to remember. Any time you have a buffer and the pH of your solution is equal to the pK_a of your buffer, you immediately know that the concentration of your acid and its conjugate base are the same. And this comes up a lot not just when you’re talking about buffers by themselves, but also when you’re doing titrations. And the point in your titration where the HA is equal to A minus is called the half-equivalence point. Half-equivalence point. And if you haven’t learnt about buffers, that’s okay. Oh, sorry, if you haven’t learn about titrations yet, that’s fully fine. Just ignore what I just said (laughs), but if you have, the moral is just that, this is a really, really important relationship that is really helpful to remember. And I said really a lot there. So there are two other possibilities for pH and pK_a. We can have a pH that’s greater than pK_a for your buffer, and you can have a pH that is less than you pK_a for your buffer. So if your pH is bigger than your pK_a, then this term up here, 10 to the pH minus pK_a, is going to be positive. And when you raise 10 to a positive number, when you raise 10 to a positive number, you get a ratio that is greater than one. So if our ratio A minus over HA is greater than one, that tells us that A minus, the numerator, is actually greater than the denominator, HA. So if you know the pH and you know it’s bigger than the pK_a of your buffer, buffer’s acid, to be more specific, then you immediately know that you have more conjugate base around than your acid. The last scenario, pH is less than pK_a, well in that case, we’re raising 10 to a negative number, because we’re subtracting a bigger number from a smaller number. And that means that our ratio, A minus over HA, is actually less than one. So that tells us that our denominator, HA, is actually bigger than our numerator. So just to wrap up, we can look at the Henderson-Hasselbalch equation and we can just look at the relationship between pK_a and pH, and depending on whether they’re equal to each other or one is bigger than the other, we can immediately know what the relationship is between our acid and its conjugate base in our solution. It’s very easy to derive. We did it in just a few minutes, so it’s okay if you don’t remember this all the time. I usually just remember that pH equals pK_a means you have the same concentration, and then, if I forget, I will derive the other relation.