Topic 8.9: Henderson-Hasselbalch Equation

The Henderson–Hasselbalch equation

You met this equation in the titration lesson, and today we’re going to own it. It starts from the K a expression for a weak acid, hydronium times the base partner, over the acid partner.

If I solve that for the hydronium, I get K a times the ratio of acid to base. Then I take the negative log of both sides. The negative log of hydronium is the pH, the negative log of K a is the p K a, and the minus sign flips the ratio over, so base ends up on top.

And that’s the Henderson-Hasselbalch equation, the pH equals the p K a plus the log of base over acid. You’ll never have to derive it on the exam. I just want you to see that it’s the K a expression, wearing a disguise. The p K a tells you which acid you’ve got, so at a given temperature it stays the same however you mix the buffer, and the ratio is the recipe, how much base partner you put in compared with acid partner.

When the partners are equal, the ratio is one, and the log of one is zero, not one, so the pH lands exactly on the p K a. That’s last lesson’s rule, that when the pH equals the p K a, half the particles still have their proton and half have lost it.

Here’s our acetate buffer with equal partners, and the pH sits right on the p K a, four point seven four. Now twice as much base as acid. The p K a is really four point seven four five before rounding, and the log of two is about point three, so the pH goes up to five point oh five. Flip it, twice as much acid as base, and it comes down by point three, to four point four four.

Ten times as much base, and since the log of ten is one, we’re a whole unit above, at five point seven four. More base pushes the pH above the p K a, more acid pulls it below, and the size of the push is the log of the ratio. Now you set it. Drag the mix until this buffer’s pH is five point two two.

Let me do one all the way through. A buffer has point two zero zero molar acetic acid and point one zero zero molar acetate, and K a for acetic acid is one point eight times ten to the minus five. Before I touch the calculator, I’m going to predict. There’s more acid than base, so the pH has to come out below the p K a, and if my answer lands above it, I’ll know I flipped something.

The p K a is the negative log of one point eight times ten to the minus five. That’s four point seven four five, and I’ll keep that extra digit until the end. Base over acid is point one over point two, with the molars cancelling.

That’s one half, and the log of a half is minus point three oh one, so four point seven four five minus point three oh one gives four point four four, below the p K a, just like I predicted. Halving the ratio moved the pH by point three. It didn’t halve the pH, because the ratio lives inside a log. Here’s one to do in your head. If a buffer has four times as much acetic acid as acetate, what’s its pH?

Mixing a buffer: count the moles

Nobody hands you the concentrations in real life. You pour two solutions together, so here’s thirty milliliters of point one zero zero molar acetic acid and twenty milliliters of point three zero zero molar sodium acetate, going into one beaker.

The tempting move is to take the labels, point three over point one, a ratio of three. But those are the concentrations inside the bottles, and we didn’t pour equal amounts. So let’s count moles. Thirty milliliters times one liter per thousand milliliters times point one zero zero moles per liter, the milliliters and liters cancel, and we get point zero zero three zero zero moles of acetic acid.

Same for the acetate, twenty milliliters times point three zero zero moles per liter, point zero zero six zero zero moles. Both end up in the same fifty milliliters, so when I divide each by that volume and take the ratio, the volume cancels, and the ratio of concentrations is just the ratio of moles, two. So now you finish it. What’s the pH of this mixture?

And the meter in the beaker agrees, five point oh five. Since only the ratio matters, a more dilute or more concentrated version of the same mix would have about the same pH.

Why a little acid barely moves the pH

Now we can finally answer the question from the start. Why did a whole teaspoon of strong acid move the buffer by only four hundredths? Here’s our liter of buffer again, with each token one millimole, a hundred acetic acids and a hundred acetates, and the teaspoon brings in five millimoles of hydronium. Each one gets caught by an acetate, so now there are ninety-five acetates and a hundred and five acetic acids, and almost no extra hydronium left over.

And that’s what the meters showed. The water ended at two point three, and the buffer at four point seven oh. The AP exam won’t ask you to calculate this, but it’s worth seeing once. The ratio went from a hundred over a hundred, which is one, to ninety-five over a hundred and five, about point nine oh five, whose log is minus point oh four, so the pH went from four point seven four to four point seven oh.

The ratio barely moved, so the pH barely moved. The water had no partner, so all five millimoles of hydronium stayed free and its pH fell more than two units. So a small amount of acid or base hardly changes the ratio, which means it hardly changes the pH, and the change is much smaller than it would be without the buffer. Let’s try it with a different buffer. If a little sodium hydroxide goes into an ammonia buffer, what happens?

Any buffer pair: ammonia, and choosing a buffer

Every pH so far was for acetic acid, but the equation works for any pair. Swap in ammonium for acetic acid, ammonia for acetate, and chloride for sodium, and the p K a jumps from four point seven four to nine point two five. Same two-to-one ratio, and now the pH is nine point five five.

So here’s a buffer to do on your own, point two five zero molar ammonia and point one zero zero molar ammonium chloride, and the p K a of ammonium is nine point two five. There’s a trap here. The equation says A minus on top, so people put chloride on top. But A minus means the base partner, the one without the extra proton, and here that’s ammonia, which has no charge, while chloride is just a spectator.

Okay, this whole one is yours. What’s the pH of this ammonia buffer? This also tells you how to choose a buffer. To hold a solution near pH nine, start from an acid whose p K a is near nine, like ammonium, so the ratio you need is close to one and both partners stay plentiful, and for pH four point seven you’d pick acetic acid.

Back to the runner: the equation in her blood

And that brings us back to the hall. That analyzer measured the pH and the carbon dioxide in the runners’ blood, and calculated the bicarbonate with Henderson-Hasselbalch. For blood, the base partner is bicarbonate and the acid partner is carbonic acid. But carbonic acid falls apart into carbon dioxide so readily that there’s very little of it at any moment, so blood chemists count the dissolved carbon dioxide instead, and with that pair the effective p K a at body temperature is six point one.

Before the race, the runners’ blood had a pH of seven point three two. What was the ratio of bicarbonate to dissolved carbon dioxide? Four minutes after the race, six point nine seven minus six point one is point eight seven, and ten to the point eight seven is about seven point four.

So the ratio fell from about seventeen to about seven. Every hydrogen ion the blood caught used up a bicarbonate, so the top of the ratio shrank. And this is where her breathing comes in. Every catch makes carbonic acid, which falls apart into carbon dioxide, and her lungs blow it away, so the bottom of the ratio doesn’t pile up. That’s one reason her pH fell only as far as it did.

An hour later the runners’ blood was back to seven point two eight, the ratio back near fifteen, as their bodies rebuilt their bicarbonate. In a beaker, every catch leaves the acid partner sitting there, but in blood it can leave through the lungs.

So let’s put it all together. A buffer is two partners, where the base partner catches added acid and the acid partner catches added base, and its pH is the p K a plus the log of base over acid. A little acid or base barely changes that ratio, and that’s why it barely changes the pH.

But the runners’ bicarbonate fell from about twenty-four to under seven in just four minutes. So what happens to a buffer when its base partner starts to run out? That’s buffer capacity, and it’s where we’re going next.