Topic 8.3: Weak Acid and Base Equilibria

Weak acids: the face in the bog, and what weak means

Take a good look at this face. The man it belongs to lived about two thousand four hundred years ago. You can still see the wrinkles on his forehead, his eyelashes and even the stubble on his chin. It’s as if he’d only just fallen asleep. People call him Tollund Man, and in 1950 two brothers cutting peat for fuel found him in a bog near Silkeborg, in Denmark.

He lay in a bog like this for more than two thousand years, and the bog kept him, because the water in the peat is cold, almost free of oxygen, and acidic, often about as acidic as orange juice. The museum that keeps him compares it to pickling. It’s the way a cucumber keeps for months in vinegar, because the acid holds back many of the microbes that make things rot.

In some bogs the water even dissolved the bones away, and left only the skin and hair. Most of that acid comes from this, sphagnum moss, which gives acid off into the water as it grows, and makes even more as it rots into peat. So let’s go down into that water, past the moss, all the way to the molecules, and see what that acid is actually doing.

These are acid groups along one of the moss’s big molecules, and each one has a hydrogen it could hand over to a water molecule. Here and there one of them does, and the water that catches it becomes hydronium, but look at all the others, because at any moment, most of them are still holding on. Chemists call acids like this weak acids, which is a strange name for something that helped keep a man’s face for twenty-four centuries.

This little group of atoms, a carbon with two oxygens and a hydrogen, is the same acid group that’s in vinegar, so we’ll do our math with vinegar’s acid, acetic acid, one of the simplest molecules that carries it. So how acidic does this make the water? And what fraction of its molecules actually let go? And at the very end, I’m going to pour water into an acid like this. You’ll be able to predict something that sounds completely backwards.

So let’s start with one molecule of acetic acid, among the water molecules it’s dissolved in. The hydrogen it can give away is this one, on the oxygen at the end. When it lets go, that hydrogen leaves its electron behind, so it’s really just a proton, and a water molecule grabs it. The water becomes hydronium, H three O plus, and the rest of the acid becomes acetate, with a minus charge. And a hydronium can hand the proton right back, so protons are hopping in both directions all the time.

Now you try it. Acetic acid has four hydrogens, but only one of them is the one it gives away, so find it and drag it onto a water molecule.

Now let’s pull back to a thousand of these molecules, at point one oh oh molar, and count how many of them have ionized at any moment, because the AP exam asks you to count particles exactly like this. There they are, these glowing pairs, only about thirteen out of a thousand, and every other molecule is still whole acetic acid.

It isn’t the same thirteen for long, because protons keep jumping on and off, but for every new pair, another pair recombines somewhere else, so the count holds steady, and that’s equilibrium for a weak acid. Now compare hydrochloric acid at the same concentration, where every single molecule hands its proton to water.

That’s the whole difference between a strong acid and a weak one. Both of these hold the same amount of acid, so weak isn’t about how much acid there is. It means the acid only partly ionizes, and at this concentration, most of it is still whole molecules.

This exact picture is a classic AP exam question, where they draw you a box of particles and ask whether it could be a weak acid. So look at these four boxes, each one a tiny sample of a different acid solution, and decide which one shows a weak acid, HA, in water.

Writing weak down: Ka and pKa

Let’s write down what we just watched, acetic acid plus water making hydronium plus acetate, with a double arrow, because the protons go both ways. A lot of students write a single arrow here out of habit, but that says every molecule reacts, and only about one in seventy-five does.

Chemists write any weak acid as HA, and what’s left when it gives up its proton, A minus, is its conjugate base. The equilibrium constant is Ka, the acid ionization constant, with the ions on top and the acid that’s still whole on the bottom, and water is left out, because it’s the solvent and its concentration barely changes.

For acetic acid at twenty-five degrees, Ka is one point eight times ten to the negative fifth, and a number that small means that at a concentration like point one oh oh molar, most of the acid stays as molecules. Chemists often report pKa instead, the negative log of Ka, which for acetic acid is four point seven four, and if a problem gives you pKa instead, you just undo the log, because Ka is ten to the negative pKa.

Here are three weak acids side by side. Formic acid, which some ants spray, has a Ka ten times bigger than acetic acid’s, and hypochlorous acid, which chlorine makes in swimming pools, has one about six hundred times smaller.

A bigger Ka means more of the acid lets go, so it’s a stronger acid, and because of the minus sign in the log, a stronger acid has a smaller pKa, like golf, where the lowest score wins. So if we made all three of these at the same concentration, which solution would have the lowest pH?

Every species in 0.100 M acetic acid

Now let’s do the calculation for that point one oh oh molar acetic acid, and find its pH, the percent ionized, and every species in the water. We’ll use an ICE table, for initial, change and equilibrium. We start with point one oh oh molar acid and essentially no ions. Some of the acid ionizes, and we’ll call that amount x, so the acid goes down by x and each ion goes up by x.

At equilibrium, that leaves point one oh oh minus x of the acid, and x of each ion. In the Ka expression that’s x squared over point one oh oh minus x, and it equals one point eight times ten to the negative fifth. Now here’s a shortcut that works most of the time, as long as we check it afterward. Ka is tiny, so x is tiny next to point one oh oh. So we can just use point one oh oh on the bottom.

That leaves x squared equals one point eight times ten to the negative fifth, times point one oh oh, so x is one point three four times ten to the negative third molar. Let’s check that the shortcut was fair. x over the starting concentration, times a hundred, is one point three percent, and anything under five percent is small enough to ignore. And that’s also the percent ionization, the fraction of the acid that let go. Since x is the hydronium concentration, the pH is the negative log of x, which is two point eight seven.

Now every species. The acid that’s left is about point oh nine nine molar, and hydronium and acetate are each one point three times ten to the negative third. There’s even a little hydroxide, because hydronium times hydroxide always equals Kw, so the hydroxide is seven point five times ten to the negative twelfth molar, and the pOH is eleven point one three.

If acetic acid were a strong acid, the hydronium would be the full point one oh oh molar and the pH would be one point oh oh, so two point eight seven, less acidic but still acidic, is just what a weak acid should give. And notice that the percent ionized and Ka are different numbers, and before the end of this lesson you’ll see the percent change while Ka stays exactly the same. And here’s that same solution again, thirteen ion pairs in a thousand, the same one point three percent the ICE table gave us.

Working backward from a pH meter

Chemists often work backward. Say a point oh five oh oh molar solution of an unknown weak acid has a pH of three point oh oh, and you want its Ka. Think about the first step before I show you. The pH gives us the hydronium directly, one point oh times ten to the negative third molar.

That’s x, so the ICE table fills itself in. A minus is also x, and the acid left is point oh five oh oh minus point oh oh one oh, or point oh four nine oh molar.

So Ka is x squared over point oh four nine oh, which is two point oh times ten to the negative fifth. Now you work out the percent ionization, x over the starting concentration. It’s two point oh percent.

In fact, you can get the percent ionization from any species you know, even from how much acid is left, because whatever’s missing from the acid is what ionized. Try this one. A point oh two oh oh molar solution of a weak acid has point oh one nine oh molar of the acid left at equilibrium, so what percent of it ionized?

Weak bases: the ammonia fountain

Now let’s turn from acids to bases, with one of the most famous demonstrations in chemistry. This upside-down flask is full of ammonia gas, with a tube running down into water, and the water holds an indicator, phenolphthalein, that turns pink when a solution is basic. We squeeze a few drops of water up into the flask. Now keep an eye on that tube.

Ammonia dissolves so easily that the gas in the flask vanishes into those drops and the pressure falls, so the air outside pushes the water up, and there it goes! The moment the water reaches the flask it turns pink, so now it’s basic. But ammonia has no oxygen at all, so where did the hydroxide come from? Let’s go down into that pink water and find out.

Instead of giving a proton away, ammonia takes one, right off a water molecule. The ammonia becomes ammonium, N H four plus, and the water is left behind as hydroxide. So the oxygen in that hydroxide came from the water, which is how a molecule with no oxygen can still make a solution basic.

Now you make one. Take a proton from one of the water molecules and give it to the ammonia. And ammonia is weak too, so in point one oh oh molar ammonia only about thirteen in a thousand molecules have taken a proton, and the rest are still plain ammonia.

Here’s that reaction written out, ammonia plus water going back and forth to ammonium and hydroxide, and in general, a weak base B makes H B plus and hydroxide. Its constant is Kb, the base ionization constant, and for ammonia it’s one point eight times ten to the negative fifth, a pKb of four point seven four. And that one point eight times ten to the negative fifth is exactly the same as acetic acid’s Ka, which is going to be handy in a second.

So let’s find the pH of point one oh oh molar ammonia, with the same ICE table, except that x is now ammonium and hydroxide. Same constant, same concentration, so we get the same x, one point three four times ten to the negative third molar, but this time x is the hydroxide. If you take its negative log and call that the pH, you get two point eight seven, which would make ammonia an acid.

That number is really the pOH, so the pH is fourteen minus two point eight seven, or eleven point one three. The hydronium is Kw over the hydroxide, seven point five times ten to the negative twelfth, and a strong base like sodium hydroxide at this concentration would have a pH of thirteen point oh oh. Now you try one. What’s the pH of a point oh two oh oh molar ammonia solution?

Partners: Ka × Kb = Kw

Back to acetic acid and acetate, which differ by one proton. Acetate can act as a base, taking a proton from water and making a little hydroxide, so acetate has its own Kb.

Now watch what happens when we multiply acetic acid’s Ka by acetate’s Kb. The acetate cancels, the acetic acid cancels, and all that’s left is hydronium times hydroxide, which is Kw. Add the two reactions and the same thing happens, because H A and A minus show up on both sides. So they drop out, leaving just water reacting with itself.

So for any conjugate pair, Ka times Kb equals Kw, and pKa plus pKb equals fourteen, at twenty-five degrees. For acetate, Kb is Kw over one point eight times ten to the negative fifth, which is five point six times ten to the negative tenth, so its pKb is nine point two six.

Their product never changes, so the two act like a seesaw, and the stronger the acid, the weaker its conjugate base. This only works for a real pair, though, so acetic acid’s Ka times ammonia’s Kb doesn’t give Kw, because they aren’t one proton apart, and nothing cancels.

Let’s see if you can do one with formic acid. Its Ka is one point eight times ten to the negative fourth, so what’s Kb for its conjugate base, formate?

The backwards prediction: pour in water

Now for the backwards prediction I promised you. Here’s a hundred milliliters of our point one oh oh molar acetic acid, with its thirteen ion pairs in a thousand. This bright box marks off a hundred milliliters of space and stays put, while we pour in water until the solution fills a liter, ten times the volume.

Before we pour, make your prediction. When the volume goes up ten times, what happens to the percent of the acid that’s ionized, and to the hydronium concentration? Go ahead and pour by dragging the water slider up to a full liter. Keep an eye on both counters as you go.

In the whole solution, thirteen ion pairs became about forty-two, so more of the acid ionized, four point two percent instead of one point three. But the box that held thirteen ion pairs now holds about four, so the hydronium concentration went down, and the pH went up.

Here are the numbers. Before, we had point one oh oh molar acid, one point three percent ionized, at a pH of two point eight seven. These are the shortcut values, and the exact numbers in the readout agree with them to two significant figures.

Ten times the volume makes the acid ten times more dilute, point oh one oh oh molar, and the square root of Ka times that concentration gives x, four point two four times ten to the negative fourth molar. That fills in the new row, four point two percent ionized, still under five, and a pH of three point three seven.

In moles, before the water went in: one point three four times ten to the negative third molar, times a tenth of a liter, is one point three four times ten to the negative fourth moles of hydronium. After: four point two four times ten to the negative fourth molar, times a full liter, is four point two four times ten to the negative fourth moles. That’s about three times as many hydronium ions, spread through ten times as much water.

And here’s why more of it ionizes. Adding the water drops every concentration to a tenth, and the reaction quotient has two of them on top and only one on the bottom, so Q falls to a tenth of Ka, and the reaction runs forward until Q catches back up. If the AP exam asks why percent ionization goes up when you dilute a weak acid, this is the sentence that earns the point.

Diluting makes Q smaller than Ka, so more of the acid ionizes, but the volume grows so much that the hydronium concentration still falls. And Ka itself never changed, because it depends only on the acid and the temperature.

Let’s dilute it ten times more, all the way to ten liters. That makes the acid point oh oh one molar. The shortcut gives x as the square root of Ka times C. That’s one point three four times ten to the negative fourth molar.

It’s thirteen percent of the acid, way over five. The more you dilute, the bigger x gets next to C. The shortcut stops working. Solve it without dropping x, and you get one point two five times ten to the negative fourth molar. That’s twelve and a half percent, an even bigger fraction.

Back to the bog

So let’s go back to the bog. The acids here are weak. They only partly ionize, and in this water, at any moment, most of their acid groups are still holding on to their protons. You can now use Ka or Kb and a concentration to work out how acidic a solution is and what fraction has let go, and you know why adding water makes more of it let go, even as the water gets less acidic.

Weak only ever meant partly ionized, and given enough acid, cold water and twenty-four centuries, weak acids helped keep this face, so that we can still look at it today. Before you go, have a look around the bog. You can even dive back down into its water.