Weak acid equilibria | Acids and bases | AP Chemistry | Khan Academy

Before we get into the topic of weak acid equilibria, let's look at a strong acid first. So let's say that H A is a strong acid and reacts with water to produce the hydronium ion and A minus, the conjugate base.

2H A—technically, the reaction comes to an equilibrium; however, the equilibrium favors the product so much that we don't draw an equilibrium arrow. We simply draw an arrow going to the right, indicating at equilibrium we have essentially all H3O plus and A minus and no H A. So we assume that a strong acid ionizes 100 percent in solution.

We can see that in our particulate diagrams. Keep in mind that these particulate diagrams are not meant to represent the entire solution, just a small portion of it, so we can get an idea about what's happening in the entire solution. We can think about the particulate diagram on the left representing the initial concentration of our acid, H A.

There are five H A particles in solution and also five water particles. Since the reaction goes to completion, the H A particles are going to react with the water particles to produce 5 hydronium (H3O plus) and 5 A minus. We can see that in our second particulate diagram, which represents the equilibrium mixture, and there are five hydroniums present and also five A minuses.

We can write the equilibrium constant expression for this reaction from the balanced equation. We look at our products and say it’d be the concentration of H3O plus, and since there's a coefficient of one, it’d be the concentration of H3O plus raised to the first power times the concentration of A minus, also raised to the first power, divided by the concentration of H A raised to the first power. Pure water is left out of the equilibrium constant expression, so that's equal to the equilibrium constant K.

Since this shows the ionization of an acid, we could call this the K_a value—the acid ionization constant or the acid dissociation constant. Since the equilibrium lies so far to the right for a strong acid, at equilibrium, there's going to be essentially all products and very, very little reactants. That means the K_a value for a strong acid will be much greater than one.

Now, let's look at a weak acid. We're going to use the same general equation here: H A plus H2O yields H3O plus plus A minus. However, if H A is a generic weak acid, we're going to include an equilibrium arrow here. The equilibrium arrow needs to be included because weak acids only partially ionize, and we can see that in the particulate diagrams.

In the particulate diagram on the left, this is supposed to represent the initial concentration of our weak acid H A, and there are five particles of H A and five particles of water. Since weak acids only partially ionize, let's say that only one H A particle reacts with one water molecule to produce one hydronium ion and one A minus anion.

So if one H A particle reacts with one H2O particle, that gives us one hydronium (H3O plus) and one A minus. So, out of the original five H A particles, only one of them ionized, leaving four H A particles still in solution at equilibrium. Therefore, H A only partially ionized and is a weak acid.

The equilibrium constant expression for a weak acid looks the same as it did for our strong acid because we use the same general equation. However, since a weak acid only partially ionizes, at equilibrium there's going to be a small amount of products and a large amount of reactants. Therefore, the K_a value for a weak acid will be less than one.

Let's look at some K_a values for some weak acids at 25 degrees Celsius. Hydrofluoric acid has a K_a value of 6.3 times 10 to the negative fourth, and acetic acid has a K_a value of 1.8 times 10 to the negative fifth. Since both of these have K_a values less than one, they're both considered to be weak acids. However, the higher the K_a value, the stronger the acid.

So, out of these two weak acids, since hydrofluoric acid has the higher K_a value, hydrofluoric acid is the stronger acid of the two. Now let's do a calculation using a weak acid. Let's say we have a 0.10 molar solution of benzoic acid, which is a weak acid, and the pH of the solution is 2.60 at 25 degrees Celsius.

Our goal is to calculate the K_a value for benzoic acid at 25 degrees Celsius. Since we're given the pH in the problem, we can go ahead and plug the pH of 2.60 into this equation. So, 2.60 is equal to the negative log of the concentration of hydronium ions.

To solve for the concentration of hydronium ions, first we move the negative sign to the left, which gives us negative 2.60 is equal to log of the concentration of hydronium ions. To get rid of the log, we take 10 to both sides. This gives us the concentration of hydronium ions is equal to 2.5 times 10 to the negative third molar.

It's also important to recognize this is the equilibrium concentration of hydronium ions. The next step is to write out the equation showing benzoic acid reacting with water. So here's the chemical formula for benzoic acid reacting with water to form the hydronium ion and the conjugate base of benzoic acid, which is the benzoate anion.

To help us find the K_a value for benzoic acid, we're going to use an ICE table where I stands for the initial concentration, C is the change in concentration, and E is the equilibrium concentration. In our problem, we were told the initial concentration of benzoic acid was 0.10 molar, and if we assume that benzoic acid hasn't reacted yet, the initial concentration of hydronium ion would be zero, and the initial concentration of benzoate would also be zero.

We just calculated the equilibrium concentration of hydronium ions, which we found to be 2.5 times 10 to the negative third molar, so we're going to go ahead and put that under hydronium ion in for the equilibrium concentration. So, if the initial concentration of hydronium ion was zero and the equilibrium concentration is 2.5 times 10 to the negative third, there was an increase in 2.5 times 10 to the negative third.

So we're going to write that under hydronium for the change in the concentration. Looking at the mole ratio of hydronium ion to benzoate, it's a one-to-one mole ratio. So if the hydronium ion increased by 2.5 times 10 to the negative third, so did the benzoate anions. I can go ahead and write here plus 2.5 times 10 to the negative third, which means the equilibrium concentration of benzoate anion would also be 2.5 times 10 to the negative third.

Since we saw an increase in the concentration of hydronium ion and benzoate, that increase in the concentration of the products came from the ionization of our weak acid. In the balanced equation, there's a 1 in front of benzoic acid, so if we write plus 2.5 times 10 to the negative third for hydronium and it's a one-to-one mole ratio from hydronium to benzoic acid, we would have to write minus 2.5 times 10 to the negative third since we lost some of the benzoic acid when it ionized.

Therefore, the equilibrium concentration of benzoic acid would be 0.10 minus 2.5 times 10 to the negative third. Since our goal is to calculate the K_a value for benzoic acid, the next step is to write out the equilibrium constant expression, which we get from the balanced equation. It would be the concentration of hydronium ions times the concentration of benzoate anions divided by the concentration of benzoic acid.

Since these are equilibrium concentrations, we can get the equilibrium concentrations from the ICE table and plug them into our equilibrium constant expression. So we can plug in the equilibrium concentration for hydronium ions, we can plug in the equilibrium concentration for benzoate anions, and we can plug in the equilibrium concentration for benzoic acid.

Here we have the equilibrium concentrations plugged into the equilibrium constant expression, and when we solve for K_a, we find that K_a for benzoic acid at 25 degrees Celsius is equal to 6.4 times 10 to the negative fifth. Notice that K_a is less than one because benzoic acid is a weak acid.

Finally, let's talk about the idea of percent ionization. To help us think about percent ionization, let's go back to our example with a strong acid. We started off with five H A particles, and all five of those H A particles ionized to produce five hydronium ions in solution and five A minus anions in solution.

Therefore, one hundred percent of our acid ionized, so we can write 100 percent ionization for this hypothetical strong acid H A. Now let's go back to our weak acid. For our weak acid, we started off with five H A particles, but only one of them ionized to produce one H3O plus and one A minus, so four remain unionized at equilibrium.

Therefore, one out of five ionized, so the percent ionization for this hypothetical weak acid would be twenty percent. Notice that the one particle of H A that ionized made one hydronium ion, so we can also use the concentration of hydronium ion at equilibrium to calculate percent ionization.

So, the equation for percent ionization is equal to the equilibrium concentration of hydronium ions divided by the initial concentration of the acid H A times 100 percent. Looking at our ICE table for our benzoic acid problem, the equilibrium concentration of hydronium ions was 2.5 times 10 to the negative third molar, and the initial concentration of our acid was 0.10 molar.

So, we can plug those into our equation. We plug in our concentrations and molar cancels, and we find for the final answer the percent ionization of benzoic acid is equal to 2.5 percent.