Henderson–Hasselbalch equation | Acids and bases | AP Chemistry | Khan Academy

The Henderson-Hasselbalch equation is an equation that's often used to calculate the pH of buffer solutions. Buffers consist of a weak acid and its conjugate base. So, for a generic weak acid, we could call that HA, and therefore its conjugate base would be A⁻.

To calculate the pH of the buffer solution, we would find the pKa of the weak acid, and to that, we would add the log of the concentration of the conjugate base divided by the concentration of the weak acid. Let's use the Henderson-Hasselbalch equation to calculate the pH of an aqueous buffer solution that consists of acetic acid and its conjugate base, the acetate anion.

Let's use this particulate diagram to help us calculate the pH of the buffer solution. Remember that the goal of a particulate diagram is not to represent every particle in the solution, but to give us an idea about what's going on in the entire solution. Also, when looking at the particulate diagrams of buffer solutions, water molecules and cations are often left out for clarity.

Let's count the number of particles of acetic acid in our particulate diagram. So, in our diagram, there are five particles of acetic acid. For the acetate anion, there are also five. Because there are five particles of both acetic acid and the acetate anion, the concentration of acetic acid is equal to the concentration of the acetate anion.

Next, let's think about the Henderson-Hasselbalch equation. Our goal is to calculate the pH of this buffer solution represented in the particulate diagram. First, we need to know the pKa of the weak acid, which is acetic acid. At 25 degrees Celsius, the Ka value for acetic acid is equal to 1.8 times 10 to the negative fifth. The Ka value is less than one because acetic acid is a weak acid.

To find the pKa of acetic acid, we take the negative log of the Ka value. So, the negative log of 1.8 times 10 to the negative fifth is equal to 4.74. We can go back to the Henderson-Hasselbalch equation and write that the pH is equal to the pKa, which we just calculated to be 4.74, plus the log of the concentration of the conjugate base, and the conjugate base is the acetate anions.

Let's write that in here, CH₃COO⁻, and that's divided by the concentration of the weak acid, which is acetic acid, CH₃COOH. We've already figured out that the concentration of acetic acid is equal to the concentration of the acetate anion; therefore, the concentration of the acetate anion divided by the concentration of acetic acid is just equal to one, and the log of one is equal to zero.

So, let's go ahead and write that in here. The log of one is equal to zero; therefore, the pH of the buffer solution is equal to 4.74 plus zero, or just 4.74. So, whenever the concentration of the weak acid is equal to the concentration of the conjugate base, the pH of the buffer solution is equal to the pKa of the weak acid.

Let's look at another particulate diagram. We still have an acetic acid-acetate buffer solution; however, this is a different buffer solution than the previous problem. So let's count our particles. For acetic acid, there are six particles, and for the acetate anion, there are only four.

Since we have more acetic acid particles than acetate particles, the concentration of acetic acid is greater than the concentration of the acetate anion. We can use the Henderson-Hasselbalch equation to think about the pH of this buffer solution. The pH is equal to the pKa, which we calculated in the previous problem for acetic acid; it's 4.74 at 25 degrees Celsius, plus the log of the concentration of the acetate anion divided by the concentration of acetic acid.

In this case, the concentration of acetic acid is greater than the concentration of the acetate anion; therefore, we have a smaller concentration divided by a larger concentration. So, we have a number less than one, and the log of a number less than one is negative. Therefore, all of this would be negative or less than zero, so we would be subtracting a number from 4.74. Therefore, we can say the pH of the solution would be less than 4.74.

Let's do one more particulate diagram of an acetic acid-acetate buffer solution. Once again, we count our particles. So, for acetic acid, this time there are four particles, and for the acetate ion, this time, there are six particles. Since we have only four particles of acetic acid and six particles of the acetate anion, the concentration of acetic acid is less than the concentration of the acetate anion, or we could say the concentration of the acetate anion is greater than the concentration of acetic acid.

Thinking about the Henderson-Hasselbalch equation once again, the pKa is equal to 4.74, and we need to think about the ratio of the concentration of the acetate anion to the concentration of acetic acid. For this example, the concentration of the acetate anion is greater than the concentration of acetic acid; therefore, the ratio would be greater than one, and the log of a number greater than one is positive or greater than zero.

Therefore, we would be adding a number to 4.74. So, for this buffer solution, the pH would be greater than 4.74. Finally, let's summarize what we've learned from our three different particulate diagrams.

In the first example, the concentration of the weak acid was equal to the concentration of the conjugate base, and therefore the pH of the buffer solution was equal to the pKa of the weak acid. In the second example, the concentration of the weak acid was greater than the concentration of the conjugate base, and therefore the pH of the buffer solution is less than the pKa of the weak acid.

In the third example, the concentration of the weak acid was less than the concentration of the conjugate base; therefore, the pH of the buffer solution was greater than the pKa of the weak acid. So, if we know the pH of a buffer solution, we can think about the Henderson-Hasselbalch equation to think about the relative concentrations of the weak acid and the conjugate base. For example, if we have a particular buffer solution and we know the pH of the buffer solution is less than the pKa of the weak acid, we know that in that buffer, at that moment in time, the concentration of the weak acid is greater than the concentration of the conjugate base.