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# Henderson–Hasselbalch equation

One way to determine the pH of a buffer is by using the Henderson–Hasselbalch equation, which is pH = p

*Kₐ*+ log([A⁻]/[HA]). In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid–base pair used to create the buffer solution. When [HA] = [A⁻], the solution pH is equal to the p*Kₐ*of the acid. Created by Jay.## Want to join the conversation?

- hi there, may i know what about basic buffer solutions? are they not required to know?

i would like to understand the basic buffer solutions, it's equilibrium equations, formulas relating to basic buffer solutions and if there is a Henderson-Hasselbalch equation for basic buffers. Thank you very much!(2 votes)- A basic buffer solution is simply one where the pH > 7. So fundamentally it's no different from the buffer system shown in this video. We use the same Henderson-Hasselbalch equation and can use the same acetic acid/acetate solution if we wanted to. To get a basic pH we just need to adjust the concentrations of the acid and conjugate base correctly.

Hope that helps.(5 votes)

- Is it correct to have the equation written with the dominant pH in the denominator? Example if at around0:36, the solution had more base then the equation would be pH=pKa+log([HA]/[A-]).(1 vote)
- Not really. The Henderson-Hasselbalch equation is derived from the equilibrium expression of a weak acid dissociation reaction.(2 votes)

- Hi, how do we immediately get the Ka for the weak acid in a buffer solution? Is it in the official formula table for AP Chemistry?(1 vote)
- Each acid has its own acid dissociation constant, Ka. Other than calculating the Ka of an acid using the equilibrium expression and concentrations, these are just constants which would be given for a specific problem.(2 votes)

## Video transcript

- [Instructor] The
Henderson-Hasselbalch equation is an equation that's often used to calculate the pH of buffer solutions. Buffers consists 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. And 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. And 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, and 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. And so 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. So 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, so 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. So 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 anion, 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. And 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.