8.9: Henderson-Hasselbalch Equation
Introduction to the Henderson-Hasselbalch Equation
This section is a relatively short one and focuses intimately on one equation: the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is useful because it helps us find the pH of a buffer. Thinking back to 8.8, a buffer is a solution that resists changes to its pH and is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. Let's take a look at the Henderson-Hasselbalch equation and get ourselves situated with it:
Breaking Down The Equation
Let's break down each piece of this one by one. pH as we know is the -log([H+]) and is oftentimes the measure that we're looking for when solving problems using the Henderson-Hasselbalch Equation. pKa is -log(Ka) and describes a logarithmic scale to describe an acids acidity (lower = more acidic). Finally, we get to the new bit, the log base 10 of the ratio of the concentrations of an ion, A-, and an acid HA. We can start by realizing that there is a unique relationship between [A-] and [HA]. They're a conjugate acid base pair! HA is a weak acid and A- is its conjugate base! This is where the Henderson-Hasselbalch equation ties into buffers because you will always have a concentration of conjugate base and a concentration of conjugate acid. It also shows why the strongest buffer is when these concentrations are equal because then log([A-]/[HA]) = 0.
Two Forms of the Henderson-Hasselbalch Equation. Image from MicrobeNotes
Example Problem #1: Directly Stated Buffer
Find the pH of a buffer with 0.5M CH3COOH mixed with 0.25M CH3COONa (Ka = 1.8 * 10^-5).
For this we can plug directly into the Henderson-Hasselbalch
Example Problem #2: Using the Hasselbalch During A Titration
Calculate the pH in the titration of 25.0 mL of 0.100M acetic acid with 0.100M NaOH after adding 15.0mL of 0.100M NaOH.
Let's start by writing out our net ionic equation for this reaction:
CH3COOH + NaOH <=> CH3COONa + H2O
CH3COOH + OH- <=> CH3COO- + H2O
Next, we can use stoichiometry to find how many mmol of each compound we have after the reaction goes forward:
CH3COOH + OH- <=> CH3COO- + H2O
Start: 2.5mmol CH3COOH, 1.5mmol OH-, 0mmol CH3COO-, 0mmol H2O
End: 2mmol CH3COOH, 0mmol OH-, 1.5mmol CH3COO-, 1.5mmol H2O
Because we have concentrations of both an acid and its conjugate base, we can find the pH of this by finding the pH of that buffer using the Henderson-Hasselbalch. Note that because we're dividing by the same volume to find concentration, they cancel out and we can just divide the mmols:
Example Problem #3: Preparing A Buffer
How many moles of CH3COONa would need to be added to 1L of water to be able to be added to a 1M solution of CH3COOH to form a buffer with a pH of 3 (Ka = 1.8 * 10^(-5))?
There's a lot happening here. Let's break it down piece by piece. First off, we're adding some mass to 1L of water, telling us that concentration will be involved. We're adding it to a 1M solution of acetic acid, so we have that concentration straight off the bat, and we are told we want our final solution to have a pH of 3. We know that CH3COO- and CH3COOH will form a buffer, so let's plug into the Henderson-Hasselbalch and see what we can find: