Henderson-Hasselbalch Equation — Overview & Examples

The Henderson-Hasselbach equation is a simple rearrangement of the equation for the equilibrium constant. Suppose that a Bronsted-Lowry acid $HA$ dissociates in solution as follows: $$\text{HA} (aq)\leftrightharpoons \text{H}^+ (aq) + \text{A}^- (aq)$$ The acid dissociation constant $K_a$ can be written as follows: $$K_a = \frac{\left[\text{H}^+\right]\left[\text{A}^-\right]}{\left[\text{HA}\right]}$$ Now, we rearrange: $$\left[\text{H}^+\right] = K_a\frac{\left[\text{HA}\right]}{\left[\text{A}^-\right]}$$ We then take the negative base-$10$ logarithm of both sides. Note that $\text{p}K_a = -\log_{10}K_a$ and that $\text{pH} = -\log_{10}\left[\text{H}^+\right]$: $$\text{pH} = \text{p}K_a + \log_{10}\frac{\left[\text{A}^-\right]}{\left[\text{HA}\right]}.$$ This is the Henderson-Hasselbach Equation. This rearrangement is very useful because it allows us to compute $\text{pH}$ or $\text{p}K_a$ from the other easily and see the relationship between $\text{pH}$ and $\text{p}K_a$ very intuitively. We see immediately that, $\text{pH} = \text{p}K_a$ if exactly half of the $\text{HA}$ has dissociated. If less than half has dissociated, then $\text{pH} < \text{p}K_a,$ and if more than half has dissociated, then $\text{pH} > \text{p}K_a.$

The above equation for acids also has an analogous equation for bases which is identical in form and is derived in exactly the same way. For a Bronsted-Lowry base $B$ with base dissociation constant $K_b$: $$\text{pOH} = \text{p}K_b + \log_{10}\frac{\left[\text{BH}^+\right]}{\left[\text{B}\right]}.$$

(Video) 17.1 Buffers | General Chemistry

by Chad's Prep

In this section of the video, Chad explains the Henderson-Hasselbalch equation. In the first part, he derives the formula. Then, he shows you how to calculate the pH of a buffer solution. Finally, he shows you how to calculate the pH when an acid or base is added to the buffer. After watching this video, you'll have a great understanding of Henderson-Hasselbalch.