# When is the Henderson-Hasselbalch Equation used?

Jun 25, 2016

The Henderson-Hasselbalch equation is used when you are doing buffer calculations.

#### Explanation:

An acid buffer is a solution of a weak acid and its salt.

The equilibrium reaction is

$\text{HA + H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-}$

${K}_{a} = \left(\left[\text{H"_3"O"^+]["A"^"-"])/(["HA}\right]\right]$

The Henderson-Hasselbalch equation is

"pH" = "p"K_a + log((["A"^"-"])/"[HA]")

There are four variables in this equation: $\text{pH", "p"K_a, ["A"^"-"]}$, and $\left[\text{HA}\right]$.

You can calculate any one of them if you know the values of the other three.

Similarly, a basic buffer is a solution of a weak base and its salt.

The equilibrium reaction is

$\text{B + H"_2"O" ⇌ "BH"^+ + "OH"^"-}$

${K}_{b} = \left(\left[\text{BH"^+]["OH"^"-"])/(["B}\right]\right]$

The Henderson-Hasselbalch equation is

"pOH" = "p"K_b + log ((["BH"^+])/(["B"]))

There are four variables in this equation: $\text{pOH", "p"K_b, ["BH"^+]}$, and $\left[\text{B}\right]$#.

You can calculate any one of them if you know the values of the other three.

You can calculate $\text{[pH](https://socratic.org/chemistry/acids-and-bases/the-ph-concept)}$ from the relationship

$\text{pH" = "14.00 - pOH}$