Question

applications of henderson equation

Answer 1

Example 1: Calculate the pH of a buffer composed of 0.1M acetic acid (CH3COOH) and 0.6M acetate (CH3COO-) knowing that the acid dissociation constant Ka is 1.8 x 10-5.

Step 1. Figure out what variables we have:

i.e we know that [HA] = [CH3COOH] = 0.1M (unionized species)

[A-]   = [CH3COO-]  = 0.6M (ionized species)

Ka = 1.8 x 10-5

Step 2. Analyze what we need for the Henderson-Hasselbalch equation

i.e. The Henderson-Hasselbalch equation is written as: pH = pKa + log ([A–]/[HA])

We already know the values for [A-] and [HA]. Thus we only need to convert the Ka into pKa.

Note- Make sure the units for the [A-] and [HA] are in molar (M) units. pKa = – log Ka = – log 1.8 x 10-5 = 4.7

Step 3. Insert all knowns into Henderson-Hasselbalch equation and calculate the unknown pH

pH = pKa + log ([A–]/[HA])
pH = 4.7 + log ( 0.6 / 0.1)
pH = 4.7 + log 6
pH = 4.7 + 0.78
pH =  5.48       (Answer)