Question

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.220 M sodium benzoate. How much of each solution should be mixed to prepare tis solution?

Answer 1

The pH of a buffer is calculated using hendersen equation

pH = pKa + log [conjugate base]/[acid]

Given pKa = 4.2 and pH = 4.0

Thus

4.0 = 4.2 + log [conjugate base]/[acid]

Hence   log [conjugate base]/[acid] = -0.2

and  l[conjugate base]/[acid] = 0.63

NO total volume of buffer = 100mL

Let the volume of conjugate base , benzoate solution be V mL ith concentration 0.220M

then the volume of 0.100M benzoic acid is (100-V) mL.

Then [conjugate base in the buffer = molarity x volume / total volume

= V mL x 0.220 /100

[acid] in buffer = (100-V) mL x 0.100M /100

Thus the ratio

[conjugate base]/[acid] = 0.63 = V x0.22 /(100-V) x0.1

Thus solvig for V , v = 22.26mL and

volume of sodium benzoate solution= 22.26mL

volume of benzoic acid solution = 77.74 mL

and total buffer volume = 100mL