Question

a) Calculate the pH of 75 mL of the undiluted buffer solution after the addition of 1 mL of 3 M HCl.

b) Calculate the pH of 75 mL of the undiluted buffer solution after the addition of 15 mL of 3 M HCl.

Buffer solution was made from 1 M of sodium acetate, and 1.1005 M of acetic acid. (pKa of acetic acid 4.75). pH of buffer = 4.79.

Answer 1

According to the given data: [NaOAc] = 1 M and [AcOH] = 1.1005 M

Pka of AcOH = 4.75

According to Henderson-Hasselbulch equation: pH = pKa + Log([NaOAc]/[AcOH])

i.e. pH = 4.75 + Log(1 M/1.1005 M)

i.e. pH = 4.75 + Log(0.91)

i.e. pH = 4.75 + (-0.04)

i.e. pH = 4.71

But the given pH of the buffer = 4.79

This calculated pH is not matching to give pH, which indicates there is some thing wrong in the given data.

Please cross-check the data, and proceed further. Here, I will provide you the procedure how to calculate pH after the addition of a strong acid like HCl.

a) The no. of mmol of NaOAc (n_NaOAc) = 75 mL * 1 mmol/mL = 75 mmol

The no. of mmol of AcOH (n_AcOH) = 75 mL * 1.1005 mmol/mL = 82.5375 mmol

Now, the no. of mmol of added HCl (n_HCl) = 1 mL * 3 mmol/mL = 3 mmol

According to Henderson-Hasselbulch equation: pH = pKa + Log{(n_NaOAc - n_HCl)/(n_AcOH + n_HCl)}

i.e. pH = 4.75 + Log{(75 - 3)/(82.5375 + 3)}

i.e. pH = 4.75 - 0.075

i.e. pH = 4.675

b) The no. of mmol of added HCl (n_HCl) = 15 mL * 3 mmol/mL = 45 mmol

According to Henderson-Hasselbulch equation:

pH = 4.75 + Log{(75 - 45)/(82.5375 + 45)}

i.e. pH = 4.75 - 0.63

i.e. pH = 4.12

i.e. pH = 4.675