pH solution calculation

Question

Calculate the pH of a solution prepared by adding 20.0mL of 0.25M acetic acid with 2.0mL of 0.75M potassium hydroxide solution.

in consideration of the Henderson-Hasselbalch equation, what would you expect to be the effect on the pH of the buffer if 50mL of carbonate-free water was added to dilute the buffer.

Answer 1

Reaction:

CH₃COOH + KOH ==> CH₃COOK + H₂O

The presence of the weak acid CH3COOH and the conjugate base CH₃COOK makes a buffer.

initial moles present:

CH₃COOH: 20.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.005 mols CH₃COOH

CH₃COOK: 2.0 ml x 1 L/1000 ml x 0.75 mol/L = 0.0015 mols CH₃COOK

Henderson Hasselbalch equation:

pH = pKa + log [CH₃COOK] / [CH₃COOH] and using a pKa = 4.75, we have...

pH = 4.75 + log (0.0015/0.005) = 4.75 + log 0.3 = 4.75 - 0.52

pH = 4.23

Adding 50 ml of carbonate-free water to this buffer should not change the pH because you are simply diluting the buffer and changing the concentrations of both the weak acid and the conjugate base by the same factor. Thus, the log of CH³COOK / CH₃COOH will be the same and the pH will not change. However, the buffer capacity will be significantly reduced.