Question

Undiluted Buffer:

Ka = 0.0000178

pKa = 4.75.

(a) Calculate the pH of the 1:5 diluted buffer if 40mL of the undiluted buffer is added to 160mL of water.

(I got a pH of 5.45, but I thought the pH wont change if its diluted?)

(b) Calculate the pH of the 1:50 diluted buffer if 20mL of the 1:5 diluted buffer is added to 180mL of water.

Please explain how you got the answers because im so confused right now. Thanks.

Answer 1

The ratio of the [salt] : [acid] will remain the same as in the original buffer.

According to the Henderson Hasselbalch equation,

pH = pKa + log [salt]/[acid]

The only thing that changes is the [H+], therefore we should be able to calculate the pH from the original pH.

We will assume that [A-] and [HA] ratio is 1 that means [A-] = [HA].

pH = pKa + log [A-]/[HA]

      = 4.75 + log 1 = 4.75 + 0 = 4.75

Original pH = 4.75
Original [H+] = 10^-4.75 = 1.7782x10^-5
(40 ml)(1.7782x10^-5) = (200 ml)(x)
x = 3.556x10^-6 M = [H+]
pH = - log 3.556 x10^-6
pH = 5.4489

b) In second case we are using 1:5 diluted buffer from above.

Original pH = 5.44
Original [H+] = 10^-5.44 = 3.63x10^-6
(20 ml)(3.63x10^-6) = (200 ml)(x)
x = 3.630x10^-7 M = [H+]
pH = - log 3.630 x10^-7
pH = 6.44