What is the pH of the buffer solution after addition of strong acid?

Question

Buffer solution:
5 mL 0.20M CH3COOH
5 mL 0.20M NaCH3COO
upon addition of strong acid: 0.05 mL 1.0M HCl

Calculate pH of solution.

Ka is given as: 1.8x10^-5

and why is 5ml x 0.20M = 0.10mmol?
shouldn't it be... 1 mmol?

Answer 1

5mL x 0.20M = 1mmol of acid and base of buffer

0.05mL x 1.0M = .05mmol of strong acid (HCl)

1-0.05= 0.95mmol of the base of the buffer left in solution
1+0.05= 1.05mmol of the acid of the buffer now in solution

pH = 4.74 (pka of weak acid) + log (0.95/1.05)

Answer 2

The pKa & Ka are given for a temperature of 298K
pKa(Ethanoic Acid) = 4.76
Ka(CH3COOH) = 10^-4.76 = 1.7378 x 10^-5
&
Ka = [H+][CH3COO-] / [CH3COOH]
Ka = [H+]^2 / [CH3COOH]
[H+]^2 = Ka x [CH3COOH]
[H+] = sq rt{Ka x [CH3COOH]}
[H+] = sq rt{1.7378 x 10^-5 x 0.20}
[H+] =sq rt{3.4756 x 10^-6}
[H+] = 1.8643 x 10^-3
pH = -log(10)1.8643 x 10^-3 = --2.73 = 2.73
This the pH of the aqueous acid without any additions.
As an equal amount of ethanoate ions is added pH is:-
Ka = [H+][CH3COO-] / [CH3COOH]
[H+] = Ka x [CH3COOH] / [ CH3COO-]
[H+] = 1.7378 x 10^-5 x 0.20 / 0.20
[H+] = 1.7378 x 10^-5
pH = 4.76

However:-
On adding the strong acid HCl
moles(HCl) = 1.0 x 0.05 / 1000 = 5.0 x 10^-5 moles
moles(Eth Ad/Eth ion) = 0.2 x 5 / 1000 = 1.0 x 10^-3 each
The H+ of the strong acid re-associates with the ethanoate ions. This reduces the ethanoate ions and increases the ethanoic acid.
moles(ethanoate) = 1.0 x 10^-3 - 5.0 X 10^-5 = 9.5 x 10^-4
moles(ethanoic acid) = 1.0 x 10^-3 + 5.0 X 10^-5 =
1.05 x 10^-3
So
Ka = [H+][CH3COO-] / [CH3COOH]
[H+] = Ka x [CH3COOH] /[CH3COO-]
[H+] = 1.7378 x 10^-5 x 1.05 x 10^-3 / 9.5 x 10^-4 =
[H+] = 1.9207 x 10^-5
pH = 4.72

NB The pH has changed very slightly from 4.76 to 4.72, this is because of the addition of the strong acid, but the pH change has been resisted.

Answer 3

a)
5 mL x 0.20 M = 0.1 mmol of acid
5 mL x 0.20 M = 0.1 mmol of base


0.05 mL x 1.0 M = 0.05 mmol of acid HCl

The HCl reacts with base
0.1 - 0.05 = 0.05 mmol of base left.
0.1 + 0.05 = 0.15 mmol of acid left

Buffer pH formula:
pH = pka + log ([Base] / [Acid])

Acetic acid: pka = 4.74

pH = 4.74 + log (0.05/0.15)
pH = 4.26
The pH decreases 0.48 units (solution is more acid)

bye and Happy new year