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Chemistry Buffer pH question

Mazza2000

Hi guys,

I was wondering whether someone could help me out with this question. Part c is what I need help on

2. a) Calculate the pH of 0.12 moldm-3 ethanoic acid (Ka = 1.7 x 10-5 moldm-3).
Answer is 2.85

b) Calculate the mass of sodium ethanoate (CH3COONa) which must be added to 500 cm3 this solution to give a buffer solution of pH = 4.60.
Answer is 3.33g

c) Calculate the pH of this solution after 0.01 moles of HCl are added.
I keep getting an answer of 1.76 by pretty sure this is wrong as the answer should be close to the original because this is a buffer solution?

Thanks in advance!

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

Reply 2

charco

Study Forum Helper

Original post by Mazza2000

ka = [H+][A-]/[HA]

(1.7 x 10^-5 x 0.12)/2.5 x 10^-5 = [A-] = 0.0816

This is in 500 ml so mol ethanoate = 0.0816/2 = 0.0408 mol
relative mass of sodium ethanoate = 82
therefore mass required = 0.0408 x 82 = 3.3 g
:smile:

Spoiler

c) Calculate the pH of this solution after 0.01 moles of HCl are added.
I keep getting an answer of 1.76 by pretty sure this is wrong as the answer should be close to the original because this is a buffer solution?

Thanks in advance!

If you add 0.01 mol HCl this is absorbed by the ethanoate ions in equal quantity (approximation)

hence mol of ethanoate is reduced by 0.01 mol
new mol of ethanoate = 0.0406 - 0.01 = 0.0306
and the mol of HA increases by 0.01 = 0.07

ka = [H+][A-]/[HA]

hence [H+] = ka[HA]/[A-]

[H+] =(1.7 x 10^-5 x 0.14)/0.0612 = 3.89 x 10^-5

pH= 4.41

Reply 3

Mazza2000

Original post by charco

Spoiler

If you add 0.01 mol HCl this is absorbed by the ethanoate ions in equal quantity (approximation)

hence mol of ethanoate is reduced by 0.01 mol
new mol of ethanoate = 0.0406 - 0.01 = 0.0306
and the mol of HA increases by 0.01 = 0.07

ka = [H+][A-]/[HA]

hence [H+] = ka[HA]/[A-]

[H+] =(1.7 x 10^-5 x 0.14)/0.0612 = 3.89 x 10^-5

pH= 4.41

Ahh ok Thank you so much for your help :smile:

(edited 6 years ago)

Reply 4

islamchadni

What is the pH of a buffer solution made by making up 1.00dm3 of solution containing 1.00 mole of ethanoic acid and 1.00 mole of sodium ethanoate? Ka for ethanoic acid=1.74 x 10-5moldm-3.

Reply 5

ChemistryWebsite

Original post by islamchadni

What is the pH of a buffer solution made by making up 1.00dm3 of solution containing 1.00 mole of ethanoic acid and 1.00 mole of sodium ethanoate? Ka for ethanoic acid=1.74 x 10-5moldm-3.

New question? Perhaps you would be better starting a new thread to get better engagement.

Reply 6

username3249896

Original post by islamchadni

What is the pH of a buffer solution made by making up 1.00dm3 of solution containing 1.00 mole of ethanoic acid and 1.00 mole of sodium ethanoate? Ka for ethanoic acid=1.74 x 10-5moldm-3.