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# Chemistry Buffer pH question watch

1. 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).

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.

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?

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3. (Original post by 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).

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.
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

Spoiler:
Show

we agree

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?

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
4. (Original post by charco)
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

Spoiler:
Show

we agree

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

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Updated: February 8, 2018
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