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Chemistry Buffers help level 4

Hi,
I needed some help with a past paper question
29. Answer ALL parts of this question. Show ALL arithmetical steps and chemical equations.

(a) Calculate the pH of a buffer solution composed of 0.15M HF and 0.15M NaF. (Ka for HF is 6.8 x 10-4).
(8 marks)

(b) Calculate the ΔpH caused by the addition of 7.0 ml of 0.10M NaOH to 100ml of the buffer solution obtained in part (a) of this question.

any help will be appreciated
Reply 1
Avatar for h3rmit
h3rmit
15
Original post by HabibiMarwan
Hi,
I needed some help with a past paper question
29. Answer ALL parts of this question. Show ALL arithmetical steps and chemical equations.

(a) Calculate the pH of a buffer solution composed of 0.15M HF and 0.15M NaF. (Ka for HF is 6.8 x 10-4).
(8 marks)

(b) Calculate the ΔpH caused by the addition of 7.0 ml of 0.10M NaOH to 100ml of the buffer solution obtained in part (a) of this question.

any help will be appreciated

What will HF (partially) dissociate into, and what will the products of dissociation do? What will NaF dissociate into, and what will the important product do?

What will NaOH do in water, and how will that affect the pH?
Reply 2
Avatar for HabibiMarwan
HabibiMarwan
OP
Original post by h3rmit
What will HF (partially) dissociate into, and what will the products of dissociation do? What will NaF dissociate into, and what will the important product do?

What will NaOH do in water, and how will that affect the pH?


for part a HF will dissociate into H+ and F- and NaF into Na+ F- these products react with each other? so the concentration doesnt change? i got the ph as 3.16 but i dont think its right
Reply 3
Avatar for h3rmit
h3rmit
15
Original post by HabibiMarwan
for part a HF will dissociate into H+ and F- and NaF into Na+ F- these products react with each other? so the concentration doesnt change? i got the ph as 3.16 but i dont think its right


Your dissociation products are right, bit in bold not quite.

If you think about the equilibrium reaction: HF ⇋\leftrightharpoons H+ + F-, as you are adding more conjugate base, and the conjugate base will react with the conjugate acid, the conjugate acid amount, and hence concentration must change. (since you get less of the conjugate acid). Same goes for the conjugate base.

I haven't done the actual numbers but I'd be happy to look through your working.
(edited 7 years ago)
Original post by HabibiMarwan
for part a HF will dissociate into H+ and F- and NaF into Na+ F- these products react with each other? so the concentration doesnt change? i got the ph as 3.16 but i dont think its right


3.17

Attention to rounding! Don't waste marks!
Reply 5
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HabibiMarwan
OP
Was my answer right?! All I did was I plugged those concentrations into the Henderson hasselbach equation. It looked too simple for 8 marks
Reply 6
Avatar for HabibiMarwan
HabibiMarwan
OP
Original post by TeachChemistry
3.17

Attention to rounding! Don't waste marks!


Was my answer right?! All I did was I plugged those concentrations into the Henderson hasselbach equation. It looked too simple for 8 marks
Original post by HabibiMarwan
Was my answer right?! All I did was I plugged those concentrations into the Henderson hasselbach equation. It looked too simple for 8 marks


Your answer to part a is correct but I suspect for all 8 marks you need to do part b as well.

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