Can someone please help me with this question about buffers?

A biochemist plans to make up a buffer solution with a pH of 5.000.
The biochemist adds solid sodium ethanoate, CH3COONa, to 400 cm3 of 0.200 mol dm–3
ethanoic acid.
Ka for ethanoic acid = 1.75 × 10–5 mol dm–3
Calculate the mass of sodium ethanoate that the biochemist needs to dissolve in the ethanoic
acid to prepare this buffer solution.
Assume that the volume of the solution remains constant at 400 cm3 on dissolving the sodium
ethanoate.

The mark scheme says that the answer is 11.48g but I don't understand how they got there and the calculations in the mark scheme are very confusing. Can someone help?

I've worked it out and got the correct answer, where have you gotten upto so far and i can try and explain it to you

All I've done is calculate the concentration of the buffer and write the equation for the buffer. I'm not sure where to go from there.

1. Rearrange the equation - [CH3COO-] = Ka[CH3COOH] / [H+]
2. [H+] = 10^-5 = 1 x 10-5
3. [CH3COO-] = 1.75 x 10-5 x 0.2 / 1 x 10-5 = 0.35 mol dm-3
4. 0.35 x 400 x 10-3 = 0.14 moles
5. 0.14 x 82 (Mr of sodium ethanoate) = 11.48g

i hope this makes a bit more sense, if you don't get anything i'll try and explain!

Thank you so much! I was trying to figure out where the 0.35 came from but now i see it. Thanks a bunch

A biochemist plans to make up a buffer solution with a pH of 5.000.
The biochemist adds solid sodium ethanoate, CH3COONa, to 400 cm3 of 0.200 mol dm–3
ethanoic acid.
Ka for ethanoic acid = 1.75 × 10–5 mol dm–3
Calculate the mass of sodium ethanoate that the biochemist needs to dissolve in the ethanoic
acid to prepare this buffer solution.
Assume that the volume of the solution remains constant at 400 cm3 on dissolving the sodium
ethanoate.

The mark scheme says that the answer is 11.48g but I don't understand how they got there and the calculations in the mark scheme are very confusing. Can someone help?

where did you get the markscheme from? i really need it

1. Rearrange the equation - [CH3COO-] = Ka[CH3COOH] / [H+]
2. [H+] = 10^-5 = 1 x 10-5
3. [CH3COO-] = 1.75 x 10-5 x 0.2 / 1 x 10-5 = 0.35 mol dm-3
4. 0.35 x 400 x 10-3 = 0.14 moles
5. 0.14 x 82 (Mr of sodium ethanoate) = 11.48g

i hope this makes a bit more sense, if you don't get anything i'll try and explain!