Buffer calculation help

500 cm3 of a buffer solution of pH = 4.70 is required.
Calculate the volume of 0.800 mol dm–3 sodium ethanoate solution and of 0.800 mol dm–3 ethanoic acid needed to make this buffer.
[Ka]
(3)

I don’t understand what you do after calculation the concentration of H+ ions. Can someone explain? Thanks

After calculating the H^+ ion concentration, you can use Ka = [H^+][A^-]/[HA] to find the ratio of A^- to HA in the buffer solution.

The ratio of [A^-] to [HA] in the buffer solution, [A^-]/[HA], is simply found by dividing both sides through by [H^+]:

[A^-]/[HA] = Ka/[H^+]

You can use the fact that concentration is directly proportional to moles to deduce that this ratio is also equal to the mole ratio of A^- to HA.

You can then say A^- and HA are in an n:1 ratio, so [A^-]/[HA] = n/1. Once you have this ratio, split the 500 cm^3 into this ratio, since the solutions used are of the same concentration. That should tell you the volumes of each solution needed.

What do you mean divide both sides by [H+]

This is the formula: Ka=[H+][A-]/[HA] so what do u divide by [H+]? At least use some numbers, I already said I don’t understand it.

So I divide Ka by [H+] but then how do I divide the other side if it’s [H+][A-]/[HA]?

If you divide both sides of Ka = [H^+][A^-] / [HA] by [H^+], you get Ka/[H^+] = [A^-]/[HA].

Now using some numbers, it gives you Ka = 1.74 x 10^-5 and you should have got [H^+] = 10^-4.7

So [A^-]/[HA] = (1.74 x 10^-5)/(10^-4.7) ≈ 0.872

That means that in the buffer solution, the ratio of A^- to HA is 0.872 to 1.

To split something by a ratio, you multiply it by the corresponding part of the ratio divided by the total of the parts of the ratio.

The total of the parts of this ratio is 1 + 0.872 = 1.872

To find the volume of A^-, the part of the ratio corresponding to it is 0.872:

Volume of solution containing A^- = 0.872/1.872 x 500 cm^3 ≈ 233 cm^3

To find the volume of HA, the part of the ratio corresponding to it is 1:

Volume of solution containing HA = 1/1.872 x 500 cm^3 ≈ 267 cm^3

So you need 233 cm^3 of sodium ethanoate and 267 cm^3 of ethanoic acid

Ok thanks so much. This made me understand it a lot better. 1 question: does diving Ka by [H+] give you the [A-]? And if so why?