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

(I looked at the mark scheme and still don’t get it)

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

(I looked at the mark scheme and still don’t get it)

(edited 10 months ago)

Original post by Den322221

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

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.

Original post by TypicalNerd

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.

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]?

Original post by Den322221

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]?

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

Original post by TypicalNerd

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

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?

Original post by Den322221

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?

Dividing Ka by [H^+] gives you the ratio of A^- to HA, rather than giving you just [A^-]. It’s because it is always assumed that the ratio can be written in the form n : 1 and it’s just easier to make the HA part of the ratio 1, since it is on the bottom of the fraction [A^-]/[HA].

The ratio of A^- to HA is conveniently what you need for finding the volumes of each solution required.

(edited 10 months ago)

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