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pH calculations question

username2207531

Hello! I've got some Chemistry homework but I think I've done one of the questions wrong.

The question:
What mass of sodium methanoate should be dissolved in 250cm3 of 0.100M methanoic acid to form a buffer solution with a pH of 5.20? (ka of methanoic acid = 1.74x10^-4.)

My workings:
[H+] = 10^-5.2 = 6.31x10^-6
[H+] = ka x [HA]/[A-]
6.31x10^-6 = 1.78x10^-4 = 0.1/[A-]
[A-] = 2.82 mol dm^-3
moles = vol x conc
moles of A- = 2.82 x 0.25 = 0.705

This is where I get confused.
Are the moles of A- the same as the moles of HCOONa???

What I did:
I worked out the Mr of HCOONa to be 68.
mass = moles x Mr
mass = 0.705 x 68 = 47.94

final answer = 47.9 g.

I'm pretty sure I did the last part wrong. To summarise, my actual question is: for a weak acid, are the moles of [A-] the same as the moles of [HA]?

Sorry for the confusing post. Thanks!

Reply 1

charco

Study Forum Helper

Original post by JustJusty

No, the moles of HA is not the same as the moles of A^-.

You seem to have the concepts correct:

ka = [H⁺][A^-]/[HA]

[H⁺] = ka * [HA]/[A^-]

10^-5.2 = 1.74 x 10^-4 x [HA]/[A^-]

6.31 x 10^-6/1.74 x 10^-4 = [HA]/[A^-]

0.0363 = [HA]/[A^-]

and as it's a buffer solution the volumes is the same so we can work in moles.

Hence A^- mol = HA mol / 0.0363

moles of HA = 0.25 x 0.1 = 0.025 mol

A^- mol =0.025 / 0.0363 = 0.689 mol

mass of 1 mol HCOONa = 68

0.689 mol = 46.9g