[flyer]

This is the problem:
How many millilitres of 0.421 M HCl must be added to 50.0 mL of a 0.055 M
solution of disodium malonate (Na2A) in order to prepare a buffer that has pH of a.) 6,00 b.) 3,2

Some facts, that are mentioned in the question:
- malonic acid - (H2A) - HO2CCH2CO2H
- pk1= 2,847
- pk2= 5,696

So, i had no problem with dealing with the example a.). (You have a buffer that contains [A2-] and [HA-]. As much HCl you add, as much [HA-] you get, and as much less [A2-] you have at the end. ). However, I cannot solve the b.), even though I know it is pretty much similar to a.). The correct answer for a.) 2,17 ml and for b.) 8,47 ml.

Because the buffer has pH of 3,2 it is logical, that you have mix of [HA-] and [H2A]. But I really don't know how to get these two concentrations. Is there also [A2-] in the mix?

Thanks for your help.

[Borek]

First, add enough acid to get solution of HA^- alone. What will happen when you add even more acid?

[flyer]

Is it like this?

A2- + H+ <=> HA-

So, you have 0,055M*0,05L=0,421M* Vx1

Vx1= 6,53 ml

If you add more acid, the HA- will go into H2A?
K1/[H3O+] = [HA-]/[H2A]

[HA-]= 0,421*Vx1 - 0,421*Vx2
[H2A]= 0,421 * Vx2

Vx2 = 2,01 ml, and the sum of HCl added is 8,54 ml. Where did I go wrong, since the solution is 8,47 ml?

[AWK]

compare with:
http://www.chemicalforums.com/index.php?topic=59755.msg213754#msg213754

[flyer]

After comparing I still think that equation K1/[H+]= [HA-]/[H2A] is right, which leads to the same result as before. Can you help me out of this?

[AWK]

You should use more precise equation showed in this post

[flyer]

Was trying with ICE method, but no real success. Should i use equation (9.13), showed on that link?

[AWK]

In this link Borek show, how to calculate ICE tables for phosphate buffer. You have an analogous problem.

[flyer]

I think it is a little bit more complex since I don't have the whole V. I tried, but couldn't figure it out, since I have only [HA-] at the start, 0 H2A and 10^(-3,2) H+. It's a mix of moles, concentrations and volumes, and i really stucked in there.

[Borek]

I would not worry too much about volume - assuming it is 59 will probably do the trick, you want to fine tune the result and the difference in concentrations for the solution being 58.9 and 59.1 is negligible. Just concentrate on the number of moles.

From my calculations you need 8.61 mL of the acid. But I checked using my programs, not on paper.

[AWK]

I think it is a little bit more complex since I don't have the whole V. I tried, but couldn't figure it out, since I have only [HA-] at the start, 0 H2A and 10^(-3,2) H+. It's a mix of moles, concentrations and volumes, and i really stucked in there.

If you do not correct calculations for the ionic strength, you can safely use moles of reagents instead of concentrations.