[Chemistry] Buffer solution question.

GertBeef

I have what I can only imagine is the most insane buffer question ever asked.

I need to be able to calculate the volume of 4M HCl to be added to tris buffer at 0.18 M so that the final pH is 7.

Now I haven't been given the pKa of tris buffer buts it 8.10.

I assume i need to use the henderson hasselback equation to find the concentrations of acid and conjagate base at pH 7. This I have done. But I can't figure what I need to do next out.

Please don't just give me the answer. That would be pointless as I wouldn't learn anything. Any hints would be greatly appreciated.

Plutonium

Most of the time in this situation you would probably just have to titrate the buffer with the 4M HCl and a few drops of phenolphtalein. There's a reason that they have to do a bunch of this stuff experimentally. I've only been given problems like this in lab situtuations.

DiscGrace

You said you know the concentration of acid in the final solution as well as in the stock solution. Do you know the volume of the tris solution before the HCl was added? If so, you can use the equation C1*V1 = C2*V2 to get the volume of HCl added (where C1 and C2 are the concentrations, before and after, and V1 and V2 are the volumes, before and after). C1 is 4M, C2 you found using Henderson-Hasselbach; V1 = x mililiters, and V2 = initial volume of tris + x. Then it's just simple algebra.

Plutonium

let me give a bit more concentrated attempt

you have the Ka for the buffer solution, I assume. If not, they have tables for this.
Take 1 liter of the solution, so the molarity = # of moles

Ka = [X][Conjugate Base + x]/[Acid - x]

X is usually so small that it goes to zero, so you can use this equation instead:

Ka = [X][Conjugate Base]/[Acid]

You now have the initial H+ concentration, which is X

Then take the pH of your titrated solution, 7 in your case.

pH = -log(Ka)
likewise
10^-pH = Ka

10^-7 = Ka

Here's the hard part and it's probably gonna involve some quadratics at least. You're gonna have to find the new concentration of all of the solutions. The HCl dissociates like this: HCl -> H+ + Cl-, so every 1 L of the 4M HCl you add, you will add 4 moles of H+ ions.

10^-7 = [[H+ initial] + 4X/(1L + XL HCl)][0.18 mol conjugate base/1L + X L HCl] / [0.18 mol acid/1L + X L HCl]

The deal is, you don't know how much HCl is added, so this is the part I'm confused about.

Whatever, once you get past this part, what you do is find the difference in the concentration of H+, then multiply it out to find how much HCl was added, because 1 mol of H+ added = 1 mol HCl added. The Chloride are just spectator ions that float around.

Do some stoicheometry and convert the change in [H+] to Liters HCl

Edit: too lazy to put it in superscript, fixed some confusing crap

Deswa

DiscGrace wrote: »

You said you know the concentration of acid in the final solution as well as in the stock solution. Do you know the volume of the tris solution before the HCl was added? If so, you can use the equation C1*V1 = C2*V2 to get the volume of HCl added (where C1 and C2 are the concentrations, before and after, and V1 and V2 are the volumes, before and after). C1 is 4M, C2 you found using Henderson-Hasselbach; V1 = x mililiters, and V2 = initial volume of tris + x. Then it's just simple algebra.

Normally, this would work. In this case they are using a buffer, so that formula will not work.

trixtah

you can make a table and use solution stoichiometry. The change can be the concentration of OH- that you can find from the pH. First make a table for adding the HCl and then make a table for equilibrium. Is that the full question? Like, all the information given?

When in doubt, convert to moles

I have this worked out in my head if you still can't get it. Even with the Henderson Hasselbalch, I like to do things stoichiometrically anyway.

GertBeef

Woah. Thanks for the replies. They are helping but i''m not there yet.
When I deduce all of this ill tell you how to do it... because you care so much. ^______^

Oh. Plutonium, the pH isnt -log(pKa) its -log([HA]). Man you wrote alot. : D