Chem

Use the Henderson-Hasselbach equation to determine the relative concentrations of acid (acetic acid) and salt (sodium acetate) necessary to attain the desired pH. You'll need to look up the pKa of acetic acid (use Google or check your textbook).

Henderson-Hasselbach equation:

pH = pKa + log([anion]/[acid])

You'll have to do some algebra to put the equation in terms of the unknown you're looking for -- ie, [anion]/[acid].


Wikipedia's got a decent summary at http://en.wikipedia.org/wiki/Buffer_solution .

Your textbook or course notes should do nicely, too. That's generally the idea of these things.

Scratch that. It's a weak acid, and I forgot how exactly you deal with the weak acid thing with any sort of precision...

i still dont get it...im trying to find the volume for both of them so they will equal out to be 60 mL in total

Well, what ratio do you have?

ok i know the ka of the acid...

ka=[H+][A-]/[HA]

since [H+] = 3.16e-5

and ka is = 1.8e-5

the ratio is around 0.570....then im stuck there

You're working backwards. Ka is the equilibrium constant for any given acid. There's a good explanation of what Ka (and hence pKa) mean at this site: http://www.science.uwaterloo.ca/~cchieh/cact/c123/wkacids.html

pKa is used to determine pH, not vice versa.





If you get stuck, you might find the following page helpful as a model problem -- but to actually learn the stuff, it's a good idea to get a good working understanding of equilibrium constants and what they actually mean...

http://chimge.unil.ch/En/ph/1ph37.htm

i understand what ur saying, but em... since Ka is a constant and doesnt change, that means the concentration changes.... H+ is calculated from the pH given above... blah im lost....

I hope I remember this stuff well enough to be of some help. It's been a while, and it's stuff I only did to move on to other things...

So, the dissociation equation for acetic acid goes like this:

CH3COOH + H2O <--> CH3COO- + H3O+

The chemical structure of the acetate anion (CH3COO-) determines its affinity for protons (H+). Water molecules also have an affinity for protons, however. At any given moment, a certain number of protons are stuck to acetate ions (making acetic acid), and a certain number are stuck to water molecules (making hydroniium - H3O+). It is this competition that makes acetic acid a weak acid. The strong acids (for instance, HCl) readily give up their protons to water (yielding Cl- and H3O+).

Okay.

So water and acetate are both fighting for protons. Since every acetate ion is identical to every other acetate ion and every water molecule is identical to every other molecule, the balance always ends up the same. It is this balance that is described by the equilbrium constant Ka.

For acetic acid in water, Ka = [H+][CH3COO-]/[CH3COOH]

Since Ka doesn't change under normal conditions, we can figure out some things about the behavior of the solution using the two equations I've posted (and this, along with other stuff, is likely review for you). (Also note that everywhere we see H+, it should really be written H3O+. We use H+ as a matter of convention and convenience.)

* If we add H+ to the solution, some of this will stick to CH3COO- to make CH3COOH, and it will do so in sufficient quantity that Ka does not change.

* If we remove H+ from the solution (by adding a base such as NaOH, for instance), some of the CH3COOH will break up into H+ and CH3COO-. Again, enough acetic acid molecules do this that Ka doesn't change.

* If we add CH3COO- to the solution (in the form of sodium acetate, say), some of the H+ will stick to the CH3COO- to make CH3COOH -- again, according to the dictates of the equilibrium constant Ka. In this case (and in every case), the acetate (CH3COO-) is acting like a weak base by sucking up some protons.

The conjugate base of every weak organic acid is a weak organic base. The former has a tendency to give up some protons, the latter has a tendency to soak up some protons.

This is the basic principle of a buffer solution. We can add elements that would radically change the pH of pure water and not much change the pH of the buffer solution (providing that the amount of CH3COOH and CH3COO- in the solution is considerably greater than the amount of acid or base that we add to the buffer).

****

Now, with a little rearrangement, we get the Henderson-Hasselbach equation:

pH = pKa + log([A-]/[HA])

or, in our case,

pH = 4.78 + log([CH3COO-]/[CH3COOH])

We know the desired pH, so we can substitute that into the equation:

4.5 = 4.78 + log([CH3COO-]/[CH3COOH])


Can you rearrange this equation in terms of the ratio of [CH3COO-] to [CH3COOH]?





(This would really be easier in person -- is there anyone you can go to, like an instructor or a teaching assistant?)

Sorry if I'm going on and on about basic principles, by the way. I'm not sure where exactly you're stuck.

im stuck in finding out the volume.... >.<

Damn it, I just realized when I worked it out quickly earlier I got a sign wrong and you were working it differently (but validly) and I thought you were way off track.

So, you've got you're ratio -- 0.57. (I used pH and pKa and got around 0.52 -- probably just using slightly different values for something or other, or it's a significant figure issue.)

Your final volume is to be 60 mL.

Let A be the volume of acetic acid and B be the volume of sodium acetate.

A+B=60
A=0.52*B

There's two equations and two unknowns. Solve either for A in terms of B and substitute it into the other. Or solve for B and substitute, doesn't matter.

yea!! I figured it out thanks alot!