[melissadavies]

You are tasked with preparing an acetate buffer solution. You are provided with a 100-mL volumetric flask, 4.0g of sodium acetate, 10-mL of 3.0M acetic acid, and enough distilled water to fill the remaining volume of the flask.

A.) The pH of the buffer solution is: 4.96 (am I right..?)

B.) The equilibrium concentration of acetate ion in the buffer solution is: 0.049 (not sure if this is right either...)

C.) The pH of the buffer solution after adding 1.0-mL of 1.0M NaOH solution is (assume the total volume doesn't change): I have no idea how to do this part...any help? Thanks so much

[Xenonman]

For C.), you'd be adding a thing that steals protons to a solution with things that release protons.
Sounds like a reaction, so a RICE table should clear things up.

I had the same answer for A.)

[melissadavies]

So for part C would it be:

CH3COOH + NaOH <----> CH3COONa + H2O
0.03           0.001                0 
-0.001       -0.001            +0.001
____________________________________
0.029         0                    0.001

Then it would be like:

CH3COOH <----> CH3COO- + H+
0.029                      0         0
-x                          +x       +x
___________________________
0.029-x                    x         x

Then I would do: x^2/0.029-x = 1.8x10^-5

I would solve for x and get take the negative log and get 3.14...this isn't right.

[Xenonman]

Your first table is perfect! ...In case you never added the sodium acetate you were given in the first post to the solution.

[Borek]

http://www.chembuddy.com/?left=buffers&right=pH-change

[melissadavies]

I'm sorry I still don't understand how to do this at all. So my table is right but how do I calculate the pH?

[Xenonman]

Your first table is perfect! ...In case you never added the sodium acetate you were given in the first post to the solution.

So not quite right for the situation you're in.

http://www.chembuddy.com/?left=buffers&right=pH-change

Reading this should clear every doubt. Like, all of them.

[melissadavies]

Ok so for the acetate ion concentration at equilibrium for part B I got 0.049 and for the acetic acid I got 0.03....I'm not sure if that's right. Then I did 4.96 + log [0.049/0.03] to get a final answer of 5.17 for part C.

[Borek]

Ok so for the acetate ion concentration at equilibrium for part B I got 0.049 and for the acetic acid I got 0.03

Looks to me me like you are off be a factor of 10.

Then I did 4.96 + log [0.049/0.03] to get a final answer of 5.17 for part C.

That's wrong for sure. Assuming you are trying to use Henderson-Hasselbalch equation, formula should not start with 4.96 but with the pKa for the acetic acid, and it should contain new concentrations of the acetic acid and acetate ion (after NaOH was added), not equilibrium concentrations of the previous solution.

How to calculate these new concentrations is explained on the page I linked to, have you seen it?

[melissadavies]

Oh, so it's 0.49M for the acetate ion concentration for part B? I knew I had to divide by the total volume....as for part C, yeah I saw the link you showed me and it helped a lot except I don't know how they calculated their [Acetate-] and their [HAcetate] concentrations. It shows:

[Acetate-]/[HAcetate] = 10

Then it shows:

[HAcetate] + [Acetate-] = 0.1M

Then somehow they got:

[HAcetate] = 0.0091M
[Acetate-] = 0.0909M

That's the only part I'm stuck on. I understand the rest of it, I just don't know how they calculated those numbers for their new concentration.

[sjb]

Oh, so it's 0.49M for the acetate ion concentration for part B? I knew I had to divide by the total volume....as for part C, yeah I saw the link you showed me and it helped a lot except I don't know how they calculated their [Acetate-] and their [HAcetate] concentrations. It shows:

[Acetate-]/[HAcetate] = 10

Then it shows:

[HAcetate] + [Acetate-] = 0.1M

Then somehow they got:

[HAcetate] = 0.0091M
[Acetate-] = 0.0909M

That's the only part I'm stuck on. I understand the rest of it, I just don't know how they calculated those numbers for their new concentration.

So, for instance you can rearrange [Acetate-]/[HAcetate] = 10 to give [Acetate-] = 10[HAcetate]; and plug into equation 2...

[Borek]

[Acetate-]/[HAcetate] = 10

Then it shows:

[HAcetate] + [Acetate-] = 0.1M

Then somehow they got:

[HAcetate] = 0.0091M
[Acetate-] = 0.0909M

These are just simultaneous equations with two unknowns, simple algebra.