March 30, 2015

Homework Help: Chemistry

Posted by Rio on Wednesday, March 31, 2010 at 4:14pm.

I have a question about buffers.

Part A
So it starts with 20ml 0.1 sodium acetate and 25ml 0.1 acetic acid.

Calculate ph of buffer is 4.74 because the acid and conjugate base have the same molarity correct? So the Ph is just pKa (1.8e-5)?

Part B
So the next part is add 5ml of 0.1 HCl and this is the part where it messes me up.

First I use an ICE chart.
Equation is
CH3COO- + HCl --> CH3COOH + H2O
So I get confused here. Should I use moles or concentrations in this ice chart or does it make a difference?

If I use concentrations, would the initial concentration be from the calculated ones after the buffer is at equilibrium, as in part a? So the initial concentration of CH3COO- is 0.1 + 1.8e-5M? and CH3COOH is 0.1 - 1.8e-5M? (This is from buffer solution in part a)

It doesn't work to use molarity because the HCl is already 0.1 M! And subtracting that from the M of acetic acid and acetate ion would make the number negative or very small.

So I try to use mol. So the amount of mol HCl is 0.0005. What I'm confused about is the mol of CH3COO- and CH3COOH.
I know the concentrations, but what volume do I mulitply it by? Initially I have 45 ml since (20ml of sodium acetate and 25ml of acetic acid). So would I just multiply by 45 ml?

And after the change which is add 0.0005 mol of HCl, does that mean the concentration is 50 ml now? So when I have the final mol of CH3COO- and CH3COOH do I divide by 50ml to get the concentration??

Examples of this question seem to show that you just use the final volume throughout so I would multiply initial concentration by 50ml to get initial # of mol?

And then I use Pka + log (conjugate base/acid) = PH

But the PH I got here is much lower the the measured Ph. I got 4.65. Is that correct?

My final concentrations I got were CH3COO- = 0.09018M and CH3COOH = 0.109982. Is this correct???

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