Writing Common Ion Concentration in Buffer Eq. Expression

  • #1
JeweliaHeart
68
0

Homework Statement


What is the pH of a buffer solution created by combining 100 mL of 0.2 M acetic acid and 400 mL of 0.10 M sodium acetate? Ka= 1.8 x 10^(-5)


Homework Equations



1.8 x 10-5=([C2H3O2-][H+])/([HC2H3O2])

The Attempt at a Solution



I know what I'm supposed to do to solve this problem, but I'm not sure how to set up the concentrations in the expression.

I think it should go like this:


1.8 x 10-5= (0.08+x)(x)/(0.04-x)

The explanation in my book has basically the same thing with just one tiny exception. In the equilibrium expression, it used only 0.08 as a concentration for C2H3O2-, without the x.

I thought the x would be necessary b/c 0.08 is only the initial concentration of the acetate ion and the acetic acid will dissociate more before reaching the equilibrium.

Any help would be nice. Thanks in advanced.
 
Last edited:

Answers and Replies

  • #2
AGNuke
Gold Member
455
9
There's a simple formula to it. Search for it. I believe it's Henderson-Hasselbalch Equation.
 
  • #3
JeweliaHeart
68
0
Oh, thanks. I just figured it out.
 
  • #4
Borek
Mentor
29,101
3,716
While you are right about the fact acid can dissociate a little bit, in most typical situations changes induced by the dissociation are so small, we can safely ignore them. Then calculating pH of a buffer is just a matter of calculating concentrations of acid and conjugate base, and plugging them into Henderson-Hasselbalch equation.

Actually you don't even need to calculate concentrations, it is enough to calculate numbers of moles of acid and conjugate base, as volume cancels out.
 

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