Writing Common Ion Concentration in Buffer Eq. Expression

AI Thread Summary
To determine the pH of a buffer solution made from 100 mL of 0.2 M acetic acid and 400 mL of 0.10 M sodium acetate, the Henderson-Hasselbalch equation is applicable. The equilibrium expression involves concentrations of the acid and its conjugate base, but for buffer calculations, the change in concentration due to dissociation can often be ignored. Instead of focusing on concentrations, calculating the moles of acetic acid and acetate is sufficient, as the total volume cancels out in the equation. The initial confusion about including 'x' in the concentration setup highlights a common misunderstanding in buffer chemistry. Ultimately, the pH can be easily calculated using the Henderson-Hasselbalch equation with the moles of acid and base.
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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.
 
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There's a simple formula to it. Search for it. I believe it's Henderson-Hasselbalch Equation.
 
Oh, thanks. I just figured it out.
 
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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