Question

A buffer is formed by adding 500mL of .20 M HC2H3O2 to 500 mL of .10 M NaC2H3O2. What would be the maximum amount of HCl that could be added to this solution without exceeding the capacity of the buffer?

A. .01 mol
B. .05 mol
C. .10 mol
D. .15 mol
E. .20 mol

[I know the correct answer is B, but I don't know why. Any help would be great.]

just tell me where to start? please?

I don't get 0.05 mol but it's close.
Calculate pH of the buffer.
I have pH = pKa + log [(base)/(acid)]
pH = 4.76 + log [(0.05/0.1) = 4.46

The buffer capacity is the number of mols of strong acid or strong base that can be added to make the buffer change by 1.00 pH units. So adding strong acid we want the pH to be 1.00 less than 4.46 or = 3.46
3.46=4.76 + log (b/a)
-1.3 = log (b/a)
0.05 = (b/a)
If we add x mols of a strong acid, then
mols base = what we start with minus strong acid = 0.05 mols acetate - x mols strong acid.

mols acetic acid = what we start with + mols strong acid = 0.1 mols acetic acid + x mols strong acid.
0.05 = (0.05-x)/(0.1+x)
solve for x mols strong acid.
I found 0.043 mols which doesn't round to 0.05. In fact, I don't think any of the choices are right although 0.05 mols is the closest. However, putting 0.05 mols stong acid gives this for the pH.
acetate = 0.05 - 0.05 = 0
acetic acid = 0.10 + 0.05 = 0.15
and pH = 4.76 + (0.0/0.15) and we run into trouble because of the zero concentration of the base. The buffering capacity is gone if we have no acetate left. I hope this helps.

yah, i wasn't allowed a calculator so we had to guess and estimate things, but thanks so much that really cleared it up.

Answer 1

You're welcome! I'm glad I could help clarify things for you. If you have any more questions or need further explanations in the future, please don't hesitate to ask. Good luck with your studies!