Posted by Sandra on .
A beaker with 100 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.10 mL of a 0.260 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
I got 0.27 as the change, but it says I made a rounding error, and I can't find my rounding error. I also got 0.26 and 0.25, but they didn't work.

Chemistry 
bobpursley,
I would assume three significant digits on this.
work it out to three digits. 
Chemistry 
Sandra,
Sorry, I forgot to say that it told me to round it to two decimal places.

Chemistry 
bobpursley,
recheck how you rounded the antilogs.

Chemistry 
Sandra,
I did everything from the beginning and I didn't round anything at all until my last step where it told me to round to two decimal places. I still can't get the answer, though

Chemistry 
DrBob222,
I can't get any of your answers.
Post your work and let me look for the error. I think the answer is close to 0.7 if I didn't make an error. 
Chemistry 
Sandra,
Here's what I did:
So I set values for:
[Acetic Acid] = y
[H3O+] = 10^5
[Acetate] = 0.1M  y
Ka = 10^4.76
And from the equation: Ka = [Acetate][H3O+]/[Acetic Acid], I found y, or the molarity of acetic acid, to be 10^6/(10^5 + 10^7.46). And from this, I also got the molarity of acetate: 0.1  y.
Then, I found the total molarity. I converted 10^5 M acetic acid to 10^6 M/0.1 L. I then took 0.260 M HCl and converted it to 0.001586 M/0.0061 L. I added the top and the bottom and it turned out to be 0.001587 moles/.1061 L. Let this value be A.
And since the Ka value is low, it goes to completion to the left. So, [Acetic Acid] = y + A. And [Acetate] = 0.1  y A. And now for the Henderson/Hasselbach equation:
pH = 4.76 + log((y + A)/(0.1  y  A))
And I plugged in:
y = 10^6/(10^5 + 10^7.46)
A = 0.001587/.1061
Then, I subtracted 5 from this and got my answer. 
Chemistry 
DrBob222,
I said I would find your error but I've changed my mind after I see how you worked the problem. Have you seen the HendersonHasselbalch equation? Have you used it? Why not use it here? It is MUCH simpler and it uses exactly the same chemistry; just a different form of it.
pH = pKa + log[(base)/(acid)] and all of that comes from Ka = (H^+)(acetate)/(acetic acid) and pH = log(H^+).
From pH = 5.00 and pKa = 4.760, find (base)/(acid) ratio. (Two unknowns there). Then from the problem you know base + acid = 0.1 (two of the same unknowns). Solve for (base) and (acid). Then set up an ICE chart, add the 6.10 mL and plug that information into a new HH equation and solve for pH. calculate delta pH. 
Chemistry 
Sandra,
Okay, so I found the ion concentrations through the HendersonHasselbalch equation, and the concentrations I got:
[Acetic Acid] = 0.036525666
[Acetate] = 0.063474334
I got the same concentrations as when I did my ice chart, and I also got the same delta pH, which was 0.266. 
Chemistry 
DrBob222,
I still don't get it.
pH = pKa + log(B/A)
5.00 = 4.76 + log(B/A)
B/A = 1.7378
A + B = 0.100
Two equation in two unknowns.
A = your number
B = your number.
You started with 100 mL; therefore, you started with 3.653 mmoles HAc (which may be the difference) and 6.347 mmoles Ac^
When I did the ice chart I ended up with this.
...........Ac^ + H^+ ==> HAc
..........6.347....0......3.653
add..............1.586...........
change.....1.586..1.586..+1.586
final.....4.761.....0......5.239
pH = 4.760 + log(4.761/5.239)
pH = 4.71845
delta pH =0.28155 to 2 s.f. is 0.28 and you data base MAY want to see 0.28
I did find an error in my original calculation in which I thought it was closer to 0.7. Interestingly enough, I check every one of my steps but one and I didn't check that one because it was too simple (just before I posted the 0.7 number). Guess where the error was? I subtracted incorrectly. But I think 0.28 is right. I hope this helps. 
Chemistry 
Sandra,
Thanks a lot! I realized that I forgot to account for the volume change when I calculated my molarities for the base and the acid, and that was enough to get the pH off by 0.01.