Question

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.

Answer 1

I would assume three significant digits on this.

work it out to three digits.

Answer 2

Sorry, I forgot to say that it told me to round it to two decimal places.

Answer 3

recheck how you rounded the antilogs.

Answer 4

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

Answer 5

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.

Answer 6

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.

Answer 7

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 Henderson-Hasselbalch 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.

Answer 8

Okay, so I found the ion concentrations through the Henderson-Hasselbalch 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.

Answer 9

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.

Answer 10

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.