Chemistry
posted by Anonymous .
A beaker with 150 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 7.90 mL of a 0.480 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

I can get you started. Let's call acetic acid HAc and the anion Ac.
pH = pKa + log (Ac/HAc)
You know pH = 5 and you know pKa, you can calculate (Ac)/(HAc). Do that first.
Then you know from the problem that
(HAc) + (Ac) = 0.1
That give you two equations and two unknowns. Solve for (HAc) and (Ac).
Next, you add 7.90 mL x 0.480 M H^+ or 3.79 millimoles acid. That will use up 3.79 millimoles Ac and form 3.79 millimoles of additional HAc. That will let you calculate a new (HAc) and a new (Ac). Plug those numbers into the HH equation and solve for the new pH.
Post your work if you get stuck. 
When I plugged the pH and the pKa into the pH = pKa + log [Ac/HAc] I got [Ac/HAc] to be 0.575.
I then rearranged the equation of [Ac/HAc] = 0.575 and solved for Ac, which gave me the following equation: Ac = 0.575HAc.
I then plugged that equation into HAc + Ac = 0.1 and found that HAc = 0.06 and that Ac = 0.04.
Then I multiplied 7.90 x 0.480 and got 3.79 millimoles.
I then found the new value for HAc: 0.06 + 3.79x10^6 = 0.06000379
I then found the new value for Ac: 0.04  3.79x10^6 = 0.03999621
I then plugged those of those values into the HH equation as follows:
pH = 4.760 + log [0.03999621/0.06000379] = 4.58
Therefore the pH change would be: 5.00  4.58 = 0.42
The only problem is that it said my answer was wrong... 
When I plugged the pH and the pKa into the pH = pKa + log [Ac/HAc] I got [Ac/HAc] to be 0.575.
I don't get that.
5.00 = 4.76 + log(B/A)
0.24 = log(B/A)
(B)/(A) = 1.7378 and
(B) = 1.7378*(A) which still gives values for (Ac) and (HAc) very close but reversed(see below).
I then rearranged the equation of [Ac/HAc] = 0.575 and solved for Ac, which gave me the following equation: Ac = 0.575HAc.
I then plugged that equation into HAc + Ac = 0.1 and found that HAc = 0.06 and that Ac = 0.04.
Using my value for (B/A), I came up with (HAc) = 0.0365 M and (Ac) = 0.0635 M
Then I multiplied 7.90 x 0.480 and got 3.79 millimoles.
I then found the new value for HAc: 0.06 + 3.79x10^6 = 0.06000379
This step isn't correct. What you must do is add moles (or millimoles), not molarity.
mmoles HAc in the 150 mL buffer is 150 mL x 0.0365 M = 5.475 mmoles HAc.
mmoles Ac = 150 mL x 0.0635 M = 9.52 mmoles Ac.
Now add 3.79 mmoles to HAc and subtract 3.79 mmoles from Ac and recalculate pH and the difference.
I then found the new value for Ac: 0.04  3.79x10^6 = 0.03999621
I then plugged those of those values into the HH equation as follows:
pH = 4.760 + log [0.03999621/0.06000379] = 4.58
Therefore the pH change would be: 5.00  4.58 = 0.42
If I didn't make an error, and you should check it closely to confirm that I did not, I found
pH = 4.76 + log(5.73/9.265)
pH = 4.55 which isn't all that far away from your final value. You should realize, also, that I plugged in millimoles for base and acid and not concentration. That is, I didn't go through th extra step of dividing base and acid millimoles by 150 mL to obtain molarity BECAUSE both numerator and denominator would be divided by 150 mL so that plus the conversion from millimoles to moles cancel. I just saved a little work, that's all. 
I also got the new pH to be 4.55. Thank you so much. This helps out a lot.

I trust that's the correct answer.