chem
posted by pablo on .
A beaker with 120mL 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 5.70mL of a 0.450M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Use the HendersonHasselbalch equation and substitute pH, pKa, and solve for ratio of (b ase)/(acid) which I will call b/a. That's equation 1.
The problem tells you that
b + a = 0.1M; that's equation 2.
Solve eqn 1 and eqn 2 simultaneously for b and for a. You should get approximately 1.74 for b/a (but you need to do it more accurately than that) which makes b = 1.74a
and b = 0.0635M with a = 0.0365 (again you should be more accurate).
Then
millimols HAc = 120 mL x 0.0365 = about 4.4
mmols base(Ac^) = 120 x 0.0635 = about 7.6
mmols HCl added = 5.70 x 0.450M = about 2.6.
........HAc + H^+ ==> Ac^ + H2O
I.......4.4..0.......7.6.......
add..........2.6.............
C......2.6..2.6....+2.6
E.......1.8...0.....10.2
Now substitute the E line into a new HH equation and solve for pH.
I love these problems.