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.

Answer 1

Use the Henderson-Hasselbalch 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.