A beaker with 120 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M, A student adds 5.60 mL of a 0.400 M
HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
First, determine the concentrations of acid and base in the buffer. You can do that by solving two equations simultaneously.
eqn 1: pH = pKa + log (base)/(acid) and substitute.
eqn 1: 5.000 = 4.740 + log (base)/(acid) and solve for base/acid
0.260 = log (b)/(a)
(b)/(a) = 1.819
eqn 2: (acid) + (base) = 0.100
Solve eqn 1 and eqn 2 simultaneously and I will let a stand for (acid) and b stand for (base) as I've done above.
b/a =1.819 or b =1.819a. Substitute this into the
a + b = 0.1
a + 1.819a = 0.1
2.819a = 0.1 and a = about 0.035 M.
Then b = about 0.065
It will make it easier if I switch to mols. You can switch back whenever you wish. You should confirm all of this. I don't always punch the right keys.
The problem says you use 120 mL of buffer so you have
millimols acetic acid(HAc) = mL x M = 120 x 0.035 = 4.2
millimoles acetate (Ac^-) (the base) = 120 x 0.065 = 7.8
Now we add 5.60mL x 0.400 M = 2.24 millimols HCl.Make an ICE chart.
....................Ac- + H^+ ==> HAc
I...................7.8....................4.2
add........................2.24.......................
C................-2.24...-2.24..........+2.24
Equil............5.56.......0............6.44
At equilibrium (base) = mmols/mL = 5.56/125.6 = ?
(acid) = 6.44/125.6 = ?
Plug those numbers into the pH = pKa + log (base)/(acid) and solve for pH. Then subtract from the initial pH of 5.00 to find the change.
Toward the last I started rounding so you should redo the whole thing and carry out the numbers to the correct number of significant figures. Nothing like solving the whole thing for you.