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.
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.
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