Michael S. answered • 08/05/26
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
pH = 4.44 (using Ka = 1.8 × 10-5 for acetic acid, so pKa = 4.74).
Work in moles rather than concentrations. Henderson-Hasselbalch uses a ratio, so the volume cancels, and that saves you from worrying about the dilution caused by the 3 cm3 you added.
Starting amounts, in 0.100 L:
CH3COOH: (0.100 L)(0.1 M) = 0.0100 mol
CH3COO-: (0.100 L)(0.05 M) = 0.00500 mol
HCl added: (0.003 L)(0.01 M) = 3.0 × 10-5 mol
What the acid does. Strong acid is consumed by the basic component of the buffer, converting acetate into acetic acid one for one:
CH3COO- + H+ gives CH3COOH
acetate: 0.00500 - 0.00003 = 0.00497 mol
acetic acid: 0.0100 + 0.00003 = 0.01003 mol
Henderson-Hasselbalch:
pH = pKa + log([base]/[acid]) = 4.74 + log(0.00497 / 0.01003) = 4.74 - 0.305 = 4.44
Now the part actually worth your attention. Work out the pH before the acid was added: 4.74 + log(0.00500/0.0100) = 4.74 - 0.301 = 4.444. After the addition it is 4.440. The pH moved by about 0.004 units.
Compare that with dropping the same 3.0 × 10-5 mol of HCl into 100 cm3 of pure water instead. That gives 2.9 × 10-4 M H+, so pH 3.54, a swing of about three and a half units from neutral. Same acid, same amount, roughly a thousand-fold difference in the pH change. That is the entire point of a buffer, and this problem is built to show it.
The reason is visible in the numbers: you added 3.0 × 10-5 mol of acid to a solution holding 5.0 × 10-3 mol of acetate ready to absorb it, about 170 times more than needed. The ratio barely shifts, and since pH depends on the log of that ratio, the pH shifts even less.
Two cautions. If your table lists pKa as 4.76, every pH above rises by 0.02; the method is unchanged. And this reasoning holds only while both buffer components remain in real excess - add enough HCl to consume most of the acetate and the buffer fails abruptly, which is why buffer capacity is a separate idea from buffer pH.
