Question

A 40.00 mL sample of 0.10 M weak acid with Ka of 1.8×10−5 is titrated with a 0.10 M strong base. What is the pH after 20.00 mL of base has been added?

Answer 1

this is a weak acid + strong base

addition of base will form conjugate of the weak acid i.e. conjugate base

this will be a buffer

Apply buffer equation

pH = pKa + log(conjujgate/acid)

pKa = -log(ka) = -log(1.8*10^-5) = 4.74472

Total volume = V1+V2 = 40 + 20 = 60 ml

mol of base added = MV = 0.1*20 = 2 mmol of base added

mol of conjugate formed = 0 + 2 mmol

mol of acid initially = 40*0.1= 4 mmol acid

mol of acid left = 4-2= 2 mmol left

note that this is the HALF equivalence point

this is a pretty unique step since

[Conjugate] = [Acid]

log(1) = 0

then

pH = pKa + log(conjujgate/acid)

pH = pKa = 4.747