Question

Consider the titration of a 24.0 mL sample of 0.105 molL−1 CH3COOH with 0.125 molL−1 NaOH. Determine each quantity:

a) inital pH = 2.86

b) the volume of base required to reach the equivalence point = 20.2 mL

c) the pH at 6.0 mL of added base = 4.37

d) the pH at the one-half equivalence point = 4.74

e) the pH at the equivalence point = 8.75

f) the pH after adding 4.0 mL of base beyond the equivalence point

I calculated the answers for a through e, but I can't get f.

Answer 1

f) the pH after adding 4.0 mL of base beyond the equivalence point

this one is simple:

HA + NaOH = H2O + NaA

so...

find moles of acid = MV = 0.105*24 = 2.52 mmol of acid

after 4 mL past equivalence point

find equivalence point first

mmol of aicd = mmol of base

mmol of base = M*V = 2.52

V = 2.52/0.125 = 20.16 mL of base required

so..

total volume at end of base = V1+Veq = 4 + 20.16 = 24.16 mL

mmol of base = MV = 24.16*0.125 = 3.02 mmol of base

so..

mmol of base left after neutralization = 3.02 - 2.52 =) 0.50 mmol of base left

Vtotal = 24 + 24.16 = 48.16 mL

so..

[OH-] = mmol of OH- / total V = 0.50 / 48.16 = 0.01038 M of OH- left

so

pOH = -log(OH-) = -log(0.01038 = 1.98380

pH = 14-1.98380

pH = 12.0162 which is basic as expected