Question

Consider the titration of a 21.0 −mL sample of 0.100 M HC2H3O2 with 0.120 M NaOH. Determine each of the following.

the volume of added base required to reach the equivalence point

the pH at the equivalence point

Answer 1

1. moles acetic acid = 0.0210 x 0.100=0.0021
moles NaOH required = 0.0021
Volume NaOH = 0.0021/ 0.120 M=0.0175 L => 17.5 mL

2. moles NaOH required to reach the equivalence point = 0.0021
volume NaOH = 0.0021/ 0.120 M=0.0175 L
total volume = 0.0175+ 0.0210 =0.0385 L
moles acetate formed = 0.0021
concentration acetate = 0.0021/ 0.0385 =0.0545 M
CH3COO- + H2O <=> CH3COOH + OH-
Kb = Kw/Ka = 5.56 x 10^-10 = x^2 / 0.0545-x
x = [OH-]=3.03 x 10^-11 M
pOH = 10.51
pH = 3.48