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

Consider the titration of a 21.0 −mL sample of 0.100 M HC2H3O2
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