Consider the titration of a 23.0 −mL sample of 0.110 MM HC2H3O2 with 0.130 M NaOH. Determine the pH after adding 4.00 mL of base beyond the equivalence point.
The balanced equation is
HC2H3O2 + NaOH -----> NaC2H3O2 + H2O
Number of moles of HC2H3O2 = M*V = 0.110 M* 23.0 mL = 2.53
mmol
At equivalence point,
number of moles of NaOH = number of moles of acid =
2.53 mmol
Volume of NaOH required to reach equivalence point
= moles/molarity = 2.53 mmol /0.130 M = 19.46 mL
4.00 mL beyond equivalence point = 19.46 + 4.00 = 23.46 mL
Number of moles of NaOH = M*V = 0.130 M* 23.46 mL = 3.05 mmol
At this point, the pH of the solution is determined by the
amount of unreacted NaOH in the solution.
Net moles of NaOH = 3.05 –2.53 = 0.52 mmol
Total volume of the solution = 23.0 + 23.46 = 46.46 mL
Conc. of NaOH = [OH–] = moles/ volume = 0.52 mmol / 46.46 mL =
0.011192 M
pOH =− log [OH–] = − log[0.011192] = 1.95109 = 1.95
pH = 14 – pOH = 14 – 1.95 = 12.05
pH of the resulting solution = 12.05
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