A 40.0 mL sample of 0.080 M propanoic acid, CH CH COOH ( )
CHEMISTRY · Multi-part · A Level · Chemistry · Acids and bases · Acids and bases
Problem
A 40.0 mL sample of 0.080 M propanoic acid, CH3CH2COOH (Ka=1.3×10−5), is titrated with 0.100 M KOH. The titration proceeds past the equivalence point.
Answer the following questions fully, showing all reasoning.
(a) Calculate the exact volume of 0.100 M KOH required to reach the equivalence point.
(b) Determine the pH of the reaction mixture after a total of 33.0 mL of KOH has been added.
(c) Determine the pH of the reaction mixture after a total of 45.0 mL of KOH has been added.
(d) Explain, in terms of your calculations, why the pH difference between the two post-equivalence points is smaller than many students expect.