CHEM 1212K Lecture Notes - Lecture 4: Sodium Acetate, Weak Base, Equivalence Point

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Weak acid and base titration problems (questions and detailed solutions) Problem #1: consider the titration of a 24. 0-ml sample of 0. 105 m ch3cooh with 0. 130 m. Solution to part a: insert values into the ka expression for acetic acid. The ka for acetic acid is 1. 77 x 10-5. 1. 77 x 10-5 = [(x) (x)] / 0. 105 x = 1. 3633 x 10-3 m ph = 2. 865. Solution to part b: calculate moles of acid present: (0. 105 mol/l) (0. 0240 l) = 2. 52 x 10-3 moles, determine moles of base required to react equivalence point: There is a 1:1 molar ratio between acetic acid and sodium hydroxide. Therefore, 2. 52 x 10-3 moles of base required: calculate volume of base solution required: 2. 52 x 10-3 mol divided by 0. 130 mol/l = 0. 0194 l = 19. 4 ml. Solution to part c: calculate moles of acid and base in solution before reaction:

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