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In: Chemistry

For the titration of 35.0 mL of 0.15 M trimethylacetic acid (C4H9COOH, Ka= 9.33 x 10-6)...

For the titration of 35.0 mL of 0.15 M trimethylacetic acid (C4H9COOH, Ka= 9.33 x 10-6) with 0.35 M KOH

a. Calculate the pH after 8.0 mL of titrant has been added

b. Calculate the pH after 15.0 mL of titrant have been added

c. Calculate the pH after 17.0 mL of titrant have been added

Solutions

Expert Solution

millimoles of acid = 35 x 0.15 = 5.25

pKa = - log Ka = - log [9.33 x 10-6] = 5.03

a) millimoles of KOH added = 8.0 x 0.35 = 2.8

5.25 - 2.8 = 2.45 millimoles acid left

2.8 millimoles salt formed.

[acid] = 2.45 / (35+8) = 0.057 M

[salt] = 2.8 / 35+8 = 0.065 M

now mixture of acid and salt act as acidic buffer

pH = pKa + log [salt] / [acid]

pH = 5.03 + log [0.065] / [0.057]

pH = 5.09

b) millimoles of KOH added = 15 x 0.35 = 5.25

all acid convert to salt

[salt] = 5.25 / 15 + 35 = 0.105 M

pH = 1/2 [pKw + pKa + logC]

pH = 1/2 [14 + 5.03 + log 0.105]

pH = 9.02

c) millimoles of KOH added = 17 x 0.35 = 5.95

5.95 - 5.25 = 0.7 millimoles KOH left

[KOH] = 0.7/35 + 17 = 0.0135 M

pOH = - log [OH-]

pOH = - log [0.0135]

pOH = 1.87

pH = 14 - 1.87

pH = 12.13

queen_honey_blossom answered 2 years ago
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