A sample of 0.100 M acetic acid (Kg = 1.8 x 10-5) in 50.0 mL of...

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Question

A sample of 0.100 M acetic acid (Kg = 1.8 x 10-5) in 50.0 mL of solution is titrated with standard 0.100 M NaOH. What is the

Answer 1

Answer #1

HOAc + H2O -----> OAc- + H3O+

I(M) 0.1 0 0

E(M) 0.1-x x x

Ka = [H3O+][OAc-]/[HOAc]

1.8 x 10-5 = x2/0.1-x

x = 0.001335

CH3COOH+ NaOH--------> CH3COONa + H2O

NaOH is a strong base ===> react completely with HOAc to form H2O and Ac-.

When we add 15 mL NaOH (1.5 mole) ==> react with 1.5 mmol of HOAc to form 1.5 mmol of OAc-.

Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

pH = 4.75 + log(1.5/3.5)

pH = 4.38

Answer 2

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