What is the pH when 100ml of 0.1 M NaOH is added to 150ml of 0.2 M HAc if pKa for acetic acid is 4.76?

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The reaction that takes place when NaOH is added to acetic acid is:

NaOH + HAc --> NaAc + H2O

100 mL of 0.1 M NaOH contains 0.01 moles of NaOH. 150 mL of 0.2 M HAc contains 0.03 M of HAc. After the reaction we are left with 0.03 - 0.01 = 0.02 moles of acetic acid.

Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.

pKa = -log(10)[Ka] = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]

as [H+] = [CH3COO-]

pKa = log(10)[CH3COOH] - 2*log(10)[H+]

For acetic acid pKa = 4.76 and -log(10)[H+] = pH

4.76 = log(10)(0.2) + 2*pH

=> 2*pH = 4.76 + 0.698

=> 5.458

pH = 2.729

The pH when 100ml of 0.1 M NaOH is added to 150ml of 0.2 M HAc is 2.729

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