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?

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

aara259's profile pic

aara259 | College Teacher | In Training Educator

Posted on

A detail is missing in the explanation. After the reaction,  the concentration of acetate is not the same as [H+] since the reaction of sodium hydroxide with acetic acid produced 0.01 moles of sodium acetate. Therefore, the molar concentration of acetate is 0.01 moles/(0.25 L: total volume)=0.04 mol/L. The molar concentration of acetic acid is 0.02 moles/0.25 L=0.08 mol/L.
Therefore
Ka = [H+][CH3COO-]/[CH3COOH]
pKa=pH-log ([CH3COO-]/[CH3COOH])=pH-log (0.5)
pH=pKa+log (0.5)= 4.76-0.3=4.46

The pH after the reaction is 4.46.

We’ve answered 318,994 questions. We can answer yours, too.

Ask a question