Question

What is the pH of the solution in the flask at the equivalance point of the ethanoic acid/sodium hydroxide titration.

Worked molarity of ethanoic acid: 0.09 mol/L

Concentration of base: 0.1 mol dm-3

Volume of ethanoic acid used: 25cm3

Volume of NaOH used: 50cm3

I believe I need to work out the Ka value to find the pH, but my calculations are seemingly incorrect. Thank you.

Answer 1

Ethanoic acid = Acetic acid = CH₃COOH

Sodium hydroxide = NaOH

[CH₃COOH] = 0.09 mol/L = 0.09 M

Volume of CH₃COOH = 25 cm³ = 25 mL = 0.025 L

[NaOH] = 0.1 mol dm-3 = 0.1 mol/L = 0.1 M

Volume of NaOH = 50 cm³ = 50 mL = 0.50 L

Now,

Molarity = Moles/Liter

or Moles = Molarity x Liter

So, Moles of CH₃COOH = 0.09 M x 0.025 L = 0.00225 moles

Moles of NaOH = 0.1 M x 0.50 L = 0.05

CH₃COOH and NaOH will react in 1:1 ratio; So, 0.00225 moles of CH₃COOH will neutralize 0.00225 moles of NaOH.

So, excess moles of NaOH left = 0.05 moles - 0.00225 moles = 0.04775 moles

Total volume = 25 mL + 50 mL = 75 mL = 0.075 L

[NaOH] = 0.04775 moles / 0.075 L = 0.64 M

NaOH is a strong base and it will dissociate completely.

So, [OH^-] = 0.64 M

pOH = - log [OH^-] = - log (0.64) = 0.19

pH = 14 - pOH = 14 - 0.19 = 13.81

NOTE:

I think you gave wrong value of NaOH concentration. So, the answer is as written above.