Question

A 0.1000 M NaOH solution was employed to titrate a 25.00-mL solution that contains 0.1000 M HCl and 0.0500 M HOAc. Please determine the pH of the solution after 27.00 mL of NaOH is added. Ka, HOAc = 1.75*10-5 .

Answer. 4.04

Just want to see how they got that.

Answer 1

Volume of NaOH = 27mL

Molarity = 0.1 M

moles of NaOH used= Molarity x volume = 27 X0.1 = 2.7 millimoles

Moles of acid HCl present = Volume X molarity = 25 X 0.1 = 2.5 millimoles

Moles of acetic acid present = Molarity X volume =0.05 X 25 = 1.25 millimoles

2.5 millimoles of HCl will neutralize 2.5 millimoles of NaOH

Millimoles of NaOH left = 0.2 millimoles

These 0.2 millimoles will react with 0.2 millimoles of Acetic acid

So millimoles of acetic acid left = 1.05 millimoles

[HOAc] = millimoles of acid / total volume = 1.05 / (25 + 27) = 0.0202 M

Millimoles of AcO- formed = 0.2

[AcO-] = 0.2 / 25 + 27 = 0.0038 M

pH = pKa + log[Salt] / [Acid]

pKa = 4.757 ( for acetic acid)

pH = 4.757 + log [0.0038 / 0.0202] = 4.03