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.

a) Please determine the pH of the solution after 27.00 mL of NaOH is added. Ka, HOAc = 1.75*10^-5

^b) what's the solution pH at the second equivalence point?

Answer 1

a) the volume of NaOH = 27mL Molarity = 0.1 M

So 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

Millimoles of NaOH left = 0.2 millimoles as 2.5 millimoles of HCl will neutralize 2.5 millimoles of NaOH

These 0.2 millimoles will react with 0.2 millimoles of Acetic acid

So millimoles of acetic acid left = 1.3 millimoles

[HOAc] = millimoles of acid / total volume = 1.3 / 25 + 27 = 0.025 M

Millimoles of AcO- formed = 0.2

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

so this will form a buffer containing weak acid and its salt, the pH of buffer can be calcualted using hendersen hassalbalch equation

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

pKa = 4.75 ( for acetic acid)

pH = 4.75 + log [0.0038 / 0.025] = 3.93

b) at second equivalent point the whole acid will be neutralized by the given base

Millimoles of NaOH used = Millimoles of HCl + millimoles of AcOH = 2.5 +1.25 = 3.75 millimoles

Volume of NaOH used =millimoles Molarity = 3.75 / 0.1= 37.5 mL

Millimoles of AcO- formed = 1.25

concentration of AcO- = 1.25 / 37.5 + 25 = 0.02 M

So we have salt which will hydrolyze as

AcO- + H2O --> AcOH + OH-

Kb = Kw / Ka = 10^-14 / 1.8 X 10^-5 = 5.55 Xx 10^-10

Kb = [AcOH] [ OH- ] / [AcO-]

Let the amount of acetate ion hydrolyzed is x

Then the amount of acid and hydroxide ions formed = x

So Kb = x^2 / 0.02 - x

x <<1

5.55 Xx 10^-10 = x^2 / 0.02

x = 3.33 X 10^-6

[OH-] = 3.33 x10^-6

pOH = 5.47

pH = 14- 5.47 = 8.53