Question

A volume of 25.0mL of a 0.10M aqueous solution of acetic acid(HC₂H₃O₂ or HOAc or CH₃CO₂H, Ka=1.8x10^-5) is titrated with 0.080M NaOH.

What is the pH Initially

After 10.0mL of NaOH has been added

At the halfway point

After 20.0mL of NaOH has been added

At the equivalence point

And after 40.0mL NaOH has been added

Answer 1

CH3COOH + NaOH -----> CH3COONa + H2O

Since acetic acid is a weak acid so,

pH = 1/2(pKa - logC) before any addition of NaOH

pH = 1/2(-log(1.8*10^-5) - log(0.1))

pH = 2.87

After adding NaOH solution it becomes buffer solution i.e., acetic acid and salt of acitic acid both are present so,

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

after adding of 10 mL NaOH,

pH = -log(1.8*10^-5) + log((0.08*10)/((0.1*25)-(0.08*10))

pH = 4.42

After adding of 20 mL of NaOH,

pH = -log(1.8*10^-5) + log((0.08*20)/((0.1*25)-(0.08*20)))

pH = 4.99

At half equivalence point [Salt] = [Acid]

so pH = pKa

pH = -log(1.8*10^-5) = 4.745

at equivalence point,

the acid is conpletely converted to salt,

number of moles of acid = number of moles of salt

number of moles of acid = 25*0.1 = 2.5 mmol

number of moles salt = 2.5 mmol

pH of the solution = 7 + 1/2(pKa + logC)

C = concentration of the salt solution

pH = 7 + 1/2(-log(1.8*10^-5) + log(2.5*10^-3/1)

pH = 8.07

After adding of 40 mL of NaOH,

number of moles of NaOH = 0.08*40 = 3.2 mmol

number of moles of CH3COOH = 25*0.1 = 2.5 mmol

number of moles of NaOH remaining = 3.2 - 2.5 = 0.7 mmol

pOH of salt solution = -log(0.7*10^-3*1000/65)

pOH = 1.968

pH = 14 - 1.968 = 12.032