Question

1.) Acetic acid is a weak monoprotic acid with Ka=1.8*10^-5. In an acid base titration, 100 mL of 0.100M acetic acid is titrated with 0.100M NaOH. What is the pH of the solution:

a.)Before any NaOH is added

b.)Before addition of 15.0mL of 0.100M NaOH

c.)At the half-equivalence point

d.)After addition of total of 65.0mL of 0.100M NaOH

e.)At equivalence point

f.)After addition of a total of 125.0mL of 0.100M NaOH

Answer 1

First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well

M = moles / volume

pH = -log[H+]

a) HA -> H+ + A-

Ka = [H+][A-]/[HA]

a) no NaOH

Ka = [H+][A-]/[HA]

Assume [H+] = [A-] = x

[HA] = M – x

Substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(0.1-x)

This is quadratic equation

x = 0.00133

since x = [H+] = 0.00133

For pH

pH = -log(H+)

pH = -log (0.00133

pH in a = 2.876

b) 15 ml NaOH

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

NOTE that in this case [A-] < [HA]; Expect pH lower than pKa

pH = pKa + log (A-/HA)

Vt = V1+V2

[A-] = mol A- / Vt = M*V1 / (Vt) =0.1*15 / (15+100) = 0.013 M

[HA] = mol HA / Vt = M*V2 /(Vt) = 0.1*100 / (15+100) = 0.086 M

Apply equation

pH = pKa + log ([A-]/[HA]) = 4.75+ log (0.013 /0.086 ) = 3.929

c) at half equivalence point

A- = HA

so

pH = pKa + log(a-/HA)

since A-/HA = 1

then

pH = pKa + log(1) = pKa

pH = 4.75

d) After

after V = 65 mL of NaOH

mmol of base = MV = 65*0.1 = 6.5

mmol of acid = MV = 100*0.1 = 10

mmol of acid left after reaction = 10-6.5 = 3.5

mmol of conjguate formed = 0 + 6.5 = 6.5

so

pH = pKa + log(a-/HA) = 4.75 + log(6.5/3.5)

pH = 5.018