Question

Find the pH during the titration of 20.00 mL of 0.1460 M benzoic acid, C6H5COOH (Ka = 6.3 ✕ 10-5), with 0.1460 M NaOH solution after the following additions of titrant.

A) 0 mL

B) 10.00 mL

C) 15.00 mL

D) 19.00 mL

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/(M-x)

6.3*10^-5 = x*x/(0.146-x)

This is quadratic equation

x =0.003

For pH

pH = -log(H+)

pH =-log(0.003)

pH in a = 2.52

b) 10 ml KOH

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

pH = pKa + log (A-/HA)

initially

mmol of acid = MV = 20*0.1460 = 2.92 mmol of acid

mmol of base = MV = 10*0.1460 = 1.46 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 2.92 -1.46 = 1.46 mmol

mmol of conjguate left = 0 + 1.46 = 1.46

Get pKa

pKa = -log(Ka)

pKa = -log(6.3*10^-5) = 4.20

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 4.20+ log (1.46/1.46) = 4.20

c) for 15 ml

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

pH = pKa + log (A-/HA)

initially

mmol of acid = MV = 20*0.1460 = 2.92 mmol of acid

mmol of base = MV = 15 *0.1460 = 2.19mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 2.92 -2.19= 0.73mmol

mmol of conjguate left = 0 + 2.19= 1.46

Get pKa

pKa = -log(Ka)

pKa = -log(6.3*10^-5) = 4.20

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 4.20+ log (1.46 /0.73) = 4.501

d) After 19 mL

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

pH = pKa + log (A-/HA)

initially

mmol of acid = MV = 20*0.1460 = 2.92 mmol of acid

mmol of base = MV = 19*0.1460 = 2.774mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 2.92 -2.774= 0.146 mmol

mmol of conjguate left = 0 + 2.774= 2.774

Get pKa

pKa = -log(Ka)

pKa = -log(6.3*10^-5) = 4.20

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 4.20+ log (2.774/1.46 ) = 4.4787