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These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

### Titration Curves

Titrations cannot be carried out between weak acids and weak bases as the inflection is not steep enough, i.e. it does not pass through a large enough range of pH at the equivalence point.

### Calculating a Titration Curve

Imagine an experiment in which 0.10M HCl is added 1mL at a time to a conical flask containing 10mL 0.10M NaOH solution.

HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)

• Calculate the pH of the NaOH(aq) before any HCl is added.
• [OH-] = [NaOH] = 0.10 mol L-1
• pOH = -log10[OH-] = -log10[0.10] = 1
• pH = 14 - pOH = 14 - 1 = 13
• Calculate the pH of the solution after 1mL 0.10 HCl has been added. (NaOH is in excess, HCl is the limiting reagent)
• Calculate moles of HCl added: n(HCl) = M x V
M = 0.10M
V = 1mL = 1 x 10-3L
n(HCl) = 0.10 x 1 x 10-3 = 1 x 10-4 mol
• Calculate moles NaOH unreacted = initial moles NaOH - moles NaOH reacted
initial moles NaOH = M x V
M = 0.10M
V = 10mL = 10 x 10-3L
initial moles NaOH = 0.10 x 10 x 10-3 = 1 x 10-3mol
moles NaOH reacted = moles HCl added = 1 x 10-4mol
moles NaOH unreacted = 1 x 10-3 - 1 x 10-4 = 9 x 10-4mol
• Calculate [OH-] = n(unreacted OH-) ÷ total volume
n(unreacted OH-) = n(unreacted NaOH) = 9 x 10-4mol
total volume = 10mL + 1mL = 11mL = 11 x 10-3L
[OH-] = 9 x 10-4 ÷ 11 x 10-3 = 0.082 mol L-1
• Calculate pOH: pOH = -log10[OH-] = -log10[0.082] = 1.09
• Calculate pH: pH = 14 - pOH = 14 - 1.09 = 12.91
Continue these calculations until 11mL 0.10 HCl is added. At this point the NaOH is no longer in excess, rather it is now the HCl that is in excess.
• Calculate the pH of the solution after 11mL HCl has been added
• moles HCl: n(HCl) = M x V
M = 0.10 mol L-1
V = 11mL = 11 x 10-3L
n(HCl) = 0.10 x 11 x 10-3 = 1.1 x 10-3mol
• Calculate moles HCl in excess
n(HCl) unreacted = total n(HCl) - n(HCl) reacted
total n(HCl) = 1.1 x 10-3 mol
n(HCl) reacted = n(NaOH) = 1 x 10-3 mol
n(HCl) unreacted = 1.1 x 10-3 - 1 x 10-3 = 1 x 10-4 mol
• Calculate [H+]: [H+] = n(H+ unreacted) ÷ total volume
n(H+) unreacted = n(HCl) unreacted = 1 x 10-4 mol
total volume = 10mL + 11mL = 21mL = 21 x 10-3L
[H+] = 1 x 10-4 ÷ 21 x 10-3 =4.76 x 10-3 mol L-1
• Calculate pH of the solution
pH = -log10[H+] = -log10[4.76 x 10-3] = 2.32

Continue these calculations until all the HCl has been added

volume HCl added in L moles (n)HCl added moles (n)NaOH present Total volume of solution [OH-] = n(NaOH) ÷ total volume pOH = -log10[OH- pH = 14 - pOH
0 0 1 x 10-3 10 x 10-3 0.10 1 13
1 x 10-3 1 x 10-4 9 x 10-4 11 x 10-3 0.082 1.09 12.91
2 x 10-3 2 x 10-4 8 x 10-4 12 x 10-3 0.067 1.18 12.82
3 x 10-3 3 x 10-4 7 x 10-4 13 x 10-3 0.054 1.27 12.73
4 x 10-3 4 x 10-4 6 x 10-4 14 x 10-3 0.043 1.37 12.63
5 x 10-3 5 x 10-4 5 x 10-4 15 x 10-3 0.033 1.48 12.52
6 x 10-3 6 x 10-4 4 x 10-4 16 x 10-3 0.025 1.60 12.40
7 x 10-3 7 x 10-4 3 x 10-4 17 x 10-3 0.018 1.75 12.25
8 x 10-3 8 x 10-4 2 x 10-4 18 x 10-3 0.011 1.95 12.05
9 x 10-3 9 x 10-4 1 x 10-4 19 x 10-3 0.0053 2.28 11.72
10 x 10-3 1 x 10-3 0 20 x 10-3 0 undefined undefined
volume HCl added in L moles (n)HCl added moles (n)HCl unreacted Total volume of solution [H+] = n(HCl) unreacted ÷ total volume pH = -log10[H+]
11 x 10-3 1.1 x 10-3 1 x 10-4 21 x 10-3 4.76 x 10-3 2.32
12 x 10-3 1.2 x 10-3 2 x 10-4 22 x 10-3 9.09 x 10-3 2.04
13 x 10-3 1.3 x 10-3 3 x 10-4 23 x 10-3 0.013 1.88
14 x 10-3 1.4 x 10-3 4 x 10-4 24 x 10-3 0.017 1.78

Plotting these points will result in a curve for strong acid & strong base as shown in the first table.