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.

IB syllabus for first examinations 2016

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)

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.

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.