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) + H_{2}O(l)
 Calculate the pH of the NaOH(aq) before any HCl is added.
 [OH^{}] = [NaOH] = 0.10 mol L^{1}
 pOH = log_{10}[OH^{}] = log_{10}[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^{3}L
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^{3}L
initial moles NaOH = 0.10 x 10 x 10^{3} = 1 x 10^{3}mol
moles NaOH reacted = moles HCl added = 1 x 10^{4}mol
moles NaOH unreacted = 1 x 10^{3}  1 x 10^{4} = 9 x 10^{4}mol
 Calculate [OH^{}] = n(unreacted OH^{})
÷ total volume
n(unreacted OH^{}) = n(unreacted NaOH) = 9 x 10^{4}mol
total volume = 10mL + 1mL = 11mL = 11 x 10^{3}L
[OH^{}] = 9 x 10^{4} ÷ 11 x 10^{3} = 0.082 mol L^{1}  Calculate pOH: pOH = log_{10}[OH^{}] = log_{10}[0.082] = 1.09
 Calculate pH: pH = 14  pOH = 14  1.09 = 12.91
 Calculate moles of HCl added: n(HCl) = M x V
 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^{3}L
n(HCl) = 0.10 x 11 x 10^{3} = 1.1 x 10^{3}mol  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^{3}L
[H^{+}] = 1 x 10^{4} ÷ 21 x 10^{3} =4.76 x 10^{3} mol L^{1}  Calculate pH of the solution
pH = log_{10}[H^{+}] = log_{10}[4.76 x 10^{3}] = 2.32
 moles HCl: n(HCl) = M x V
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 = log_{10}[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 = log_{10}[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.

