Titration Curves

General Type	Example	Typical Titration Curve	Features of Curve
Strong acid & Strong Base	HCl added to NaOH		Curve begins at high pH typical of strong base and ends at low pH typical of strong acid. There is a large rapid change in pH near the equivalence point (pH =7).
Strong base & strong acid	NaOH added to HCl		Curve begins at low pH typical of strong acid, and ends at high pH typical of strong base. There is a large rapid change in pH near the equivalence point (pH=7).
Weak acid & Strong base	NaOH added to acetic acid (CH3COOH)		Curve begins at a higher acidic pH and ends at high basic pH. The pH change at the equivalence point (pH > 7)is not so great.
Strong acid & Weak base	Ammonia (NH3) added to HCl		Curve begins at low pH and ends at a less high basic pH. The pH change at the equivalence point (pH < 7) is similar to that for Strong base & Weak acid.
Weak acid & Weak base	Ammonia (NH3) added to Acetic acid (CH3COOH)		Curve begins at higher acidic pH and ends at low basic pH. There is not a great pH change at the equivalence point (pH ~ 7) making this a very difficult titration to perform.

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