Acid-base Titration Curves |
Examples of Titration Curves
| General Type |
Example |
Typical Titration Curve |
Features of Curve |
|---|
| Strong Acid and 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 and 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 and 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 and 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 and Weak Acid. |
| Weak Acid and 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 |
Plotting these points will result in a curve for strong acid and strong base titration as shown in the first table.
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