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