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pH
7
1
13
25
cm3 of base
strong base
weak base
strong acid
weak acid
Titration Curve Calculations
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
Vol NaOH/cm3
Moles NaOH added
Moles excess acid
[H+] / mol.dm-3
pH
0 0 2.50 x 10-3 0.100 1.00
5
10
15
20
24.9
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
Adding NaOH will react some HCl in a 1:1 ratio.Moles NaOH added = Moles HCl removedThe remaining acid is now in a larger volume.
e.g. 5cm3 of NaOH = 5x10-4 moles Remaining moles of HCl = 2.5x10-3 – 5x10-
4
= 2.0 x10-3
[H+] = moles = 2.0x10-3 = 0.0667 volume 30x10-3
pH = -log10[H+] = -log10 0.0677 = 1.18
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
Vol NaOH/cm3
Moles NaOH added
Moles excess acid
[H+] / mol.dm-3
pH
0 0 2.50 x 10-3 0.100 1.00
5 5 x 10-4 2.00 x 10-3 0.0667 1.18
10
15
20
24.9
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
Vol NaOH/cm3
Moles NaOH added
Moles excess acid
[H+] / mol.dm-3
pH
0 0 2.50 x 10-3 0.100 1.00
5 5 x 10-4 2.00 x 10-3 0.0667 1.18
10 1.0 x 10-3 1.50 x 10-3 0.0429 1.37
15 1.5 x 10-3 1.00 x 10-3 0.0250 1.60
20 2.0 x 10-3 5.00 x 10-4 0.0111 1.95
24.9 2.49 x 10-3 1.00 x 10-5 0.000200 3.70
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
When 25cm3 NaOH is been added, all the HCl is neutralised, so leaving water as the only source of H+ and OH- ions. Use Kw to calculate [H+].
Kw = [H+] [OH-] = 1.00 x 10-14 at 298K
As [H+] = [OH-] in pure water, Kw = [H+]2
[H+] = √Kw = 1.00 x 10-7
pH = -log10[H+] = 1.00x10-7 = 7.00
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
Volume NaOH/ cm3
Moles NaOH added
Moles excess NaOH
[OH-]/ mol.dm-3
[H+] / mol.dm-3
pH
25.1
30
40
50
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
Now NaOH is in excess, and so need to calculate the moles of excess NaOH and so [OH-] in the increasingly larger volume. [H+] can be calculated from Kw.
25.1cm3 NaOH = 2.51x10-3 moles NaOHExcess NaOH = 2.51x10-3 - 2.50x10-3 = 1.00x10-5
[OH-] = moles = 1.00x10-5 = 2.00 x 10-4
volume 50.1x10-3
[H+] = Kw = 1.00x10-14 = 5.00 x 10-10
[OH-] 2.00x10-4
pH = -log10[H+] = -log105.00 x 10-10 = 10.30
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
Volume NaOH/ cm3
Moles NaOH added
Moles excess NaOH
[OH-]/ mol.dm-3
[H+] / mol.dm-3
pH
25.1 2.51 x 10-3 1.00 x 10-5 2.00 x 10-14 5.00 x 10-11 10.30
30
40
50
Strong acid v Strong base(25cm3 of 0.1 moldm-3 HCl + 0.1mol.dm-3 NaOH)
Volume NaOH/ cm3
Moles NaOH added
Moles excess NaOH
[OH-]/ mol.dm-3
[H+] / mol.dm-3
pH
25.1 2.51 x 10-3 1.00 x 10-5 2.00 x 10-14 5.00 x 10-11 10.30
30 3.00 x 10-3 5.00 x 10-4 9.09 x 10-3 1.10 x 10-12 11.96
40 4.00 x 10-3 1.50 x 10-3 0.0231 4.33 x 10-13 12.36
50 5.00 x 10-3 2.50 x 10-3 0.0333 3.00 x 10-13 12.52
Strong acid v Strong base pH
7
1
13
25
cm3 of base
Weak acid v Strong base25cm3 of 0.1 moldm-3 CH3COOH (Ka=1.74x10-5) +
0.1mol.dm-3 NaOH
Vol NaOH
/cm3
Moles NaOH added
Moles excess acid
Moles salt (A-) made
[H+] / mol.dm-3
pH
0
5
10
15
20
24.9
Weak acid v Strong base25cm3 of 0.1 moldm-3 CH3COOH (Ka=1.74x10-5) +
0.1mol.dm-3 NaOH
Before NaOH added:
Ka = [H+] [A-] [HA]
In the solution [H+] = [A-]
so [H+] = √(Ka [HA] ) = √ (1.74x10-5x 0.1) = 1.32 x 10-3
pH = -log10[H+] = -log10 1.32x10-3 = 2.88
Weak acid v Strong base25cm3 of 0.1 moldm-3 CH3COOH (Ka=1.74x10-5) +
0.1mol.dm-3 NaOHVol
NaOH/cm3
Moles NaOH added
Moles excess acid
Moles salt (A-) made
[H+] / mol.dm-3
pH
0 0 2.50 x 10-3 = moles H+ 1.30 x 10-3 2.88
5
10
15
20
24.9
Weak acid v Strong base25cm3 of 0.1 moldm-3 CH3COOH (Ka=1.74x10-5) +
0.1mol.dm-3 NaOHAdding NaOH will react some of the acid in a 1:1 ratio, but also
makes salt ions (1:1).
Moles NaOH added = Moles acid removed = Moles Salt made
e.g. 5cm3 of NaOH = 5x10-4 moles
Remaining moles of acid = 2.5x10-3 – 5x10-4 = 2.0 x10-3
Moles salt (A-) = moles NaOH = 5 x10-4
Ka = [H+] [A-] so [H+] = Ka [HA] = 1.74x10-5x 2.0x10-3
[HA] [A-] 5x10-4
[H+] = 6.96 x 10-5
pH = -log10[H+] = 4.16
Weak acid v Strong base25cm3 of 0.1 moldm-3 CH3COOH (Ka=1.74x10-5) +
0.1mol.dm-3 NaOHVol
NaOH/cm3
Moles NaOH added
Moles excess acid
Moles salt (A-) made
[H+] / mol.dm-3
pH
0 0 2.50 x 10-3 = moles H+ 1.30 x 10-3 2.88
5 5 x 10-4 2.00 x 10-3 0.5 x 10-3 6.96 x 10-5 4.16
10
15
20
24.9
Weak acid v Strong base25cm3 of 0.1 moldm-3 CH3COOH (Ka=1.74x10-5) +
0.1mol.dm-3 NaOHVol
NaOH/cm3
Moles NaOH added
Moles excess acid
Moles salt (A-) made
[H+] / mol.dm-3
pH
0 0 2.50 x 10-3 = moles H+ 1.30 x 10-3 2.88
5 5 x 10-4 2.00 x 10-3 0.5 x 10-3 6.96 x 10-5 4.16
10 1.0 x 10-3 1.50 x 10-3 1.0 x 10-3 2.61 x 10-5 4.58
15 1.5 x 10-3 1.00 x 10-3 1.5 x 10-3 1.16 x 10-5 4.94
20 2.0 x 10-3 5.00 x 10-4 2.0 x 10-3 4.35 x 10-6 5.36
24.9 2.49 x 10-3 1.00 x 10-5 2.49 x 10-3 6.99 x 10-8 7.16
Weak acid v Strong base pH
7
1
13
25
cm3 of base
pH
7
1
13
25
cm3 of base
strong base
weak base
strong acid
weak acid
Titration curves
