1. 12 장 Acid-Base Titration Things to learn :
strong acid – strong base titration
weak acid – strong base titration
strong acid – weak base titration
prediction of titration curve
acid-base indicator:
원리와 선택기준

2. Strong Acid - Strong Base Titration
In strong acid – strong base titration, there are three regions of the titration curve that represent different kinds of calculations :
before equivalence point
at equivalence point
after equivalence point

3. Example : Consider the titration of 50. 00 ml of 0. 020 M KOH with 0
Example : Consider the titration of ml of M KOH with 0.10 M HBr
KOH + HBr H2O + KBr
Before titration :
Moles of HO- present = (0.050 l)(0.020 mol/l)
= moles
pH = 12.3

4. pH of solution =-log [H+] = -log
What happens when 3.00 ml of HBr is added ?
No. of moles in 3.00 ml HBr = (0.003)(0.10) =
Moles of HO- unconsumed = – =
Since the volume in the flask in now 53 ml,
[HO-] in the flask = (0.007 mol)/(0.053 l)
= M
pH of solution =-log [H+] = -log
= 12.12

5. pH = 7.00 What happens at the equivalence point ?
At the equivalence point, the H+ is just sufficient to react with all the HO- to form water. The pH is determined by the dissociation of water.
Since there is mol HO- in the flask, mol H+ must be added to reach equivalence point
Volume of H+ added = =10 ml
Let : x = [H+] = [HO-]
Kw = [H+] [HO-] = x2
x = x 10-7
pH = 7.00
pH at the equivalence point in any strong acid – strong base titration will be 7.00 at 25oC

6. What happens after the equivalence point ?
When ml of HBr is added to the solution :
moles of excess H+ = ( l)(0.10 mol/l)
= mol
Concentration of excess H+ =
= x 10-4 M
pH = 3.08
mol l

7. Weak Acid - Strong Base Titration
Strong + weak complete reaction
In weak acid – strong base titration, the titration curve consists of four regions :
before any base is added
HA H A-
- before the equivalence point : solution consists of a mixture of unreacted HA and A-
- at the equivalence point : all the HA has been converted to A-
A H2O HA + HO-
- beyond the equivalence point : excess strong base is added and the pH of the solution is determined by the amount of strong base Ka Kb

8. Before the addition of NaOH :
Example : Consider a 50.0 ml solution of M HA with pKa = 6.15 which is treated with 0.10 M NaOH
Before the addition of NaOH :
HA H A Ka =
Let x = [H+] =[A-]
Ka = = =
x = 1.19 x 10-4
pH = -log (1.19 x 10-4)
= 3.93
[H+][A-]
[HA] x2 x2
What happens when 3.0 ml of 0.10 M NaOH is added ?

9. What happens at the equivalence point ?
When NaOH is added, a mixture of HA and A- is created a buffer
Moles of NaOH added = (0.003 l)(0.10 M) =
Concentration of A- = = 5.66 x M
Moles of HA left = (0.050 l)(0.020 M) –
=0.0007
Concentration of HA = = M
Using the Henderson-Hasselbalch equation
pH = pKa + log = log
= 5.78
0.0003
0.053
0.0007
0.053
[A-]
5.66 x 10-3
[HA]
0.0132
What happens at the equivalence point ?

10. At the equivalence point, sufficient amount of NaOH has been added to react with all the HA
Moles of HA present = (0.050 l)(0.020 M)
= mole
Volume of NaOH added = (0.0010)/(0.10 M)
= 0.01 l = 10 ml
Concentration of A- = = M
HA HO H2O A-
HA HO-
Since Kw = KaKb Kb = = x 10-8
Let x = [HA] = [HO-]
mol
0.060 l Kb Ka Kw Ka

11. What happens after the equivalence point ?x2 x
[HO-][HA]
[A-]
Kb = = = x10-8
x = 1.54 x 10-5 M
Using Kw = [H+][HO-] = 1x 10-14
[H+] =
pH = -log[H+] =
The pH at the equivalence point in not 7.00.
In weak acid – strong base titration, the pH at the equivalence point is always higher than 7
1x 10-14
1.54 x 10-5
What happens after the equivalence point ?

12. Now there is excess NaOH in the solution
Since NaOH is a strong base, we can say that the pH is determined by the concentration of the excess NaOH
When ml of NaOH is added there is an excess of ml of NaOH.
[HO-] = = 1.66 x 10-4 M
pH = -log[H+]
= -log
= 10.22
(0.10 M)( l) l Kw
[HO-]

13. Weak Base - Strong Acid Titration
B H+ BH+
Since it is a strong acid, therefore reaction goes essentially to completion
In weak base – strong acid titration, the titration curve consists of four regions :
before any acid is added:
B + H2O BH HO-
before the equivalence point – solution is a buffer
B HA BH A-
BH H2O B H3O+
pH = pKa (for BH+) + log Kb Ka
[B]
[BH+]

14. - at the equivalence point
B HA BH A-
BH H2O B H3O+
pH is obtained by considering the acid dissociation
[BH+]  original [B] because of dilution
Since the solution at equivalence point contains BH+ , thus pH should be below 7
after the equivalence point
There is excess HA in the solution. Since HA is a strong acid, it determines the pH (contribution from the hydrolysis of BH+ is comparatively small and can be neglected)

15. Titration in Diprotic Systems
Treatment is an extension from the monoprotic system
Example :
Consider the titration of 10 ml of 0.10 M base, B, with 0.10 M HCl. The base is dibasic with pKb1 = 4.00 and pKb2 = Calculate the pH at each point along the titration curve
Before an acid is added :
B + H2O BH HO-
Let x = [BH+] = [HO-] . Thus
Kb1 =
= = 1.00 x 10-4
x = 3.11 x 10-3
[H+] = pH=11.49
Kb1
[BH+] [HO-]
[B] x2 x Kw
[HO-]

16. If 1.5 ml of HCl has been added :
When acid is added and before the first equivalence point, we have a buffer containing B and BH+
B H BH+
BH H BH22+
BH BH+ + H+
BH B + H+
If 1.5 ml of HCl has been added :
Moles of HCl added = ( l)(0.10 M)
= 1.5 x 10-4
= moles of BH+ formed
[ BH+] = = x 10-2 M
Kb1
Kb2
Ka1
Ka2
1.5 x 10-4
0.0115

17. Moles of B left = (0.010)(0.10) – (1.5 x 10-4) = 0.00085 =
[B] = = 7.39 x 10-2 M
Using the Henderson-Hasselbalch equation:
pH = pKa2 + log
= p( ) + log
= log (5.667) =
0.0115
[B]
[BH+]
7.39 x 10-2 Kw
1.304 x 10-2
Kb1

18. At the first equivalence point, all the B has been converted to BH+ which is both an acid and a base
Moles of B present = (0.010)(0.10) = 0.001
Volume of acid used = = 10 ml
[BH+] = = M
Using : [H+] =
where K1 = Ka1 and K2 = Ka2
[H+] =3.16 x10-8
pH = 7.50
0.001 mol
0.10 M
0.001 mol
0.020 l
K1K2[BH+] + K1Kw
K1 +[BH+]

19. At regions between the first and second equivalence points, a buffer containing BH+ and BH2+ is formed :
BH H BH2+
BH BH H+
When 15 ml of HCl is added :
Moles of BH22+ = (0.005 l)(0.10 M) =
[BH22+] = = M
Moles of BH+ left = –
[BH+] = = 0.02 M
Ka1
mol
0.025 l
mol
0.025 l

20. At the second equivalence point, all the BH+ has been converted to BH2+
BH+ + H BH22+
BH BH+ + H+
pH of the solution is determined by the acid dissociation
Moles of BH+ present at first equivalence point = 0.001
Volume of acid added between first and second equivalence point = (0.001 mol)/((0.10 M)
= 10 ml
Ka1
Using the Henderson-Hasselbalch equation:
pH = pKa1 + log
= log 1 = 5.00
[BH+]
[BH22+]

21. Moles of BH22+ = moles of BH+ = 0.001
Ka1 = =
x = 5.72 x 10-4
pH = -log (5.72 x 10-4) = 3.24
0.001 mol
0.030 l
[BH+][H+]
[BH22+] Kw
Kb2 x2 x
Beyond the second equivalence point, the pH is determined by the excess HCl
If the total volume of HCl added is 25.0 ml,
Moles of excess HCl = (0.005 l)(0.10 M) =
[H+] = = 1.43 x 10-2 M
pH = 1.85
0.0005
0.035

22. Titration of 10. 0 ml of 0. 100 M base (pKb1 = 4. 00, pKb2 = 9
Titration of 10.0 ml of M base (pKb1 = 4.00, pKb2 = 9.00) with M HCl
(b) Titration of 10.0 ml of M nicotine (pKb1 = 6.15, pKb2 = 10.85) with M HCl

23. Finding the End Point Titrations are commonly performed to determine :
how much analyte is present
the equilibrium constants of the analyte
How would one determine the end point ?
Autotitrator
pH is measured by electrodes immersed in the analyte solution
DpH/ DV and (DpH/ DV)/ DV are computed
when DpH/ DV is maximum and (DpH/ DV)/ DV = end point
(DpH/ DV)/ DV
DpH/ DV

24. Alkalinity : pH 4.5로 되게 하는데 필요한 산의 량.
Acidity와 Alkalinity
Alkalinity : pH 4.5로 되게 하는데 필요한 산의 량.
주요 수질지수
= [OH-] + [HCO3-] + 2[CO3--]
pOH vs Alkalinity
세기함수 대 크기함수
온도, T : 열용량, Q 에 대응
Acidity : pH 8.3되게 하는데 필요한NaOH량
pH 4.5, 8.3 : H2CO3의 1, 2차 당량점
용도: Alkalinity - Hardness = 잉여 탄산 나트륨 :
농업용수 평가기준

25. (ii) Indicator Hln In- + H+
An acid-base indicator is itself an acid or base whose various protonated species have different colours : eg phenolphthalein
Hln In H+
pH = pK1 + log
The indicator changes color over a pH range
Generally only one color is observed if the ratio of the concentrations of the two forms is 10:1
acid color
base color
[In-]
[Hln]

26. When only the color of the unionized form is seen : =
pH = pK1 + log = pK1 (- 1)
When only the color of the ionized form is seen :
= pH = pK 1
[In-]
[Hln] 10 1 10
[In-] 10
[Hln] 1
So the pH in going from one color to the other has changed from (pK ) to (pK1 + 1) most indicators require a transition range of about two pH units
The pH range over which the color changes is called the transition range
0.7< pH< 2.7
8.0< pH< 9.6

27. Use only a few drops of dilute indicator solution for each titration
Choosing an Indicator
An indicator with a color change near pH 5.54 would be useful in determining the end point of the titration.
The closer the point of color change is to pH 5.54 the more accurate will be the end point
The difference between the observed end point (when there is a color change) and the true equivalence point is called the indicator error
Use only a few drops of dilute indicator solution for each titration

28. Various Indicators

29. 12-7 Practical Notes Acids or Bases Primary Standards : pure material
How about NaOH, KOH, or HCl?
Why not?
NaOH : Carbonate impurity & absorbs CO2
- Hygroscopic
- Should know few primary acids and bases
: Table 12-5

30. Titn in Non-aqueous solvent
Why?
1. Insoluble
2. Unstable (water reactive)
3. Too weak in water
Leveling Effect?
In water H3O+ and OH- are the strongest acid and base!
Any stronger one converts to?
HCl, HClO4 which is stronger?
Not in water but in HOAc
What is Micelle?

31. Acid-Base Titration <Summary>
strong acid – strong base titration
weak acid – strong base titration
strong acid – weak base titration
prediction of titration curve
acid-base indicator:
원리와 선택기준
New terms:
Gran plot, indicator error, acidity, alkalinity,
leveling effect, micelle