1. Chapter 10 Acid-Base Titrations

2. 10-1 Titration of Strong Base with Strong Acid
Standard solution(titrant): Strong acid (H+) Analyte: Strong base (OH-) Titration reaction: OH- + H+ → H2O Reaction product: Neutral salt and H2O
Titration curve: VBase vs. pH
Initial point(no acid added), :the pH is determined from the concentration of strong base.
only [OH-][H+]=Kw/[OH-]
Region 1: Before the equivalence point, :the pH is determined by excess (unreacted) OH- in the solution.
excess [OH-][H+]=Kw/[OH-]
Region 2: At the equivalence point, :the pH is determined by dissociation of water.
only water dissociation[H+]=[OH-]=1.0×10-7M
Region 3: After the equivalence point, :the pH is determined by excess H+ in the solution. excess [H+]only [H+][H+]

3. Titration Curve
Ex) Consider the titration of mL of M KOH with M HBr
Titration reaction:
- Since K is very large, each addition of H+ reacts completely with OH-
- At equivalence point, all OH- has been consumed.
(⇒ sudden increase in H+ concentration)
☞ What volume of H+ titrant is need to reach the equivalence point (Ve) ?
mol OH-
mol H+ or
mmol OH-
mmol H+

4. Region 1: Before the Equivalence Point (excess OH-)
Ex. The titration of mL of M KOH with 3.00 mL of M HBr
mmoles of OH- = original mmoles of OH- - mmoles of H+ added
Volume is mL (=50.00 mL mL)
(10-1) ⇒
Region 2: At the Equivalence Point (only water)
Ex. Addition of mL of M HBr
Just enough H+ has been added to consume OH-
pH determined by dissociation of water
⇒ pH at the equivalence point for any strong acid with strong base is 7.00
Region 3: After the Equivalence Point (excess H+)
Ex. The titration of mL of M KOH with mL of M HBr
mmoles of H+ = mmoles of H+ added - original mmoles of OH-
Volume is mL (=50.00 mL mL)

5. Rapid change in pH near equivalence point
Titration Curve Ve
Rapid change in pH near equivalence point
Equivalence point is where slope is greatest (or Second derivative is 0)
pH at equivalence point is 7.00, only for strong acid-base
Note) Titration Curve of strong acid with strong base pH
Equivalence point
Vb(mL)

6. 10-2 Titration of Weak Acid with Strong Base
Standard solution(titrant): Strong base (OH-) Analyte: Weak acid (HA) Titration reaction: HA + OH- → H2O + A- Reaction product: Basic salt and H2O
Region 1: Initial point(no acid added), : the pH is determined by Ka of HA
Titration curve: VBase vs. pH ④
Region 2: Before the equivalence point (buffer region), : the pH is determined by Henderson Hasselbach equation ③ ② ①
Region 3: At the equivalence point, : the pH is determined by Kb of A–
① HA only
② HA/A- buffer solution
③ A- only (A- + H2O = HA + OH-)
Region 4: After the equivalence point (excess OH-), : the pH is determined by the excess [OH-] present.
④ excess OH-

7. Titration Curve
Ex) Titration of mL of M MES (pKa = 6.27) with M NaOH
Titration reaction:
- Since K is very large, each addition of OH- reacts completely with HA
- At equivalence point, all HA has been consumed.
(⇒ sudden increase in OH- concentration)
☞ What volume of OH- titrant is need to reach the equivalence point (Ve) ?
mmol HA
mmol OH-

8. Region 1: Before Base Is Added (only HA present)
HA A- + H+
Initial(M)
Equili.(M) (0.02-x) (x) (x)
Region 2: Before the Equivalence Point (buffer solution)
- Adding OH- creates a mixture of HA and A-  Buffer
- Calculate pH from [A-]/[HA] using Henderson-Hasselbach equation
Ex. The titration of mL of M MES with 3.00 mL of M NaOH
HA + OH- → A- + H2O
Initial(mmol) * *
Change(mmol)
Equili.(mmol) (0.7) (0) (0.3)
Ex. The point at which volume of titrant is ½Ve (5.00 mL of M NaOH)
HA + OH- → A- + H2O
Initial(mmol)
Change(mmol)
Equili.(mmol) (0.5) (0) (0.5)
(Vb=½Ve ⇒ pH=pKa)

9. Region 3: At the Equivalence Point (only A- present)
- Exactly enough NaOH to consume HA
- Just A- (basic salt) present
Ex. The titration of mL of M MES with mL of M NaOH
HA + OH- → A- + H2O
Initial(mmol)
Change(mmol)
Equili.(mmol) (0.0) (0.0) (1.0)
Volume is mL (=50.00 mL mL)
Concentration of A- present ([A-])
A- + H2O = HA + OH-
Initial(M)
Change(M) -x x +x
Equili.(M) ( x) (x) (x)
⇒ pH will always be above 7.00 for titration of a weak acid
because acid is converted into conjugate base at the equivalence point.

10. Region 4: After the Equivalence Point (excess OH-)
Ex. The titration of mL of M MES with mL of M NaOH
mmoles of OH- = mmoles of OH- added - original mmoles of HA
Volume is mL (=50.00 mL mL)

11. Titration Curve
Rapid Change in pH near Equivalence Point (Steepest part of curve)
Equivalence point is where slope is greatest (Maximum slope, inflection point or second derivative is 0)
Vb = ½Ve  pH = pKa
:Minimum slope
pH at equivalence point above 7.00
(∵ Only A- present)

12. Effect of pKa or acid strength on titration curve
Inflection point or maximum slope decreases with weaker acid
⇒Equivalence point becomes more difficult to identify
Weak acid  small slope change in titration curve Difficult to detect equivalence point
Strong acid  large slope change in
titration curve Easy to detect equivalence point
Effect of acid concentration on titration curve
Curves change depends on acid concentration
Inflection point or maximum slope decreases with lower acid concentration
⇒Equivalence point becomes more difficult to identify
⇒Eventually can not titrate acid at very low concentrations
High concentration  large slope change in
titration curve Easy to detect equivalence point
Low concentration  small slope change in
titration curve Difficult to detect equivalence point At low enough concentration, can not detect change

13. Differences in pH Curves for Strong and Weak Acid with Strong Base7
Strong Acid
Weak Acid
equivalence point: pH>7
equivalence point: pH=7
Weak Acid(HA)
Strong Acid(H+) Ve
For same concentration of H+ and HA;
Before equivalence point : different shape of two curves (different pH)
At equivalence point : different pH
After equivalence point : same shape of two curves (same pH)

14. 10-3 Titration of Weak Base with Strong Acid
- Standard solution(titrant): Strong acid (H+) - Analyte: Weak base (B) - Titration reaction: B + H+ → H2O + BH+ - Reaction product: Acidic salt and H2O
Region 1: Initial point(no Base added), : the pH is determined by Kb of B ①
Titration curve: VAcid vs. pH ②
Region 2: Before the equivalence point (buffer region), : the pH is determined by Henderson Hasselbach equation ③ ④
Region 3: At the equivalence point, : the pH is determined by Ka of BH+
① B only
② B/BH+ buffer solution
③ BH+ only (BH+ + H2O = B + H3O+)
④ excess H3O+
⇒ The pH at the equivalence point is below 7.00 (∵Only BH+ present).
Region 4: After the equivalence point (excess H+), : the pH is determined by the excess [H+] present.

15. Titration of Pyridine with HCl
Example
Titration of Pyridine with HCl
Consider the titration of mL of M pyridine(B) with M HCl.
The volume of H+ titrant is need to reach the equivalence point (Ve);
mmol H+
mmol B
pH at Va=4.63 mL ?
Solution
Va=4.63 mL : before equivalence point ⇒ B + BH+ buffer
B H+ → BH+
Initial(mmol)
Final(mmol)

16. 10-4 Titration in Diprotic System
Principals for monoprotic systems apply to diprotic and triprotic Systems.
- Multiple equivalence points and buffer regions
- Multiple inflection points (2 or 3 equivalence points) in titration curve
A Typical Case
Curve a) Titration curves of 10.0 mL of 0.100M B(pKb1=4.00, pKb2=9.00) with 0.100M HCl
Base(B) reaction:
Titration reaction:
Kb1
Kb2
1st step: B + H+ → BH+
B BH BH22+ K2 K1
2nd step: BH+ + H+ → BH22+
Two equivalence points
B → BH → BH22+
• pH calculations (for curve a):
Point A: only B ⇒by Kb1 (base form)
B/BH+
BH+/BH2+
Point A~C: B/BH+ buffer solution
Point B: [B]=[BH+] ⇒pH=pK2
Point C(1st eq): only BH+(intermediate form)
Point C~E: BH+ /BH22+ buffer solution
Point D: [BH+]=[BH22+] ⇒pH=pK1
Point E~: excess H+
Point E(2nd eq): only BH22+ ⇒by K1(acid form)
⇒(curce a) Observe two obvious, steep eq. points(C, E).

17. • pH calculations (for curve a):
Ex) Titration curves of 10.0 mL of 0.100M B(pKb1=4.00, pKb2=9.00) with 0.100M HCl
B BH BH22+ K2 K1
Kb2
Kb1
Base reaction:
2nd step: BH+ + H+ → BH22+
1st step: B + H+ → BH+
Titration reaction:
1st eq.
B → BH+ → BH22+
B/BH+
BH+/BH2+
2nd eq.
• pH calculations (for curve a):
Point A: only B present⇒by Kb1 (base form)
B+H2O=BH++OH-
x x x
[OH-]=3.11×10-3M ∴pH=11.49
Point B: [B]=[BH+] ⇒pH=pK2
∴pH=10.00
Point C(1st eq): only BH+(intermediate form) present
∴pH=7.50
Point D: [BH+]=[BH22+] ⇒pH=pK1 ⇒
∴pH=5.00
Point E(2nd eq): only BH22+ present⇒by K1(acid form)
BH2+ = BH+ + H+
⇒ [H+]=5.72×10-4M ∴pH=3.24
x x x

18. Blurred End Point B → BH+ → BH22+
Curve b) Titration curves of 10.0 mL of 0.100M B(pKb1=6.15, pKb2=10.85) with 0.100M HCl
Two equivalence points
B → BH+ → BH22+
Curve (a): show two clear end points
Curve (b): do not show both end points ⇒ no perceptible break at 2nd end point(point J). ∵ For point J, BH22+ is too strong an acid(or BH+ is too weak a base →The pH between points I and J requires the systematic treatment of equilibrium.
For polyprotic acid or base;
When the pH is too low or too high or when pKa values are too close together, end points are obscured.
Ex mL of M nicotine (pKb1 = 6.15, pKb2 = 10.85) with M HCl.
Nicotine (B)

19. : Phosphoric Acid (H3A) with 0.1 M OH-
Titration Curve of Triprotic Acid
: Phosphoric Acid (H3A) with 0.1 M OH-
pKa1=2.15
pKa2=7.2
pKa3=12.15
(3rd end point)
:Do not observe in aqueous solution
(2nd end point)
(1st end point)

20. 10-5 Finding the End Point with a pH Electrode
Equivalence point (Chapter 1)
The point at which an added titrant is stoichiometrically
equal to the number of moles of analyte present
- Stoichiometric point (theoretical point)
End point (Chapter 1)
The point at which the physical property of solution changes
(Ex. sudden changes of color by indicator)
- Occurs from the addition of a slight excess of titrant
- Experimental point
End Point Determination(This chapter)
1) pH electrode
- Derivatives
- Glan plot
2) Indicator
- acid/base indicators

21. Titration with pH meter
Using Derivatives to Find the End Point
Titration curve of H6A:
3rd end point(~90 μL eq. point): H4A2-+OH-→H3A3-+H2O 4th end point(~120 μL eq. point): H3A3-+OH-→H2A4-+H2O
Titrant (NaOH)
V vs. pH
Analyte
4th End point
Titration curve :3rd and 4th end point ⇒ Difficult to find out end points.
3rd End point
Titration with pH meter
3rd Eq. Ve
4th Eq. Ve
First derivative :where the slope is greatest(maximum slope)
Second derivative :where the 2nd derivative is zero
Auto titrator

22. End point from derivatives (3rd end point (~90 μL))
-First derivative
Titration Data (curve) 평균 평균
→ (85.5, 0.155)
(⇒ one point in Fig. 10-5(b) near 3rd end point)
-Second derivative
→ (86.0, 0.071)
(⇒ one point in Fig. 10-5(c) near 3rd end point)
(86.0, 0.071)
(Second derivative)
3rd end point
4th end point

23. Using a Gran Plot to Find the End Point7,8
Gran plot uses data from before the end point (typically from 0.8 Ve or 0.9 Ve up to Ve) to locate the end point.
HA = H+ + A Ka = ([H+]γH+[A-]γA-)/[HA]γHA
It will be necessary to include activity coefficients in this discussion because a pH electrode responds to hydrogen ion activity, not concentration.
moles of OH- delivered
total volume
original moles of HA – moles of OH-

24. A graph of Vb10-pH versus Vb is called a Gran plot.
Gran plot equation:
A graph of Vb10-pH versus Vb is called a Gran plot.
The beauty of a Gran plot is that it enables us to use data taken before the end point to find the end point.

25. Using a Glan Plot to Find the End Point
where: Vb = volume of strong base added
Ve = volume of base needed to reach equivalence point
γA- , γHA = activity coefficients ≈ 1
Plot is a straight line - Gran plot  Graph of Vb10-pH vs Vb
(If ratio of activity coefficients is constant)
⇒ Slope = -Ka∙ γHA/γA-
⇒ X-intercept = Ve (must be extrapolated)
Measure end point with data before reach end point
Only use linear region of Gran Plot
- Changing ionic strength changes activity coefficients
 added salt to maintain constant ionic strength
⇒Vb is 10-20% of volume prior to Ve

26. ⇒ Plot Vb10-pH vs Vb Slope: Ka x-intercept: Ve
Never goes to zero, approximation that every mole of OH- generates one mole of A- is not true as Vb approaches Ve
Challenge : Show that when weak base, B, is titrated with a strong acid, the Gran function is
(10-6)
where Va is the volume of strong acid and Ka is the acid dissociation constant of BH+.

27. 10-6 Finding the End Point with Indicator
Indicators: compound added in an acid-base titration to allow end point detection
- Acid-Base Indicators (HIn) are weak acids or bases.
: Change color with pH (Different protonated species have different colors)
Thymol Blue
pK1 = 1.7
pK2 = 8.9

28. Transition range of indicator color
pK1=1.7
HIn In- + H+
(10-7)
(R)
(Y)
R color(HIn) ;
Y color(In-);
pK1-1
pK1+1 pH
[Y-]:[R]
Color
0.7
1:10
Red
pK1-1
1.7
1:1
Orange
pK1
2.7
10:1
Yellow
pK1+1
pK1
Transition
range
R color
R+Y
Y color
Transition range1 : pH = pK1 ± 1
Transition range2 : pH = pK2 ± 1

29. pH<0
0<pH<8.2
pH>13
8.2<pH<12
Phenolphthalein

30. Choosing an Indicator
- An indicator with a color change near equivalence point pH would be useful for an end point.
⇒The closer the two match, the more accurate determining the end point will be.
Transition range: 5.2~6.8
Bromocresol purple color change brackets the equivalence point and is a good indicator choice
Transition range: 3.8~5.4
Bromocresol green will change color
Significantly past the equivalence point resulting in an error.

31. The Effect of Acid Strength on indicatorPT
BTB
BCG
The Effect of Concentration on indicator
A: 0.05 M HCl
B: M HCl

32. 10-7 Practical Notes
Primary Standards-Acids: to prepare standard solution of NaOH or KOH
- Solid NaOH/KOH are not primary standards!
NaOH/KOH must be protect from the atmosphere; otherwisw absorb CO2 to form HCO3- (OH- + CO2 → HCO3-)
Standard NaOH/KOH solution should be stored in plastic container (Never keep in the glass)

33. Primary Standards-Bases: to prepare standard solution of HCl or H2SO4
CO32- + H+ = HCO3-
HCO3- + H+ = H2CO3

34. Titration curve of Na2CO3
Na2CO3 (sodium carbonate): primary standard
• End points (2 step)
         - 1st end point (pH ~8.4)
         - 2nd end point (pH ~4.0)
: used for standardization
   (∵pH changes are larger than 1st end point.)
CO32-
CO32-/HCO3-
1st end point
HCO3-/H2CO3
2nd end point

35. 10-8 Kjeldahl Nitrogen Analysis
☞ Determining N in organic substance
① Digestion
H2SO4
Organic C, H, N
NH4+ + CO2 + H2O
(10-8)
boiling
- Catalysts (Hg, Cu, Se compounds): catalyze the digestion
- Adding K2SO4: increase bp of H2SO4  speed the reaction
- H2SO4 +H2O2 or K2S2O8 +NaOH in a microwave bomb (pressurized vessel): alternative digestion
② Neutralization NH4+ + OH- → NH3(g) + 2H2O
(10-9)
③ Distillation
NH3 + H+(excess, standard solution) → NH4+ -
(10-10)
④ Titration
H+(remaining) + OH-(standard solution) → H2O
(10-11)
BOX 10-3 Kjeldahl Nitrogen Analysis Behind the Headlines
•Illegal food from China(Nonprotein nitrogen) :Adding melamine & cyanuric acid to food ingredients→die from kidney failure.
Protein source
Weight % N
Meat
16.0
Blood plasma
15.3
Milk
15.6
Flour
17.5
Egg
14.9
☜Proteins: ~16 wt% N

36. Protein concentration (A typical protein contains 16.2 wt% N).
Example
Kjeldhal Analysis
Protein concentration (A typical protein contains 16.2 wt% N).
Digestion & Distillation
H2SO4
OH-
Protein (N)
(0.500 mL)
NH4+
NH3(g)
boiling
Neutralization
NH3(g) + HCl ( M, 10.00mL) → NH4+ + Cl-
Titration
HCl (remaining) + NaOH (0.0198M, 3.26mL) → NaCl + H2O
⇒ Find the content of protein (mg)/mL.
Solution
Neutralized HCl mmol = ( M)(10.00 mL) - ( M)(3.26 mL)
= mmol
= NH3 mmol
N mass = ( mmol)( mg/mmol) = mg N
Protein mass (~16.2 wt% N) from titration result = mg/0.162 mg = 12.9 mg protein
∴ Protein mass (mg/mL) = 12.9 mg/0.5 mL = 25.8 mg protein/mL

37. 10-9 The Leveling Effect
If an acid stronger than H3O+ is dissolved in water, it cannot have a
stronger acid group than H3O+; it is said to be “leveled” by the water.
Similarly, the stronger base is also “leveled” by the water.
Strong acids behave with the same strength in water. This is called the leveling effect-they all dissociate to H3O+
Ex) HClO4 is a stronger acid than HCl, but in water they have the same strength. If these acids were dissolved in acetic acid solvent, then the difference in acid strength could be observed.
☞ In acetic acid solvent (non-aqueous solvent, less basic than H2O): ⇒ HClO4 is a stronger acid than HCl
H2O: leveling solvent
Acetic acid : differentiating solvent

38. Titration in Non-Aqueous Solvent
Amphiprotic: alchols,…
Acidic: carboxylic acids,…
Basic: amines,…
Protic
Non-Aqueous Solvent
Aprotic:
Benzene, dioxine, acetone,….
A titration curve for several acids against tetrabutylammonium hydroxide in methyl isobutyl ketone solvent. Again, we see that HClO4 is a stronger acid than HCl.
We can use non-aqueous solvents to titrate species that we can’t in water. For instance, species that are too weak (Low K) to titrate with H+ in water, might titrate with HClO4 in other solvents.
Ex) Titration of too weak base(B, Urea(Kb=1.3×10-14))
Titration with HClO4 in H2O
☞ K for titration reaction is not large enough→end point can’t be recognized.
B + H3O BH+ + H2O
Titration with HClO4 in CH3COOH
☞ HClO4 is stronger than H3O+ in acetic acid solvent→K for titration reaction is large enough→give a distinct end point.
B + HClO BH+ + H2O

39. ■ 알카리 혼합물 정량 : PT의 변색범위(pH 8.0~9.6) : BCG의 변색범위(pH 3.8~5.4) pH 분석물
VPT와 VBCG의 관계
NaOH
VPT=VBCG
Na2CO3
VPT=½VBCG
NaHCO3
VPT=0, VBCG>0
NaOH + Na2CO3
VPT>½VBCG
Na2CO3 + NaHCO3
VPT<VBCG
VPT : PT 당량점의 부피
VBCG : BCG 당량점의 부피
: PT의 변색범위(pH 8.0~9.6)
: BCG의 변색범위(pH 3.8~5.4)
NaOH
VPT
VBCG
Na2CO3
VPT
VBCG
NaHCO3
VPT=0 VBCG>0 pH
7.0
Na2CO3+NaHCO3
VPT
VBCG
NaOH+Na2CO3
VPT
VBCG