1. TITRIMETRIC METHODS OF ANALYSIS
CHAPTER 6
TITRIMETRIC METHODS OF ANALYSIS

2. Titrimetric methods include a large and powerful group of quantitative procedures based on measuring the amount of a reagent of known concentration (standard solution-titrant) that is consumed by the analyte. Titrimetry is a term which includes a group of analytical methods based on determining the quantity of a reagent of known concentration that is required to react completely with the analyte.

3. Basic principle aA +bB → Products A: Titrant (standard solution)
B: Analyte (unknown )
a and b are number of moles of each

4. Requirments inTitration
1- stoichiometric reaction
2- rapid rate reaction
3- quantitative reaction (99.9%complete at stoichiometry)
4-have a defined end or equivalence point

5. Equivalence and end Points
Equivalence point:
Related to the amounts of reactants consumed.
End point
Related to the physical sign that associate with the condition of chemical equivalence.
Ideally, the end point and equivalent one coincide. (variation may be to color change of indicator)

6. Standard solutions Standard Solution Standardization
- A primary standard
- A secondary standard
Standardization
A process in which concentration of a volumetric solution is determined by titrating it with a known mass of a primary standard.

7. Titration: This is performed by adding a standard solution from a buret or other liquid- dispensing device to a solution of the analyte until the point at which the reaction is believed to be complete.
Solution
of NaOH
of HCl
5 mL

8. PRIMARY STANDARD HIGH PURITY ATMOSPHERIC STABILITY
INDEPENDENT OF HUMIDITY
MODEST COST
LARGE MOLAR MASS
NON HYROSCOPIC

9. MnO4- + 5Fe2+ + 8H+ → 5Fe3+ + Mn2++4H2O
Types of titrimetry
1- Precipitation:
Ag+ + Cl- → AgCl (s)
2- Acid Base titration:
OH- + HA → A- + H2O
3- Complex formation
EDTA2- + Ca2+ → EDTA Ca.xH2O
4- Oxidation Reduction
MnO4- + 5Fe2+ + 8H+ → 5Fe3+ + Mn2++4H2O

10. Calculations with Molarity
The mole: is the formula weight of a substance expressed in grams
Number of moles = weight in Grams/formula weight
Number of m.moles = weight in milligrams/formula weight
Molarity= Moles/Liter or mmoles/ml

11. Moles= Volume (L) . (M)
Weight (g) = Volume (L) . (M) . form weight (g)
m.Moles= Volume (ml) . (M)
Weight (mg) = Volume (ml) . (M) . form weight (g)

12. aA + bB → Products Calculating the results of Titration
Needs: the volume and molarity of the titrant
aA bB → Products
(Titrant) (substance titrated)
No. of m.moles of A = Volume of A titrated (ml) . Molarity of A (M)
Or mmolesA = mlA . (MA)
No. of m.moles of B is obtained by multiplying the No. of mmoles of A by the combining ratio (b/a)

13. mmolesB = mmolesA . (b/a) = mlA . (MA). (b/a)

14. If the weight of substance is needed: from no. of mmoles
Wt (mg) = mmoles . Form wt (g)
mgB = (mlA) . (MA). (b/a) . (form WtB)
If % of B is needed in a sample = (WtB/Wtsample).100
The same is B% =(mgB/mgS).100
B% = (mlA) . (MA). (b/a) . (form WtB)(100)/mgS

16. Calculating the molarity of a solution from a standardizing titration
This means that from the weight of primary standard substance (B )when dissolved and titrated with other solution, the molarity of the solution (A) can be calculated as follows:
aA bB → Products
(SolutionA) (primary standard)
(mlA) (MA)/a = (mlB) (MB)/b
(mlB) (MB) = mmolesB = mgB /Form WtB
(MA)/a = mgB/ (mlA) (b/a) (Form WtB)

18. How many ml of 0.25 M NaOH will react with 10.0 ml of 0.10 M H2SO4.
Exercises
Excercize1
How many ml of 0.25 M NaOH will react with 10.0 ml of 0.10 M H2SO4.

19. (½) x MNaOH x VmL(NaOH) = M(H2SO4) x VmL(H2SO4)
Solution:
H2SO4 + 2 NaOH g Na2SO4 + 2 H2O
mmol NaOH = 2 mmol H2SO4
(½) x MNaOH x VmL(NaOH) = M(H2SO4) x VmL(H2SO4)
0.25 x VmL = 2 x 0.10 x 10
VmL = 8 ml

20. Exercise2:
g sample containing NaHCO3 (FW = mg/mmol) was dissolved and titrated with M HCl requiring mL. Find the percentage of sodium bicarbonate in the sample.

21. Solution
write the equation in order to identify the stoichiometry
NaHCO3 + HCl g NaCl + H2CO3
Now it is clear that the number of mmol of sodium bicarbonate is equal to the number of mmol HCl from the equation:
 M1.V1 (HCl)=M2.V2 (NaHCO3)
mmol NaHCO3 = mmol HCl
mmol = M x VmL
mmol NaHCO3 = ( mmol/ml ) x mL = mmol
Now get mg bicarbonate by multiplying mmol times FW
mg NaHCO3 = mmol x (84.01 mg/mmol) =
% NaHCO3 = ( x 10-3 g/ g) x 100 = 78.14%

22. Exercize3
A g of pure sodium carbonate (FW = 106 mg/mmol) was titrated with approximately 0.1 M HCl requiring ml. Find the molarity of HCl.
Solution
The first thing writing the equation of the reaction.
Na2CO3 + 2 HCl g 2 NaCl + H2CO3

23. mmol HCl = 2 mmol Na2CO3 or from the equation
2 mmol HCl react with 1 mmol carbonate
mmol HCl = 2 mmol Na2CO3 or
MHCl X Vml 1/2 = mmol Na2CO3
substitute for mmol carbonate by mg carbonate/FW carbonate. This gives
MHCl x = 2 x mg/ (106 mg/mmol)
MHCl = M

24. Back-Titrations
In back titration an accurately known amount of a reagent is added to analyte in such a way that some excess of the added reagent is left. This excess is then titrated to determine its amount.
  aA bB → Products + A(excess)
(Titrant1) (analyte)
Then back titration of A(excess) with Titrant2 C
cC dA → Products
(Back titration) (excess)
Titrant2

25. mmol reagent(Titrant1) taken = mmol reagent(net) reacted with analyte + mmol reagent(excess) titrated with Titrant2
 then the analyte can be determined since mmol reagent added are known, and mmol reagent titrated can be calculated from back titration.
A added = A net A excess
mmol reagent taken = mmol reagent net reacted + mmol reagent back titrated
mmol reagent net reacted = mmol reagent taken – mmol reagent back titrated

26. Finally, the number of mmol reagent net reacted with analyte can be related to the number of mmol analyte from the stoichiometry of the reaction between the two substances as normal titration up on the equation.
  aA bB → Products

27. . Now net calculations will be as follows:
  aA bB → Products + A(excess)
(Titrant1) (analyte)
cC dA → Products (Back titration) (excess)
Titrant2
Net reacted
Net reacted

28. Importance of back titration: 
Back-titrations are important especially in some situations like:
When the rate of reaction is slow.
When the titration reaction doesn't have a good end point.

29. Example:

30. Example:

31. Exercise
g of Na2CO3 was mixed in 30 mL of boiled water. Next, 40 mL of HCl solution was added turning the solution colorless to indicate the presence of excess acid. It then required ml of M NaOH to back titrate the excess hydrogen ion in solution to faint pink phenolphthalein endpoint. Calculate the concentration of the unknown HCl solution.

32. First work out how many excess moles of HCl that were neutralized by the NaOH
HCl + NaOH > NaCl + H2O
1 mole HCl needs 1 mole NaOH to be neutralized, a 1:1 ratio
So moles HCl neutralized = moles NaOH added
moles NaOH = Molarity x Litres
= M x L
= moles HCl were in excess.

33. Then we need moles HCl needed to react all the Na2CO3
Na2CO3 + 2HCl > 2NaCl + CO2 + H2O
1 mole Na2CO3 needs 2 moles HCl in order to fully react.
moles Na2CO3 = Wt / FWt
= g / g/mol
= moles Na2CO3
So moles HCl used up by Na2CO3 = 2 x moles Na2CO3
= 2x moles= moles

34. total moles HCl in solution =
= moles HCl
Molarity = moles / L
= / L
= M HCl

35. Calculations with Normality:
The mole: is the formula weight of a substance expressed in grams
The Normal or equivalent: is equivalent weight of a substance expressed in grams
Normality (N) = Number of equivalents per liter
= eq/liters or meq/ml
Number of equivalents = weight in Grams/equivalent weight

36. Equivalents = Volume (L) . (N)
Weight (g) = Volume (L) . (N) . eq. weight (g)
m.eq= Volume (ml) . (N)
Weight (mg) = Volume (ml) . (N) .eq. weight (g)

37. aA + bB → Products Calculating the results of Titration
Needs: the volume and Normality of the titrant
aA bB → Products
(Titrant) (substance titrated)
Using Normality in calculations of titrations, never mind a equal b or not, at equivalence in titrations:
No. of eq. of A = No. of eq. of B
NA . VA = NB . VB

38. No. of meq. of A = No. of meq. of B
NA . mlA = NB . mlB

39. How to calculate the equivalent weight:
The definition of an equivalent in terms of molarity depends on the reaction a substance undergoes.
1- In acid base-reactions, one equivalent is the number of grams of a substance that supplies or combines with a hydrogen atom.
2- In oxidation-reduction, one equivalent is the number of grams of a substance that supplies or combines with an electron.
3- In complex formation, one equivalent is the number of grams of a substance that supplies or react with one ion.

40. Examples:
1eq of HCl = 1 mole = 35.5g of HCl
1eq of H2SO4 = 1/2 mole = 98/2 = 49g of H2SO4
1 eq of Fe2+ = 1mole (Fe2+ → Fe3+ + e) = 55.85g
1eq of KMnO4 = 1/5 mole, (KMnO4 + 5e+ ..→ Mn2+ +…)
= /5= 31.6g
Its seen that the equivalent weight of a substance is related to its formula weight:
eq Wt of HCl = form Wt / 1=35.5/1=35.5 g/eq
eq Wt of H2SO4= form Wt/=98/2=49 g/eq
eq Wt of Fe2+=form Wt/5= 55.85/1=55.85 g/eq
eq of KMnO4 =158.04/5=31.6 g/eq

41. Now we can compute generalrules for calculating the equivalent Wt from the molecular Wt:
1- in acid base reactions or substances: eq Wt= form Wt/number of H+ replaced or denoted: i.e
H2SO4→2H+ + SO42-, eq Wt of H2SO4= form Wt/=98/2=49 g/eq
H3PO4→ 3H+ + PO43-, eq Wt of H3PO4 = 98/3=32.67 g/eq

42. 2- Precipitate-formation and complex formation titrations:
eq. Wt of a metal ion= form Wt/chagre on the ion
Eq Wt of an anion = form Wt/no of metal-ion equivalents it reacts with.
Examples:

43. 3- Redox Reactions: eq Wt = form Wt/number of electrons gained or lost in the reaction.
i.e

44. Examples and exercises:
1- HCl was standardized with N solution of NaOH. If 25 ml of HCl required 32.2 ml of NaOH solution. What is the concentration of HCl.

45. Calculate the percentage of Tin in the alloy.
2- A 200 g sample of a metal alloy is dissolved and then the Tin is reduced to Tin (II). Titration of the Tin (II) required ml of o.1 N K2Cr2O7.
Calculate the percentage of Tin in the alloy.

46. 3- A 1. 0 g sample required 28. 16 ml of 0
3- A 1.0 g sample required ml of 0.1 N NaOH for titration to a phenolphthalein-indicator.
Calculate the percentage of phosphoric acid (H3PO4) in the sample.

47. 4- A150 mg sample of pure Na2CO3 requires 30
4- A150 mg sample of pure Na2CO3 requires ml of HCl solution for titration:
2H+ + CO32- → CO2 (g) + H2O
2HCl + Na2CO3 → CO2 (g) + H2O+ 2NaCl
Calculate the normality of HCl.