pharmaceutical analytical chemistry - … new 30...o repeat the drying and weighing process until a...

1

CHAPTER 10 Gravimetric Analysis and Precipitation Equilibria

Learning Objectives

● Steps of a gravimetric analysis

● Gravimetric calculations

● The solubility product, the common ion effect

● The diverse ion effect

Dr. Najjar 17/18

Philadelphia U

niversity

GRAVIMETRIC ANALYSIS

Key Concepts

Gravimetric analysis is the quantitative isolation of a

substance by precipitation and weighing of the precipitate.

A precipitating reagent is the reactant used to precipitate the

analyte.

Usually, the precipitation reaction is selective for the analyte.

The precipitate should be pure, filterable and suitable for

weighing.

2

Ba+2 + CrO4-2 BaCrO4(s)

Reagent Analyte Solid Product

Philadelphia U

niversity

Dr. Najjar 17/18

How to perform a successful Gravimetric Analysis?

3


Philadelphia U

niversity

Dr. Najjar 17/18

o Weigh the sample to be analyzed.

o Dissolve the sample in a suitable solvent, eg, water.

o Add an excess of the precipitating reagent to

precipitate the analyte.

o Digest the precipitate to make larger and more pure crystals.

o Filter the mixture to separate the precipitate from the solution.

o Wash the precipitate to remove any impurities.

o Dry the precipitate by heating to remove water.

o Weigh the cooled precipitate.

o Repeat the drying and weighing process until a constant mass for the

precipitate is achieved.

o Calculate the percent by mass of analyte in the sample.

o Cool the precipitate in a desiccator to prevent the precipitate absorbing moisture from the air.

Important notes:

o Suitable solvent is not only the solvent that dissolve the analyte, but it could be

used to eliminate interfering materials and to maintain low solubility of the

precipitate. Solution conditions (polarity, pH, ionic strength) could be adjusted to

achieve these goals.

o During the precipitation process, supersaturation occurs, followed by nucleation

and precipitation. Supersaturation should be minimized to prevent the formation

of large numbers of nuclei and as a result producing more total crystals of

smaller size.

o The initial nucleus will grow to form a crystal by depositing other precipitate

particles. Again, the greater the supersaturation, the more rapid the crystal

growth rate which increases the chances of imperfections in the crystal and

trapping of impurities.

4


where Q is the concentration of the mixed reagents before precipitation occurs, S is the solubility of the precipitate at equilibrium.

Philadelphia U

niversity

Dr. Najjar 17/18

Important notes:

o Several steps are commonly taken to maintain favorable conditions for

precipitation:

o Very insoluble precipitates are not the best candidates for gravimetric analysis!

They supersaturate too easily.

o Don’t add too much excess precipitating agent. This will increase adsorption.

5


Philadelphia U

niversity

Dr. Najjar 17/18

Important notes:

o Digestion is a process to improves the purity and crystallinity of the precipitate.

In this process, the small particles were dissolved and re-precipitate on larger

crystals, this is usually done by heating the solution gently with contentious

stirring.

o When coagulated particles are filtered, they retain the adsorbed primary and

secondary ion layers along with solvent. Washing the particles with water

increases the extent of solvent (water) molecules between the layers, causing

the secondary layer to be loosely bound, and the particles revert to the colloidal

state. This process is called peptization. Adding an electrolyte will result in a

closer secondary layer and will promote coagulation.

o The electrolyte must be one that is volatile at the temperature to be used for

drying or ignition, like HNO3 or NH4NO3.

o Drying removes the solvent and wash electrolytes. The drying can usually be

done by heating at 110 to 120°C for 1 to 2 h.

o There are a number of ways in which a foreign material may be coprecipitated:

Occlusion and Inclusion, Surface Adsorption, Isomorphous Replacement, and

Postprecipitation.

6


Philadelphia U

niversity

Dr. Najjar 17/18

7


General calculation of the percent by mass of analyte in a sample:

Weight the balanced chemical equation for the precipitation reaction

Calculate the moles of precipitate:

moles = mass ÷ molecular mass

Calculate moles of analyte from the balanced chemical equation using the mole ratio of (analyte : precipitate)

Calculate mass of analyte:

mass = moles x molecular mass

Generally, mass of analyte = mass of precipitate × GF

where

Calculate percent by mass of analyte in sample:

= (mass analyte ÷ mass sample) x 100

Philadelphia U

niversity

Dr. Najjar 17/18

c. Find the moles of Ca2+(aq).

From the balanced chemical equation, the mole ratio of Ca2+ : CaC2O4(s) is 1 : 1. So, n(Ca2+

(aq)) = n(CaC2O4(s)) = 0.019mol

d. Calculate the mass of calcium in grams mass (Ca) = n x MM = 0.019 x 40.08 = 0.76g

e. Calculate the percentage by mass of calcium in the original sample: %Ca = (mass Ca ÷ mass sample) x 100 = (0.76 ÷ 2.00) x 100 = 38%

Example

A 2.00g sample of limestone was dissolved in hydrochloric acid and all the

calcium present in the sample was converted to Ca2+. Excess ammonium

oxalate solution, (NH4)2C2O4 was added to the solution to precipitate the

calcium ions as calcium oxalate, CaC2O4. The precipitate was filtered,

dried and weighed to a constant mass of 2.43g.

Determine the percentage by mass of calcium in the limestone sample.

8

a. Write the balanced chemical equation for the precipitation reaction: Ca2+

(aq) + C2O42-

(aq) -----> CaC2O4(s)

b. Calculate the moles of calcium oxalate precipitated.

n(CaC2O4(s)) = mass ÷ MM = 2.43 ÷ (40.08 + 2 x 12.01 + 4 x 16.00) n(CaC2O4(s)) = 2.43 ÷ 128.10 = 0.019 mol


Philadelphia U

niversity

Dr. Najjar 17/18

Example

9


What is the %KCl in a solid if 5.1367 g of this solid gives 0.8246g AgCl when reacted with AgNO3?

Cl- + Ag+ AgCl(s)

Moles of precipitate (AgCl) = 0.8246/143.3 = 0.00575 mole

Moles of analyte (Cl or KCl) = 0.00575 x (1 mole Cl / 1 mole AgCl) = 0.00575 mole

Mass of analyte (KCl) = 0.00575 x 74.55 = 0.429 g

Percentage by mass of analyte = (0.429/5.1367)x100 = 8.35%

Philadelphia U

niversity

Dr. Najjar 17/18

Orthophosphate (PO43−) is determined by weighing as

(NH4)PO4·12MoO3. Calculate the percent P in the sample and

the percent P2O5 if 1.1682 g precipitate were obtained from a

0.2711 g sample. Perform the calculation using the gravimetric

factor. (molar mass for P is 30.97 g/mol, for P2O5 is 141.95

g/mol and for (NH4)PO4·12MoO3 is 1876.5 g/mol)

Example

10

“Insoluble” substances still have slight solubility

“Note: the solid does not appear in Ksp expression”

another example is Ag2CrO4

PRECIPITATION EQUILIBRIA:

THE SOLUBILITY PRODUCT

Philadelphia U

niversity

Dr. Najjar 17/18

11



Philadelphia U

niversity

Dr. Najjar 17/18

12



(Ksp for Fe(OH)3 is 4x10-38)

Philadelphia U

niversity

Dr. Najjar 17/18

13

CHAPTER 11 Precipitation Titration

Learning Objectives

● Calculating precipitation titration curves

● Indicators for precipitation titrations

Dr. Najjar 17/18

Philadelphia U

niversity

PRECIPITATION TITRATION

Key Concepts

A titration in which the reaction between the analyte and titrant

involves a precipitation.

The solids produced from the titration reactions should have very

low Ksp values.

Titration curves

pX vs. volume of titrant

14

12

sp(s)42(aq)

2

4(aq)

17-

sp(s)(aq)(aq)

13-

sp(s)(aq)(aq)

10

sp(s)(aq)(aq)

101.2K CrOAgCrO2Ag

108.3K AgII Ag

105.0K AgBrBr Ag

101.82K AgClClAg

Philadelphia U

niversity

Dr. Najjar 17/18

For the reaction A+(aq) + B-

(aq) ⇌ AB(s)

The concentrations of A+ and B- are varying through the titration process. When titrating B- (analyte) with A+ (titrant).

a) in the beginning of titration, concentration of A+ is zero and concentration of B- is calculated from its salt.

b) After addition of few milliliters of A+, B- will react with A+ forming AB precipitate, concentrations of A+ will be calculated according to Ksp. B

- will be excess.

c) At the end point, concentrations A+, B- will be calculated from Ksp.

d) After the end point A+ is the excess and B- will be calculated from Ksp.

Titration curve in the precipitation titration is a plot of

pA and/or pB versus volume of titrant (A)

15


Philadelphia U

niversity

Dr. Najjar 17/18

Example: Calculate pCl for the titration of 100.0mL of 0.1000 M Cl− with 0.1000 M

AgNO3 for the addition of 0.00, 20.00, 99.00, 99.50, 100.00, 100.50, and

110.00mL AgNO3 (assume Ksp = 1.0x10-10)

a) at 0.0mL addition

pCl = -log(0.1000) = 1.000

b) after 20.0mL addition

mmol Cl− = 100.0 mL × 0.1000 mmol/mL = 10.00 mmol

mmol Ag+ = 20.00 mL × 0.1000 mmol/mL = 2.000 mmol

Cl− left = 10.00 − 2.00 = 8.00 mmol/120.0 mL = 0.0667 M

pCl = −log 0.0667 = 1.18

c) after 99.00 mL addition

mmol Ag + = 99.00 mL × 0.1000 mmol/mL = 9.900 mmol

Cl − left = 10.00 − 9.90 = 0.10 mmol/199.0 mL = 5.0×10−4 M

pCl = −log 5.0×10−4 = 3.26

d) After 99.50mL addition


Cl− left = 10.00 − 9.95 = 0.05 mmol/199.5 mL = 2.5×10−4M

pCl = −log 2.5×10−4 = 3.60

16


Philadelphia U

niversity

Dr. Najjar 17/18

e) after 100.00 mL addition

at 100.00 mL, all the Cl− is reacted with Ag+

[Cl−] =

pCl = −log 1.0×10−5 = 5.00

f) after 100.50 mL addition


Ag+ left = 10.05 − 10.00 = 0.05 mmol/200.5 mL = 2.5×10−4 M

[Cl−] = Ksp/[Ag+] = 1.0×10−10/2.5×10−4 = 4.0×10−7 M

pCl = −log 4.0×10−7 = 6.40

g) after 110.00 mL addition


Ag+ left = 11.00 − 10.00 = 1.00 mmol/210 mL = 4.76×10−3 M

[Cl−] = Ksp/[Ag+] = 1.0×10−10/4.76×10−3 = 2.1×10−8 M

pCl = −log 2.1×10−8 = 7.67

17

510 101101 spK


Philadelphia U

niversity

Dr. Najjar 17/18

Example: 20 mL of 0.05M CaCl2 was titrated with 0.10M AgNO3. Calculate the

concentration of Cl- and Ag+ in the solution after addition of 0.0mL

AgNO3, 5.0mL AgNO3 , at the equivalence point and after addition of 30mL

AgNO3. (assume Ksp = 1.82x10-10)

a) at 0.0mL addition

b) after 5.0mL addition

Firstly, calculate the new rough concentrations after addition

18


Philadelphia U

niversity

Dr. Najjar 17/18

Secondly, calculate the exact concentrations from Ksp

c) at the equivalence point.

at the equivalence point, volume of titrant will be calculated as following

mole Ag = mole Cl ⇨ (M.V)Ag = (M.V)Cl ⇨ 0.1xVAg = 0.1x20 ⇨ Vag = 20mL

at this point [Ag+] and [Cl] only comes from dissociation of AgCl

19


Philadelphia U

niversity

Dr. Najjar 17/18

d) after addition of 30mL AgNO3 , Ag is excess now

Firstly, calculate the new rough concentrations after addition

Secondly, calculate the exact concentrations from Ksp

20


Philadelphia U

niversity

Dr. Najjar 17/18

SUMMARY OF

CALCULATIONS

For the reaction A+(aq) + B-

(aq) ⇌ AB(s)

When titrating B (analyte) with A+ (titrant).

a) In the beginning of titration, concentration of A+ is zero and concentration of B- is calculated from its salt.

b) Before the equivalence point

CB= (original moles of B – moles of A) / total volume

CA= Ksp/CB

c) At the equivalence point, CB=CA=

d) After the equivalence point

CA= (moles of A added – original moles of B) / total volume

CB=Ksp/CA

21

spK

Philadelphia U

niversity

Dr. Najjar 17/18

TITRATION CURVE

22

Philadelphia U

niversity

Dr. Najjar 17/18

END POINT AND

INDICATORS

Three methods are well known for precipitation titrations involving Ag+ as

titrant (Argentometry). These are:

Mohr method (direct method)

in this method, excess Ag cations after the equivalence point will react with the

indicator (CrO42-) forming red precipitate. This method carried at pH 7-10, to

prevent formation of chromic acid (H2CrO4).

23

Philadelphia U

niversity

Dr. Najjar 17/18

EXAMPLE

(MOHR)

35.7mL of 0.10M AgNO3 were needed to reach the end point for titration

of 75.0mL of unknown chloride solution using mohr method. Calculate the

concentration of Cl- in the unknown solution.

moleAg = moleCl

(M.V)Ag = (M.V)Cl

35.7x0.1 = MClx75

3.57 = MClx75 ⇨ MCl = 0.0467M

24

Philadelphia U

niversity

Dr. Najjar 17/18

END POINT AND

INDICATORS

Fajans method (direct method)

in this method, the indicator (Fluorescein) adsorbed on the surface of the

colloidal particles of the precipitate. This will change its color from yellow-green

to red. The use of this method is limited because of the relatively few

precipitation reactions in which a colloidal precipitate is formed rapidly.

Volhard method (back titration method)

in this method, the analyte is treated with excess amount of AgNO3, and the

remaining AgNO3 is titrated then with SCN-. At the end point, the titrant SCN- will

react with the indicator Fe3+ forming red color complex. pH in this method

should be acidic to prevent the formation of Fe(OH)3.

25

Philadelphia U

niversity

Dr. Najjar 17/18

EXAMPLE

(VOLHARD)

25.0mL of 0.10M AgNO3 were added to 30 mL of unknown chloride

solution. The resulted mixture was then titrated with 19.2mL of 0.10M

thiocyanate solution (SCN-). Calculate the concentration of Cl- in the

unknown solution.

moleAg = moleCl + moleSCN

(M.V)Ag = (M.V)Cl + (M.V)SCN

25x0.1 = MClx30 + 19.2x0.1

2.5 -1.92 = MClx30 ⇨ MCl = 0.0193M

26

Philadelphia U

niversity

Dr. Najjar 17/18

END POINT AND

INDICATORS

Another Precipitation Titration Methods

1. Barium determination with sulphate:

BaCl2 + H2SO4 BaSO4(s) + 2HCl

Indicator is sodium rhodizonate, which disappearance red colour of solution.

2. Lead determination with chromate:

Pb(NO3)2 + K2CrO4 PbCrO4(s) + 2KNO3

Indicator is AgNO3 solution. At equivalence point forms red precipitate.

3. Zinc determination with K4[Fe(CN)6]:

3ZnCl2 + K4[Fe(CN)6] Zn3K2[Fe(CN)6](s) + 2KCl

Indicator is UO2(NO3)2, which forms brown precipitate with K4[Fe(CN)6].

27

Philadelphia U

niversity

Dr. Najjar 17/18

28

CHAPTER 14 Redox Reactions and Titrations

Learning Objectives

● Oxidation numbers

● Balancing redox reactions

● Iodimetry and iodometry

● Quantitative calculations

Dr. Najjar 17/18

Philadelphia U

niversity

OXIDATION REDUCTION

REACTION

Oxidation is defined as a loss of electrons. e.g. iron rusting: When iron metal is exposed to oxygen in the air the iron combines with the oxygen to

form iron (II) oxide and further reacts to form iron(III) oxide

Reduction is defined as a gain of electrons. e.g. Formation of copper from copper (II) oxide: When hydrogen gas is passed over heated copper (II)

oxide, the copper (II) oxide is reduced to copper metal.

Redox reactions are a combination of reduction and oxidation, which

occur simultaneously. e.g. Displacement of hydrogen from acidic solution with magnesium.

29

eFeFe

2eFeFe

32

2

Cu2e Cu2

Oxidation 2eMgMg

Reduction H2e2H

Redox HMg2HMg

2

2

2

2

Philadelphia U

niversity

Dr. Najjar 17/18

OXIDATION NUMBERS

Oxidation numbers are hypothetical numbers assigned to an individual atom or

ion present in a substance using a set of rules. Oxidation numbers can be positive,

negative, or zero.

It is VERY IMPORTANT to remember that oxidation numbers are always reported for

one individual atom or ion and not for groups of atoms or ions.

atom/ion Fe3+ H in H2 Mg Cl in NaCl C in CO2 N in NH4+

Oxidation # 3+ 0 0 1- 4+ 3+

When the oxidation number for an atom or ion increased then we said it was oxidized,

and we can also consider it as reducing agent

If the oxidation number decreased, then the atom (or ion) reduced and it is considered

as oxidizing agent

30

agent reducing isit oxidized, wasFe occuresoxidation number oxidation in increase

3 2 .#

Fe Fe

2

32

ox

agent oxidizing isit reduced, wasCu occuresreduction number oxidation in decrease

0 2 .#

Cu Cu

2

2

ox

Philadelphia U

niversity

Dr. Najjar 17/18

OXIDATION NUMBERS

RULES

The oxidation number for an atom in its elemental form is always zero.

• A substance is elemental if both of the following are true:

• only one kind of atom is present

• charge = 0

• Examples:

• S8: The oxidation number of S = 0

• Fe: The oxidation number of Fe = 0

The oxidation number of a monoatomic ion = charge of the monatomic ion.

• Examples:

• Oxidation number of S2- is -2.

• Oxidation number of Al3+ is +3.

The oxidation number of all Group 1A metals = +1 (unless elemental). (Li, Na, K,…)

The oxidation number of all Group 2A metals = +2 (unless elemental). (Mg, Ca, Ba,…)

Hydrogen (H) has two possible oxidation numbers (unless elemental) :

• +1 when bonded to a nonmetal

• -1 when bonded to a metal

31

Philadelphia U

niversity

Dr. Najjar 17/18

OXIDATION NUMBERS

RULES

Oxygen (O) has two possible oxidation numbers (unless elemental):

• -2 in all other compounds...most common

• -1 in peroxides (O22-)....pretty uncommon

The oxidation number of fluorine (F) is always -1 (unless elemental).

The sum of the oxidation numbers of all atoms (or ions) in a neutral compound = 0.

The sum of the oxidation numbers of all atoms in a polyatomic ion = charge on the

polyatomic ion.

32

Philadelphia U

niversity

Dr. Najjar 17/18

OXIDATION NUMBERS

RULES

Examples: Find oxidation numbers for each atom in A) Na2SO4, B) Cr2O72-

A) Na2SO4 : no elemental form, no monoatomic ion, Na from the 1A group so its oxidation number is 1+. No group 2A, no hydrogen, O is in its most common form 2-.

The sum of the oxidation numbers of all atoms (or ions) in a neutral compound = 0.

2xNa + 1xS + 4xO = 0

2x(1+) + S + 4x(2-) = 0 ⇨ 2 + S – 8 = 0 ⇨ S = 6+

B) Cr2O72-

: no elemental form, no monoatomic ion, no group 1A, no group 2A, no hydrogen, O is in its most common form 2-.

The sum of the oxidation numbers of all atoms in a polyatomic ion = charge on the polyatomic ion.

2xCr + 7xO = 2-

2xCr + 7x(2-) = 2- ⇨ 2Cr – 14 = 2- ⇨ Cr = 6+

Find the oxidation number for each atom in the following: Ba(NO3)2, NF3, (NH4)2SO4

33

Philadelphia U

niversity

Dr. Najjar 17/18

OXIDATION REDUCTION REACTION

AND OXIDATION NUMBERS

Example:

In the reactions given below, identify the species undergoing oxidation and

reduction and balance the reaction:

Cu(s) + Ag+(aq) ➝ Cu2+(aq) + Ag(s)

The reaction could be divided to two reactions

1- Cu(s) ➝ Cu2+(aq). In this reaction ox.# of Cu was changed from 0➝2+

2- Ag+(aq) ➝ Ag(s). In this reaction ox.# of Ag was changed from 1+➝0

Reaction 1 is oxidation, Cu was oxidized and it is called reducing agent, while reaction

2 is reduction, Ag was reduced and it is called oxidizing agent.

We can re-write the reactions as following:

Cu(s) ➝ Cu2+(aq) + 2e-

Ag+(aq) + e- ➝ Ag(s)

To balance the reaction we should equalize number of electrons, this will be happened

when we multiply reaction 2 by 2

Cu(s) ➝ Cu2+(aq) + 2e- ⇨ Cu(s) ➝ Cu2+(aq) + 2e-

2x { Ag+(aq) + e- ➝ Ag(s) } ⇨ 2Ag+(aq) + 2e- ➝ 2Ag(s)

Cu(s)+ 2Ag+(aq) ➝ Cu2+(aq)+ 2Ag(s)

34

Philadelphia U

niversity

Dr. Najjar 17/18



Example:

In the reactions given below, identify the species undergoing oxidation and reduction and

balance the reaction:

35

(s)(aq)3(aq)2(s) CuAlClCuClAl

Philadelphia U

niversity

Dr. Najjar 17/18



Questions:

In the reactions given below, identify the species undergoing oxidation and reduction and

balance the reaction:

36

(g)2(S)32(l)2(s)2(aq)2(aq)(aq)(s) HOFeOHFe b) HMgClHClMg a)

Philadelphia U

niversity

Dr. Najjar 17/18

BALANCE REDOX EQUATIONS IN

ACIDIC AND BASIC MEDIUM

Acidic Conditions

1-Separate the half-reactions

2-Balance elements other than O and H

3-Add H2O to balance oxygen

4-Balance hydrogen with protons

5-Balance the charge with e

37

(aq)3

3

(aq)(aq)2(aq)

2

72 NOCrHNOOCr

(aq)3(aq)2

3

(aq)(aq)

2

72

NOHNO

CrOCr

(aq)3(aq)2

3

(aq)(aq)

2

72

NOHNO

Cr2OCr

(aq)3(l)2(aq)2

(l)2

3

(aq)(aq)

2

72

NOOHHNO

O7H2CrOCr

(aq)(aq)3(l)2(aq)2

(l)2

3

(aq)(aq)(aq)

2

72

3HNOOHHNO

O7H2Cr14HOCr

e

e

23HNOOHHNO

O7H2Cr614HOCr

(aq)(aq)3(l)2(aq)2

(l)2

3

(aq)(aq)(aq)

2

72

Philadelphia U

niversity

Dr. Najjar 17/18



6-Scale the reactions so that they have an equal amount of electrons

7-Add the reactions and cancel out the electrons and common terms

Basic Conditions

Same as in acidic conditions but complete by

8-Add OH- to balance H+

9-Combine OH- ions and H+ ions that are present on the same side to form water

10-Cancel common terms

38

ee

ee

69HNO3OH3HNO323HNOOHHNO 3

O7H2Cr614HOCrO7H2Cr614HOCr 1

(aq)(aq)3(l)2(aq)2(aq)(aq)3(l)2(aq)2

(l)2

3

(aq)(aq)(aq)

2

72(l)2

3

(aq)(aq)(aq)

2

72

(aq)3(l)2

3

(aq)(aq)2(aq)(aq)

2

72

(aq)(aq)3(l)2

3

(aq)(l)2(aq)2(aq)(aq)

2

72

NO3O4H2CrHNO35HOCr

9HNO3O7H2CrOH3HNO314HOCr

)((aq)3(l)2

3

(aq))((aq)2(aq)(aq)

2

72 5OHNO3O4H2Cr5OHHNO35HOCr aqaq

(aq)(aq)3(l)2

3

(aq)(aq)2(l)2(aq)

2

72 5OHNO3O4H2Cr3HNOO5HOCr

(aq)(aq)3

3

(aq)(aq)2(l)2(aq)

2

72 5OHNO32Cr3HNOOHOCr

Philadelphia U

niversity

Dr. Najjar 17/18



Example: Balance the following reaction in acidic medium

39

(g)2

2

(aq)(aq)

2

42(aq)4 COMnOCMnO

ee

ee

ee

10CO10O5C O8HMn210H612MnO

2CO2OC5 O4HMn58HMnO2

2CO2OC O4HMn58HMnO

CO2OC O4HMn8HMnO

CO2OC O4HMnMnO

CO2OC MnMnO

COOC MnMnO

(g)2(aq)

2

422

2

(aq)(aq)4

(g)2(aq)

2

422

2

(aq)(aq)4

(g)2(aq)

2

422

2

(aq)(aq)4

(g)2(aq)

2

422

2

(aq)(aq)4

(g)2(aq)

2

422

2

(aq)(aq)4

(g)2(aq)

2

42

2

(aq)(aq)4

(g)2(aq)

2

42

2

(aq)(aq)4

O8HCO10Mn2H61O5C2MnO 2(g)2

2

(aq)(aq)

2

42(aq)4

Philadelphia U

niversity

Dr. Najjar 17/18



Example: Balance the following reaction in basic medium

40

ClCrOClOOHCr 2

433

O3HCl66HClO 6H01CrO2 OH22Cr(OH)

O3HCl66HClO1 35HCrO OHCr(OH)2

O3HCl66HClO 35HCrO OHCr(OH)

O3HCl6HClO 5HCrO OHCr(OH)

O3HClClO CrOOHCr(OH)

ClClO CrOCr(OH)

23

2

423

23

2

423

23

2

423

23

2

423

23

2

423

3

2

43

ee

ee

ee

OH5Cl2CrO4OHClO 2Cr(OH)

4OHOHCl4H2CrO4OHClO 2Cr(OH)

OHClH4CrO2ClO 2Cr(OH)

2

2

433

2

2

433

2

2

433

Philadelphia U

niversity

Dr. Najjar 17/18

POPULAR REDUCING AND

OXIDIZING TITRANTS

The two most common reducing agents used as titrants are thiosulfate ion

S2O32- and iron (II) Fe2+

The most common oxedizing agents used as titrants are Iodine I2, Potassium

permanganate KMnO4 , Cerium(IV) and potassium dichromate K2Cr2O7

41

2eOSO2S

eFeFe

2

64

2

32

32

O7H2Cr6e14HOCr

O4HMn5e8HMnO

2I2eI

2

32

72

2

2

4

2

Philadelphia U

niversity

Dr. Najjar 17/18

IODOMETRY AND IODIMETRY

Iodometry involves indirect titration of iodine liberated from reaction of iodide ion

with the analyte, whereas iodimetry involves direct titration using iodine as the

titrant.

Iodine has brawn to pale yellow color, it gives a dark

blue color in the presence of starch. Usually starch is

added to the solution to indicate the appearance or

disappearance of iodine.

In iodometry, excess I react with the analyte (which is usually an oxidizing agent),

the reaction forms I2, this I2 is then titrated with strong reducing agent like S2O32.

I2 is not soluble in water, therefore it is forming I3 in the solution.

For example:

2 Cu2+ + 4 I → 2 CuI + I2

I2 + I → I3 (yellow)

I3 + 2 S2O3

2 → S4O62 + 3 I

at the end point mole of Cu2+ = moles of S2O32

42

Philadelphia U

niversity

Dr. Najjar 17/18

While titration with S2O32 , the yellow color of the solution start decreasing and

before its disappearance, starch is added before the total disappearance of I3, the

color will change to dark blue before resuming titration.

In iodimetry, we may use I2 as a titrant in a burette for direct titration or producing

it in the reaction vessel (by reaction between IO3and I) for back titration

methods. The produced I2 will react with the analyte, the excess amount of I2 is

then titrated with reducing agent like S2O32 . Starch is added before the total

disappearance of I2 to indicate the sharp end point. (note: in all cases, I2 present

in the solution as I3)

For example:

IO3 + 8I + 6H+ → 3I3

+ 3H2O

C6H8O6 + I3 → C6H6O6 + 3I + 2H+

I3 + 2S2O3

2 → 3I- + S4O62

at the end point, moles of I3 (3 IO3

moles) = moles of C6H8O6 + ½ moles of S2O32

43

OO

OO

HO

HO

H

OO

OHHO

HO

HO

H


Philadelphia U

niversity

Dr. Najjar 17/18

44


Philadelphia U

niversity

Dr. Najjar 17/18

Example:

36.40 mL of 0.3300 M Na2S2O3to titrate the I3 in a 15.00 mL sample, calculate the

molarity of I3 in the sample solution.

I3 + 2S2O3

2 → 3I + S4O62

Solution:

moles of S2O32 = M.V = 0.3300 M x 36.40 mL = 12.012 mmol

2 moles of S2O32 react with 1 mole of I3

12.012 mmole react with ???

⇨ moles of I3 = 12.012 x 1 / 2 = 6.006 mmole

⇨ molarity of I3 = moles/volume = 6.006 mmol / 15.00 mL = 0.4004 M

45


Philadelphia U

niversity

Dr. Najjar 17/18

Example:

A 5.00 mL sample of filtered orange juice was treated with 50.00 mL of excess 0.01023 M I3–. After the

oxidation was complete, 13.82 mL of 0.07203 M Na2S2O3 was needed to reach the starch indicator

end point. Report the concentration of ascorbic acid in grams per 100 mL.

Solution:

0.01023x0.050 = moles C6H8O6 +(0.5 x 0.07203x0.01382)

moles of C6H8O6 = 1.377x10-5

mass of C6H8O6= moles x M.wt = 1.377x10-5 x 176.13 = 0.002425g

0.002425g was found in 5 mL, So how much ????? are in 100mL

concentration of ascorbic acid in grams per 100 mL= 100x0.002425/5 =0.0485g/100mL

46

The amount of ascorbic acid, C6H8O6, in orange juice was determined by oxidizing the ascorbic acid to dehydroascorbic acid, C6H6O6, with a known excess of I3

–, and back titrating the excess I3

– with Na2S2O3.


Philadelphia U

niversity

Dr. Najjar 17/18

Example:

A 0.20g sample containing copper is analyzed iodometrically. Copper(II) is reduced to copper(I) by iodide

2Cu2+ + 4I → 2CuI + I2

What is the percent copper in the sample if 20.0 mL of 0.100M Na2S2O3 is required for titration of the liberated I2.

I2 + I → I3

I3 + 2S2O3

2 → S4O62 + 3 I

Solution:

Over all reaction is 2Cu2+ + 4I + 2S2O32 → 2CuI + S4O6

2

moles of S2O32= M.V = 0.1x20 = 2mmol

moles of Cu2+ = moles of S2O32 = 2mmol

mass of copper = moles x atomic mass =(2/1000) x 63.54 = 0.12708g

% copper = [mass copper/mass sample] x 100%

=[0.12708/0.2]x100=63.54%

47


Philadelphia U

niversity

Dr. Najjar 17/18

PERMANGANATE AS

TITRANT

KMnO4 was used as an oxidizing titrant in many redox titrations. It is self

indicating material, its color was changed from purple to colorless during

the reaction.

A difficulty arises when permanganate titrations performed in the presence of chloride

ion. The oxidation of chloride ion to chlorine by permanganate at room temperature is

normally slow and could be prevented by addition of the Zimmermann–Reinhardt

reagent (Z-R reagent) which also sharpens the end point.

Example:

The concentration of hydrogen peroxide solution can be found by a redox titration with

acidified potassium permanganate solution. It was found that 10 cm3 of the hydrogen

peroxide solution reacted with 20 cm3 of 0.02M potassium permanganate solution

when titrated. Calculate the concentration of the hydrogen peroxide solution.

2 MnO4–(aq) + 6 H+

(aq) + 5 H2O2(aq) ➝ 2 Mn2+(aq) + 4 H2O(l) + 5 O2(aq)

moles of MnO4– = (M.V)MnO4

= 0.02x20 = 0.4 mmole

moles of H2O2 =(5/2) moles MnO4– = (5/2)x 0.4 = 1mmole

(M)H2O2 = (M/V)H2O2

=1mmole/10mL

⇨ (M)H2O2 =0.1M 4

8

Philadelphia U

niversity

Dr. Najjar 17/18

49

Learning Objectives

● Oxidation numbers

● Balancing redox reactions

● Iodimetry and iodometry

● Quantitative calculations

Dr. Najjar 17/18

Philadelphia U

niversity

CHAPTER 9 Complexometric Titrations

COMPLEXOMETRIC

TITRATIONS

Key Concepts

The technique involves titrating metal ions with a complexing agent or chelating agent

(Ligand)

Ligands or complexing agents or chelating agents can be any electron donating entity,

which has the ability to bind to the metal ion and produce a complex ion, Ex: H2O, NH3, Cl,

Br, I…….

Ethylenediaminetetraacetic acid, or EDTA, is an aminocarboxylic acid. In its fully

deprotonated form (Y4−), it has six binding sites—four negatively charged carboxylate

groups and two tertiary amino groups—that can donate six pairs of electrons to a metal

ion. The resulting metal–ligand complex, in which EDTA forms a cage-like structure around

the metal ion, is very stable. The actual number of coordination sites depends on the size

of the metal ion, however, all metal–EDTA complexes have a 1:1 stoichiometry.

50

2M

Philadelphia U

niversity

Dr. Najjar 17/18

COMPLEXOMETRIC

TITRATIONS

Factors influencing EDTA reactions:

The nature and activity of metal ion.

The pH at which the titration is carried out.

The presence of interfering ions such as CN-, Citrate, Tartrate, F- and other complex forming agents.

Organic solvents also increase the stability of complex.

Effect of pH:

51

Philadelphia U

niversity

Dr. Najjar 17/18

EDTA TITRATIONS

52

The basic form of EDTA (Y4-) reacts with most metal ions to form a 1:1 complex.

Fraction (a) of the most basic form of EDTA (Y4-) is defined by the H+ concentration and acid-base equilibrium constants

}][][][][][]{[ 23456

654321543214321321211

654321Y KKKKKKKKKKKHKKKKHKKKHKKHKHH

KKKKKK4

a

] [Y] [HY] Y[H] Y[H] Y[H] Y[H] Y[H

] [Yα

EDTA

] [Yα

432

2345

2

6

4

Y

4

Y 44

where [EDTA] is the total concentration of all free EDTA species in solution

aY4- is depended on the pH of the solution

Dr. Najjar 17/18

53

The concentration of Y4- and the total concentration of EDTA is solution [EDTA] are related as follows:

⇨ The formation constant will be

at fixed pH conditional formation constant (K’f) could be calculated as following

EDTAα] [Y 4Y

4

EDTA TITRATIONS

Dr. Najjar 17/18

54

Example: Calculate the molar Y4- concentration in a 0.0200M EDTA solution buffered to a pH of 10.00

0.0060M 0.0200 x 0.3

EDTAα] [Y 4Y

4

Example

Calculate the equilibrium concentration of Ni2+ in a solution with an analytical NiY2- concentration of 0.0150M at pH 8.0 (Kf=4.2x1018) .

NiY2- ⇌ Ni2+ +

0.015-x x x

K’f=4.2x1018 x 4.2x10-3 = 1.76x1015 = 0.015/x2

x = [Ni2+] = √(0.015/1.76x1015)=2.9x10-9M

EDTA

Y)HYHYHHY(Y 4

-

3

-2

2

-3-4

EDTA TITRATIONS

Philadelphia U

niversity

Dr. Najjar 17/18

55

Question: What is the concentration of free Fe3+ in a solution of 0.10 M FeY at (a) pH

2.0, (b) pH 8.0? (Kf=1.3x1025, aY4-(pH 2)=2.6x10-14, aY4-(pH 8)=4.2x10-3).

Answer: [Fe3+] = 5.4x10-7 at pH 2.0 and [Fe3+] = 1.4x10-12 at pH 8.0

Note that the metal –EDTA complex becomes less stable as pH decreases, K’f decreases

In order to get a “complete” titration (K’f ≥106), EDTA requires a certain minimum pH for the titration of each metal ion.

By adjusting the pH of an EDTA titration: metal ion (e.g. Fe3+) can be titrated without interference from others (e.g. Ca2+)

Minimum pH for Effective Titration of Metal Ions

EDTA TITRATIONS

Philadelphia U

niversity

Dr. Najjar 17/18

COMPLEXOMETRIC EDTA

TITRATION CURVES

Complexation titration curve shows the change in pM, where M is the metal ion,

as a function of the volume of EDTA.

Calculating the Titration Curve

Step 1: Calculate the conditional formation constant for the metal–EDTA complex.

Step 2: Calculate the volume of EDTA needed to reach the equivalence point.

Step 3: Calculate pM values before the equivalence point by determining the

concentration of unreacted metal ions.

Step 4: Calculate pM at the equivalence point using the conditional formation constant.

Step 5: Calculate pM after the equivalence point using the conditional formation

constant.

56

Philadelphia U

niversity

Dr. Najjar 17/18

COMPLEXOMETRIC EDTA

TITRATION CURVES

Let’s calculate the titration curve for 50.0 mL of 5.00×10–3 M Cd2+ using a titrant of 0.0100 M EDTA at pH 10.0 (assume Kf 2.9x1015 and aY4-(pH 10) is 0.3)

Step 1: Calculate the conditional formation constant for the metal–EDTA complex.

K’f = Kf x aY4- = 2.9x1015 x 0.3 = 8.7x1014

Step 2: Calculate the volume of EDTA needed to reach the equivalence point.

at the equivalence point

moles EDTA = moles Cd2+

MEDTA×VEDTA = MCd×VCd

⇨ Veq = VEDTA = MCdVCd / MEDTA

= (5.00×10−3 M)(50.0 mL) / (0.0100 M) = 25.0 mL

57

Philadelphia U

niversity

Dr. Najjar 17/18

COMPLEXOMETRIC EDTA

TITRATION CURVES

Step 3: Calculate pM values before the equivalence point by determining the concentration of unreacted metal ions.

Before the equivalence point, Cd2+ is present in excess and pCd is determined by the concentration of unreacted Cd2+.

For example, after adding 5.0 mL of EDTA, the total concentration of Cd2+ is

CCd = (initial moles Cd2+ − moles EDTA added) / total volume

= (MCdVCd − MEDTAVEDTA)/ (VCd + VEDTA)

= ((5.00×10−3 M)(50.0 mL) − (0.0100 M)(5.0 mL)) / (50.0mL + 5.0mL)

= 3.64×10−3 M

pCd = -Log(3.64×10−3) = 2.44

58

Philadelphia U

niversity

Dr. Najjar 17/18

COMPLEXOMETRIC EDTA

TITRATION CURVES

Step 4: Calculate pM at the equivalence point using the conditional formation constant.

at equivalence point all Cd2+ and EDTA converted to CdY2-,

[CdY2−] = initial moles Cd2+ / total volume = MCdVCd / (VCd + VEDTA)

= (5.00×10−3 M)(50.0 mL) / (50.0 mL + 25.0 mL) = 3.33×10−3 M

(volume of equivalence was calculated and found to be 25.0 mL)

Now calculate Ccd

CdY2− ⇌ Cd2+ + Y4−

3.33×10−3 -x x x

K’f = 8.7x1014 = [CdY−2] / CCdCEDTA = (3.33×10−3 −x)/ (x.x)

⇨ CCd = x =√(3.33×10−3 / 8.7x1014) = 1.95x10-9

pCd = -Log(1.95x10-9)= 8.7

59

Philadelphia U

niversity

Dr. Najjar 17/18

COMPLEXOMETRIC EDTA

TITRATION CURVES

Step 5: Calculate pM after the equivalence point using the conditional formation constant. At this stage CdY2− formed and excess of EDTA present, their concentrations should be calculated. For example when 30.0 mL of EDTA added:

[CdY2−] = initial moles Cd2+ / total volume = MCdVCd / (VCd + VEDTA)

= (5.00×10−3 M)(50.0 mL) / (50.0 mL + 30.0 mL) = 3.125×10−3 M

[EDTA] = (moles of EDTA added-initial moles Cd2+) / total volume

= MEDTAVEDTA -MCdVCd / (VCd + VEDTA)

= ((0.0100 M)(30.0 mL) − (5.00×10−3 M)(50.0mL)) / (50.0 mL + 30.0 mL)

= 6.25×10−4 M

Now calculate Ccd

CdY2− ⇌ Cd2+ + Y4−

3.125×10−3 -x x 6.25×10−4 +x

K’f = 8.7x1014 = [CdY−2] / CCdCEDTA = (3.125×10−3 −x)/ (x)(6.25×10−4 +x)

⇨ CCd = x =3.125×10−3 / (8.7x1014 x 6.25×10−4) = 5.75x10-15

pCd = -Log(5.75x10-15)= 14.2

60

Philadelphia U

niversity

Dr. Najjar 17/18

COMPLEXOMETRIC EDTA

TITRATION CURVES

61

Philadelphia U

niversity

Dr. Najjar 17/18

DETERMINATION OF EDTA

TITRATION END POINT

Metal Ion Indicator: a compound that changes color when it binds to a metal ion

- Similar to pH indicator, which changes color with pH or as the compound binds H+

For an EDTA titration, the indicator must bind the metal ion less strongly than EDTA

- Needs to release metal ion to EDTA

62

End Point indicated by a color change from red to blue

Philadelphia U

niversity

Dr. Najjar 17/18

DETERMINATION OF EDTA

TITRATION END POINT

Common Metal Ion Indicators

63

Philadelphia U

niversity

Dr. Najjar 17/18

Example: The concentration of a solution of EDTA was determined by standardizing against a solution of Ca2+ prepared using a primary standard of CaCO3. A 0.4071-g sample of CaCO3 was transferred to a 500-mL volumetric flask, dissolved using a minimum of 6 M HCl, and diluted to volume. After transferring a 50.00-mL portion of this solution to a 250-mL Erlenmeyer flask, the pH was adjusted by adding 5 mL of a pH 10 NH3/NH4Cl buffer. After adding calmagite as an indicator, the solution was titrated with the EDTA, requiring 42.63 mL to reach the end point. Report the molar concentration of EDTA in the titrant. (M.wt. of CaCO3 =100.09 g/mol)

Solution:

The primary standard of Ca2+ has a concentration of

[0.4071 g CaCO3 / 100.09 g CaCO3] / (0.5000 L) = 8.135×10−3 M Ca2+

The moles of Ca2+ in the titrant is

8.135×10−3 M Ca2+ × 0.05000 L Ca2+ = 4.068×10−4 mol Ca2+

which means that 4.068×10–4 moles of EDTA are used in the titration.

The molarity of EDTA in the titrant is

4.068×10−4 mol EDTA / 0.04263 L EDTA = 9.543×10−3 M EDTA

64

COMPLEXOMETRIC

EDTA TITRATION

Philadelphia U

niversity

Dr. Najjar 17/18

pharmaceutical analytical chemistry - … new 30...o repeat the drying and weighing process until a...

Documents