Practical Pharmaceutical Analytical Chemistry-1 (PC...

Practical

Pharmaceutical Analytical Chemistry-1

(PC 205)

By Staff of

Pharmaceutical Analytical Chemistry Department

Faculty of Pharmacy

Mansoura University

2

SSuubbjjeecctt PPaaggee

CChhaapptteerr II:: IInnttrroodduuccttiioonn 4

Classification of Analytical Chemistry Branches ……………..….. 5

Standard Solutions ………………………………………………….……….…………. 7

Methods of expressing the concentration ……….…………………... 8

CChhaapptteerr IIII:: AAcciidd--BBaassee TTiittrraattiioonn 11

Acid-base indicators ……………………………………………………..………….. 12

Steps for the determination of a sample using titration ….….. 13

Determination of HCl sample ………………………….…………................. 14

Types of titration methods ………………………….……………………………

Calculations for volumetric analysis (titration) …………..…………

15

16

Determination of NH4Cl sample ……..…………….………………………….

Determination of NH4Cl & HCl mixture ……….……….…………………

Acid-base titration curve and choice of acid-base

indicators ………………..................................……………..……….…………………….

20

24

27

Determination of NaOH & Na2CO3 mixture …………………………... 30

Determination of Na2CO3 & NaHCO3 mixture ………………………. 32

Determination of Aspirin Tablets (Back titration) ….…………. 40

Determination of Boric acid (H3BO3) ………………………………......... 42

Determination of Borax (Na2B4O7) ……..………………………………..… 44

Determination of HCl & CH3COOH mixture …………………………. 45

CChhaapptteerr IIIIII:: PPrreecciippiittaattiioonn TTiittrraattiioonn 49

Introduction …………………………………………………………………………...……… 50

Mohr’s Method ………………………………………………………………………….. 51

Volhard’s Method ………………………………………………….………………….. 56

Determination of NaCl (by Volhard’s method) ……..……………… 59

Determination of KBr (by Volhard’s method) ……..…………….… 63

Determination of NaCl (by Fajan’s method) 65

Contents

3

SSuubbjjeecctt PPaaggee

CChhaapptteerr IIVV:: CCoommpplleexxoommeettrriicc TTiittrraattiioonn 67

Introduction ………………………………………………...……………..……...………….. 68

Determination of Magnesium Sulfate (MgSO4.7H2O) ……. 75

Determination of Calcium Chloride (CaCl2.6H2O) ………….… 77

Determination of Potash Alum ( KAl(SO4)2.12H2O ) ………… 78

Determination of (Ca2+/Mg2+) mixture ………………………….....……... 81

APPENDIX

Instructions for using the glassware.

4

CChhaapptteerr II

IInnttrroodduuccttiioonn

5

Analytical

Chemistry

Qualitative Analysis

(=What?)

Quantitative Analysis

or Determination

or Estimation

or Assay

(=How much?)

Analysis of Anions

Volumetric Analysis

(Titration)

Gravimetric Analysis

Instrumental Analysis

Acid-base titration (Aqueous & Non-aqueous)

Precipitation titration

Complex-formation titration

Redox titration

Analysis of Cations

6

Volumetric Analysis (Titration)

Definition of Volumetric Analysis (Titration):

It is a method of quantitative analysis that depends on measuring

the volume of a solution of known concentration (standard solution)

needed to react completely with a sample of unknown concentration.

The standard solution is placed in the burette and called "Titrant".

The sample of unknown concentration is placed in a conical flask.

- 0

- 25 or 50 ml

Burette (containing standard solution)

Conical Flask (containing sample of unknown

concentration + indicator)

Burette

holder

7

Standard Solutions:

It is a solution of known concentration & composition.

Types of standard solutions:

1. Primary Standard Solution:

♦ It is a solution whose concentration remains constant for a long

period of time.

♦ It is prepared directly by dissolving known weight of the primary

standard substance in a certain volume of solvent.

♦ Example: Na2CO3.

2. Secondary Standard Solution:

♦ It is a solution whose concentration can be only be known

approximately.

♦ The exact concentration of it is determined by a process called

"Standardization or Standardisation".

♦ Standardization must be done after the preparation of

secondary standard solutions to determine their exact

concentration and correct any error if present. That is done by:-

Titrating the secondary standard solution against a primary

standard solution OR against a previously standardized

secondary standard.

♦ Examples: NaOH and HCl.

Remember: Solute + Solvent = Solution

--------------------------------------------------------

8

Methods of expressing the concentration of standard solutions:

The most common methods are: Molarity (M) & Normality (N).

1. Molarity or Molar concentration (M):

= The number of moles of the solute per 1 liter of the solution.

Example: 0.5 M NaOH means 0.5 mole of NaOH / 1 L solution.

N.B. Molar solution is a solution that contains 1 mole of the solute

per 1 liter of the solution (i.e.: Molar solution = 1 M solution).

Remember:

Definition of Mole (mol.):

For molecules:

The Mole means gram-molecular weight {i.e. Molecular wt. expressed in grams}.

Example: M.W. of NaOH = 40 1 mole of NaOH ≡ 40 g NaOH.

(i.e. The weight of 1 mole of NaOH is 40 g)

For atoms:

The Mole means gram-atomic weight {i.e. Atomic wt. expressed in grams}.

Example: A.W. of Na = 23 1 mole of Na ≡ 23 g Na.

(i.e. The weight of 1 mole of Na is 23 g)

2. Normality or Normal concentration (N):

= The number of equivalent weights of the solute per 1 liter of the solution.

Example: 2 N NaOH means 2 equivalent weights of NaOH / 1 L solution.

N.B. Normal solution is a solution that contains 1 equivalent weight of

the solute per 1 liter of the solution (i.e.: Normal solution = 1 N solution).

General definition of the equivalent weight of a substance:

Very Important

9

It is the weight of the substance that is equivalent in its reactive

power to 1 mole of hydrogen (i.e. the weight of the substance that will

combine with or replace 1 mole of hydrogen).

The specific definition and calculation of equivalent weight differ

according to the type of the reaction. Now, it will be studied in acid-base

reactions while other reactions will be discussed later.

Definition of Equivalent Weight in acid-base reactions:

It is the weight of the substance that will release, react with or

be chemically equivalent to 1 mole of hydrogen ions (H+) in that reaction.

How to calculate it?

a) For acids: Equiv. wt. = acidtheofmoleculeoneinHereplaceablofno

acidtheofWM.

..

ex: equiv. wt. of H2SO4 = M.W. of H2SO4/2 = 98/2 = 49 g H2SO4.

b) For bases: Equiv. wt. =

basetheofmoleculeoneinOHereplaceablofno

basetheofWM.

..

ex: equiv. wt. of NaOH = M.W. of NaOH/1 = 40/1 = 40 g NaOH.

c) For salts: Equiv. wt. = salttheofmoleculeonetoequivalentHofno

salttheofWM

.

..

ex: equiv. wt. of CaCO3 = M.W. of CaCO3 / 2

because CaCO3 + 2 HCl CaCl2 + H2O + CO2

(i.e. 1 CaCO3 is equivalent to 2 H+).

N.B. In case of salts the number by which we divide

the M.W. of the salt is usually the sum of +ve charges on

the cation radical OR the sum of –ve charges on the anion

radical of the salt.

10

2+ 2- 3+ 3-

CaCO3 Na3PO4

Equiv. wt. = M.W. of CaCO3/2 Equiv. wt. = M.W. of Na3PO4/3

Advantage of using normal concentration:

The main advantage of using normal concentration over molar ones is

the simple 1:1 ratio involved in all reactions of the same type and that

facilitates our calculations.

ex. 1 equiv. wt. of NaOH ≡ 1 equiv. wt. of HCl

≡ 1 equiv. wt. of H2SO4

≡ 1 equiv. wt. of H3PO4

≡ 1 mole of H+

In other words: Solutions having the same normality react in 1:1 ratio.

ex. 1 L of 1 N NaOH ≡ 1 L of 1 N HCl ≡ 1 L of 1 N H2SO4 ≡ 1 L of 1 N H3PO4 ,

10 ml of 0.2 N NaOH ≡ 10 ml of 0.2 N HCl ≡ 10 ml of 0.2 N H2SO4 ≡ 10 ml of 0.2 N H3PO4

Remember: [=] means equal while [ ≡ ] means equivalent.

Very Important

i.e. In a certain type of

reactions, equivalent weights

of all substances are

equivalent in their reactive

powers because they are all

referred to a common

standard [Hydrogen].

11

CChhaapptteerr IIII

AAcciidd--bbaassee ttiittrraattiioonn

12

Acid-Base Titrations

It's a type of titrations that involve reactions between acids and bases.

Remember: ▪ pH = - log [H+] ……………[H+] means the molar concn. of H+.

▪ pH > 7 basic, pH =7 neutral, pH < 7 acidic.

Acid-base indicators :

♦ They are indicators used to indicate the end point of acid-base titrations

(i.e. indicate the completeness of the reaction by changing their colors).

♦ Each indicator has its specific {pH range} which is the range of pH

within which the indicator changes its color.

Examples of acid-base indicators:

1. Methyl Orange (M.O.):

pH range ≈ (3 – 4.5) approximately.

We use 2-3 drops of it in our experiments.

Color:

Red Orange Yellow

pH 0 3 4.5 14

Acidic

to M.O.

E.P.

by M.O.

Basic

to M.O.

2. Phenolphthalein (ph.ph.):

pH range ≈ (8.5 – 10) approximately.

We use 10 drops of it in our experiments.

Color:

Colorless * Pink

pH 0 8.5 10 14

Acidic

to ph.ph.

E.P.

Basic

to ph.ph.

13

*Color of ph.ph. at E.P.:

-If the solution before starting titration is Colorless (i.e. acidic to

ph.ph.) E.P. is First Pink Color persistent for 30 seconds.

-If the solution before starting titration is Pink (i.e. basic to

ph.ph.) E.P. is First Colorless.

Remember:

ph.ph. (small letters) is phenolphthalein indicator

while

pH (p small, H capital) is a measure of acidity or alkalinity { = –log [H+] }

Steps for the determination of a sample

using titration

(3 steps to determine the sample concentration)

1st Preparation of the standard solution (usually HCl, H2SO4 or NaOH)

and standardization of it by titration against a primary standard

2nd Titration of the sample with the prepared standard solution and

observation of the E.P.

3rd Calculation of the sample concentration.

N.B. We say:

Titrate …………………… with or against or ≠ ……………………

in the flask in the burette (Titrant)

small f

14

Exp.: Determination of HCl sample

a) Preparation of 0.1 N NaOH & standardization of it.

b) Titration:

1- Transfer 10 ml of HCl sample into a clean conical flask (by bulb pipette).

2- Add 10 drops ph.ph. indicator.

3- Titrate ≠ 0.1 N NaOH (in burette).

Color change at E.P.:

from Colorless to first pink persistent for 30 sec.

c) Calculations: will be discussed later.

Repeat the experiment on another 10 ml sample

using 2 drops M.O. indicator.


from Red to Orange

15

TTyyppeess ooff ttiittrraattiioonn mmeetthhooddss

AA)) DDiirreecctt ttiittrraattiioonn::

It means stepwise addition of the standard solution (titrant)

from the burette into the solution being analyzed until the end

point is reached.

i.e. Sample of unknown concn. ≠ Standard (titrant)

Vol. taken from the sample [e.g. 10 ml] ≡ (E.P.) ml of the standard

BB)) RReessiidduuaall ttiittrraattiioonn OORR BBaacckk ttiittrraattiioonn::

In this type, a known excess of a standard (more than sufficient to

react with the sample) is added to the sample. Then after the reaction is

complete, the excess unreacted standard is back titrated against

another suitable standard.

i.e. Sample + known excess of standard I

titration of the remaining unreacted part

of standard I ≠ standard II (titrant)

E.P. ≡ remaining unreacted part

Sample ≡ reacted part ≡ (known excess added of 1st stand. – E.P.)

The amount of the first standard that reacted with the sample can be

calculated by subtracting the volume consumed in the back titration (E.P.)

which is equivalent to the unreacted amount of the first standard from the

initially added volume (the known added excess).

0

E.P. {

Standard

(titrant)

10 ml sample of

unknown concn.

Part reacts with the sample

Part remains unreacted

16

CCaallccuullaattiioonnss ffoorr vvoolluummeettrriicc aannaallyyssiiss ((ttiittrraattiioonn))

GGiivveenn::

1. Volume of the sample (= 10 ml for example).

2. Volume of the standard equivalent to the sample (End point “E.P” ).

3. Concentration of the standard (x N or x M).

RReeqquuiirreedd::

Determination of the sample concentration in g/L or g%.

SSoolluuttiioonn::

The concentration of the sample can be calculated as follows:

Volume of the standard

equivalent to the sample

(x) N or (x) M

17

MMaaiinn sstteeppss ooff tthhee ccaallccuullaattiioonnss

At the beginning of calculations, you should write “balanced” chemical

equations that represent the reactions of the experiment and from

these equations find the relation between the sample and the standard.

Then the concentration of the sample can be calculated by making the

2 following steps:

11sstt SStteepp:: CCaallccuullaattiioonn ooff EEqquuiivvaalleennccee FFaaccttoorr ((FF))

DDeeffiinniittiioonn ooff eeqquuiivvaalleennccee ffaaccttoorr::

It is the weight of the sample that is equivalent to 1 mL of the standard.

(F) is written as follows:

each ml of (x) N standard ≡ a x M.W. of sample x N of standard

b x 1000

≡ (F) g sample

≡ (F x 1000) mg sample

Where:

a = no. of moles of the sample equivalent to 1 mole of the standard.

i.e. 1 standard ≡ a sample ………… {from the reaction equations}

b = differs according to the type of the reaction:

In acid-base titration: b = no. of replaceable H+ or OH- present in

or react with ONE molecule of the

standard. (ex.: for H2SO4 → b =2).

In precipitation titration: b = no. of univalent cation atoms present in or

equivalent to ONE molecule of the standard.

(ex.: for AgNO3 → b =1, Hg(NO3)2 → b =2).

In complexometric titration: b is absent as we use molar concentrations.

In redox titration: b = electron transfer or no. of electrons lost

or gained by ONE molecule of the

standard. (ex.: for KMnO4 → b = 5).

18

Notes

1. b is absent if we use Molar concentration instead of Normal as in

the case of complexometric titrations.

2. In back (residual) titration, there are 2 standards and we can

calculate the factor (F) on any one of them but it is preferred to

calculate it on the 1st standard.

3. The concentration of standard solutions used in volumetric analysis

is usually expressed in Molarity (M) [British Pharmacopeia “BP”

and European Pharmacopeia “EP”] or Normality (N) [United

States Pharmacopeia “USP”].

22nndd SStteepp:: CCaallccuullaattiioonn ooff tthhee CCoonncceennttrraattiioonn

IInn ddiirreecctt ttiittrraattiioonn::

Concn. = E.P. x F x 1000

= .......... g/L Volume taken from the sample

= E.P. x F x 100

= .......... g% Volume taken from the sample

IInn bbaacckk ((rreessiidduuaall)) ttiittrraattiioonn::

Concn. = (Known excess added from 1st standard - E.P.) x F x 1000

Volume taken from the sample

= .......... g/L

= (Known excess added from 1st standard - E.P.) x F x 100

Volume taken from the sample

= .......... g%

Notes

1. Dilution factor and difference in normality or molarity should be

considered in the calculation of concentration.

2. If the original sample is Solid, {Concn. = purity}, {unit of concn.

should be g% which means g analyte/100 g of sample powder} and {we

divide by the weight taken from the sample instead of the volume}.

19

EExxaammppllee

Titration of 10 ml H2SO4 sample against 0.1 N NaOH, if the end point is

6 mL, calculate the concentration of H2SO4 sample (M.W. of H2SO4 = 98).

Solution:

2 NaOH + H2SO4 → Na2SO4 + 2 H2O

1 NaOH ≡ ½ H2SO4

standard sample

Equivalence factor (F):

each ml of 0.1 N NaOH ≡ ½ x M.W. of H2SO4 x 0.1

≡ 0.0049 g H2SO4 1 x 1000

Concentration:

Concn. of H2SO4 = 6 x 0.0049 x 1000

= 2.94 g/L 10

= 0.294 g%

AAnnootthheerr eexxaammppllee

Titration of 10 ml NaOH sample against 0.1 N H2SO4, if the end point is

4 mL, calculate the concentration of NaOH sample (M.W. of NaOH = 40).

Solution:

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

1 H2SO4 ≡ 2 NaOH

standard sample


each ml of 0.1 N H2SO4 ≡ 2 x M.W. of NaOH x 0.1

≡ 0.004 g NaOH 2 x 1000

Concentration:

Concn. of NaOH = 4 x 0.004 x 1000

= 1.6 g/L 10

= 0.16 g%

20

Exp.: Determination of NH4Cl sample

# NH4Cl is acidic because it is a salt of strong acid (HCl) & weak base (NH4OH).

# NH4Cl is determined by 2 methods:

Back (Residual) titration method.

Formol method [specific method for ammonium salts].

AA)) DDeetteerrmmiinnaattiioonn ooff NNHH44CCll ssaammppllee bbyy bbaacckk ttiittrraattiioonn

PPrriinncciippllee ::

NH4Cl + NaOH boiling

NaCl + NH3 + H20

It depends upon boiling the ammonium salt with a known excess of

standard alkali (NaOH) producing ammonia which is removed by boiling.

Then the remaining unreacted part of NaOH is back titrated against

standard HCl using M.O. or ph.ph. indicator.

NaOH + HCl NaCl + H20

NNootteess::

1) Boiling is necessary for the reaction between NH4Cl and NaOH to

remove the produced ammonia.

2) This method requires Blank Determination which is a separate

determination in which we repeat the same steps of the experiment

but without the sample and the E.P. in this case is called

Blank Reading.

(sample) known excess of st. NaOH

(50 ml)

standard titrant

remaining unreacted

part

21

Blank determination is required in this experiment because

when NaOH is heated and then cooled, its strength is decreased due to:

1- Interaction of NaOH with the glass of the flask.

2- Absorption of atm. CO2.

More explanation about blank determination:

Main e

xperiment

Sample + 50 ml 1 N NaOH

Boil

Cool

titrate the remaining

unreacted part of 1 N NaOH

≠ 1 N HCl

E.P.

50 ml of 1 N NaOH

x ml

part

reacted

with the

sample

remaining

unreacted part

(x ml)

Titr. ≠ 1 N HCl

[E.P. = x ml]

loss due to reactn. with glass &

CO

2

Blank

dete

rmination

50 ml 1 N NaOH

Boil

Cool

titrate the remaining

unreacted part of 1 N NaOH

≠ 1 N HCl

Blank reading

50 ml of 1 N NaOH

remaining unreacted part

Titr. ≠ 1 N HCl

the volume consumed from

titrant (1 N HCl) is called

[Blank reading] ex: 49.8, 49.9...etc

loss due to reactn. with glass &

CO

2

The part reacted with the sample = (Blank reading – E.P.)

22

PPrroocceedduurree ::

1) Transfer 10 ml of NH4Cl sample into a clean conical flask. (by bulb pipette 10 ml)

2) Add 50 ml of 1 N NaOH. (by bulb pipette 25 ml "2 times")

3) Boil for 10 min. inserting a funnel in the neck of the flask to avoid

evaporation of the solution.

After boiling, ensure that no more NH3 is produced indicated by

the disappearance of ammonia odor.

4) Cool.

5) Add 2 drops M.O. indicator.

6) Titrate ≠ 1 N HCl.

{ Color change at E.P.: from Yellow to Orange }

CCaallccuullaattiioonnss ::

NH4Cl + NaOH boiling

NaCl + NH3 + H20


The equivalence factor (F) can be calculated on NaOH (1st standard) or on

HCl (2nd standard) but it is preferred to be calculated on the 1st standard.

Calculation of equivalence factor (F):

1 NaOH ≡ 1 NH4Cl

each ml of 1 N NaOH ≡ 1 x M.W. of NH4Cl x 1

≡ ..................... g NH4Cl 1 x 1000

Calculation of concentration:

Concn. of NH4Cl = (Blank reading – E.P.) x F x 1000

= ................... g/L 10

Results:

E.P. =

Concn. =

1st standard

2nd standard

sample standard

23

BB)) DDeetteerrmmiinnaattiioonn ooff NNHH44CCll ssaammppllee bbyy ffoorrmmooll mmeetthhoodd


4 NH4Cl + 6 HCHO (CH2)6N4 + 4 HCl + 6 H20

It depends upon the addition of HCHO to the ammonium salt

resulting in the formation of Hexamine (Hexamethylene tetramine)

in addition to an amount of HCl equivalent to the sample analysed.

Then the produced HCl is titrated ≠ standard NaOH using ph.ph.

indicator.

4 HCl + 4 NaOH 4 NaCl + 4 H2O

4 NH4Cl ≡ 4 HCl ≡ 4 NaOH

(sample) neutral formalin

Hexamine or

Hexamethylene tetramine

Equivalent to NH4Cl sample

24

Exp.: Determination of NH4Cl & HCl mixture

Our target is to determine the concentration of each component of the mixture


a) For HCl only: (E.PHCl)

HCl only is determined by direct titration of the mixture against

standard NaOH using M.O. indicator.

(NH4Cl does not interfere because it needs boiling to react with NaOH).

b) For the total mixture [NH4Cl & HCl]: (E.Ptotal)

The determination of the total depends on the addition of neutral

formalin (HCHO) to the mixture where it reacts with NH4Cl leading

to the production of an equivalent amount of HCl {Formol method}.


Then both (HCl) produced from NH4Cl and (HCl) originally present in the

mixture are directly titrated ≠ standard NaOH using ph.ph. indicator.

NH4Cl & HCl HCl both are titrated ≠ standard NaOH

NH4Cl only ≡ (E.Ptotal - E.PHCl)


a) For HCl only: (E.PHCl)

1) Transfer 10 ml of the mixture into a clean conical flask.


3) Titrate ≠ 0.1 N NaOH.

(sample) neutral formalin

Hexamine or

Hexamethylene tetramine

Equivalent to NH4Cl sample

25

{ Color change at E.P.: from Red to Orange }

b) For the total mixture [NH4Cl & HCl]: (E.Ptotal)

1) Transfer another 10 ml of the mixture into a clean conical flask.

2) Add 5 ml neutral formalin. (by measuring cylinder )

!!!!!! Caution: Formalin is very toxic !!!!!!

3) Leave for 5 minutes. (to make the reaction completed)

4) Add 10 drops ph.ph. indicator.



from Colorless to first pink persistent for 30 sec.


E.PHCl ≡ HCl only

E.Ptotal ≡ HCl + NH4Cl

(E.Ptotal - E.PHCl) ≡ NH4Cl only

For HCl :

HCl + NaOH NaCl + H20

1 NaOH ≡ 1 HCl


each ml of 0.1 N NaOH ≡ 1 x M.W. of HCl x 0.1

≡ .......... g HCl 1 x 1000

Concentration:

Concn. of HCl = E.PHCl x F x 1000

= .......... g/L 10

For NH4Cl :


sample standard

26

4 HCl + 4 NaOH 4 NaCl + 4 H2O

4 NH4Cl ≡ 4 HCl ≡ 4 NaOH

1 NaOH ≡ 1 NH4Cl

# Note that formalin (HCHO) is not standard


each ml of 0.1 N NaOH ≡ 1 x M.W. of NH4Cl x 0.1

≡ .......... g NH4Cl 1 x 1000

Concentration:

Concn. of NH4Cl = (E.Ptotal - E.PHCl) x F x 1000

= .......... g/L 10

Results:

E.PHCl =

E.Ptotal =

Concn. of HCl =

Concn. of NH4Cl =

sample standard

27

AAcciidd--BBaassee TTiittrraattiioonn CCuurrvvee

aanndd CChhooiiccee ooff AAcciidd--BBaassee IInnddiiccaattoorrss

♦ Introduction:(Definitions of some terms)

The Equivalence Point

[Theoretical]

The End Point

[Experimental]

- It is the point at which the added

amount of the standard solution is

chemically equivalent to the amount

of the substance being determined.

- It is the point at which the visual

change of the indicator takes

place in the titration.

- Its real position can only be

theoretically calculated.

- Its position can be determined

experimentally.

- It is expressed by the volume of

the titrant added.

- It is expressed by the volume of

the titrant added.

The Titration Error

It is the difference between the equivalence point and the end point.

The more accurate the titrimetric method, the smaller the titration error.

The Titration Curve of Acid-Base titrations:

It is the curve obtained by plotting the pH of

the titrated solution (y-axis) against the volume

of the titrant added (x-axis).

The titration curve is characterized by a Sigmoid

shape and the midpoint of the vertical part of the

curve corresponds to the equivalence point.

pH

mls added of the titrant

Equivalence

point

28

How to predict the shape of the titration curve?

The pH-range of the indicator:

It is the pH-range ( pH-interval ) within which the indicator changes

its color.

Ex.: - The pH-range of M.O. ≈ (3 – 4.5) approximately.

- The pH-range of ph.ph. ≈ (8.5 – 10) approximately.

♦ Choice of acid-base indicators:

For accurate titrimetric method, the indicator should be selected so as

to make the titration error small as possible. To achieve that the pH-range

of the indicator should fall in the vertical part of the titration curve (i.e.

the part marked by sharp change of pH) because the equivalence point is

pH of

the titrated

solution

mls added

of the titrant

End of the curve

predicted from

pH of the titrant

Middle of the curve

predicted from

pH of the products

(= pH at equivalence point)

Start of the curve

predicted from

pH of the sample

If higher concn.

If lower concn.

29

located in this part so the difference between (the equivalence point) and

(the end point indicated by the selected indicator) will be small as possible.

Example:

Titration of NaOH with standard HCl using M.O. or ph.ph. indicator:

NaOH + HCl NaCl + H2O

(pH = 7)

The calculated equivalence point

corresponds to pH=7.

But since the vertical part of this

titration curve ranges from about

3.5 to 10.5 so either M.O. (3 – 4.5)

or ph.ph. (8.5 – 10) can be used as

indicator for this titration.

pH

pH range of ph.ph.

7

pH range of M.O.

mls added of

standard HCl M.O. End

Point (Experimental)

Equivalence Point

(Theoretical)

Titration Error (very small)

30

Exp.: Determination of NaOH & Na2CO3 mixture


According to the two-indicators method:

a) First step: (E.Pph.ph.)

Titration of 10 ml of the mixture ≠ standard HCl using ph.ph. indicator.

E.Pph.ph. ≡ OH- + 2

1 CO3

--

b) Second step: (E.PM.O.)

Titration of another 10 ml of the mixture ≠ standard HCl using M.O. ind.

E.PM.O. ≡ OH- + CO3--

Reaction equations:

For NaOH:

NaOH + HCl → NaCl + H2O

For Na2CO3:

Na2CO3 + HCl → NaCl + NaHCO3 ...................Half neutralization

(pH>10) (pH ≈ 8.3)

NaHCO3 + HCl → NaCl + CO2 + H2O ...............Complete neutralization

(pH ≈ 3.8) .

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O


1 E.Pph.ph.:



3) Titrate ≠ 0.2 N HCl.

{ Color change at E.P.: from Pink to Colorless }

31

2 E.PM.O.:






E.P1 (ph.ph.) ≡ OH- + 2

1 CO3

--

E.P2 (M.O.) ≡ OH- + CO3--

(E.P2 - E.P1) ≡ 2

1 CO3

--

2(E.P2 - E.P1) ≡ CO3--

[E.P2 - 2(E.P2 - E.P1)] ≡ OH-

For NaOH :


1 HCl ≡ 1 NaOH


each ml of 0.2 N HCl ≡ 1 x M.W. of NaOH x 0.2

≡ .......... g NaOH 1 x 1000

Concentration:

Concn. of NaOH = [E.P2 - 2(E.P2 - E.P1)] x F x 1000

= .......... g/L 10

For Na2CO3 :


2 HCl ≡ 1 Na2CO3

1 HCl ≡ 2

1 Na2CO3


sample standard

sample standard

sample standard

32

each ml of 0.2 N HCl ≡ 2

1 x M.W. of Na2CO3 x 0.2

≡ .......... g Na2CO3

1 x 1000

Concentration:

Concn. of Na2CO3 = 2(E.P2 - E.P1) x F x 1000

= .......... g/L 10

Results:

E.P1 =

E.P2 =

Concn. of NaOH =

Concn. of Na2CO3 =

33

Exp.: Determination of Na2CO3 & NaHCO3 mixture



1) First step: (E.Pph.ph.)

Titration of 10 ml of the mixture ≠ standard HCl using ph.ph. indicator.

E.Pph.ph. ≡ 2

1 CO3

--

2) Second step: (E.PM.O.)

Titration of another 10 ml of the mixture ≠ standard HCl using M.O. ind.

E.PM.O. ≡ CO3-- + HCO3

-

Reaction equations:

For Na2CO3:

Na2CO3 + HCl → NaCl + NaHCO3 ...................Half neutralization (pH>10) (pH ≈ 8.3)

NaHCO3 + HCl → NaCl + CO2 + H2O ...............Complete neutralization (pH ≈ 3.8) .


For NaHCO3:

NaHCO3 + HCl → NaCl + CO2 + H2O

N.B.:

The reaction between HCl & Na2CO3 or between HCl & NaHCO3 is

an example of displacement titration where HCl (strong acid) displaces

carbonic acid (very weak acid) in its salts.

34


1 E.Pph.ph.:





2 E.PM.O.:






E.P1 (ph.ph.) ≡ 2

1 CO3

--

E.P2 (M.O.) ≡ CO3-- + HCO3

-

2E.P1 ≡ CO3--

(E.P2 - 2E.P1) ≡ HCO3-

For Na2CO3 :


2 HCl ≡ 1 Na2CO3

1 HCl ≡ 2

1 Na2CO3



1 x M.W. of Na2CO3 x 0.2

≡ .......... g Na2CO3

1 x 1000

sample standard

sample standard

35

Concentration:

Concn. of Na2CO3 = 2E.P1 x F x 1000

= .......... g/L 10

For NaHCO3 :

NaHCO3 + HCl → NaCl + CO2 + H2O

1 HCl ≡ 1 NaHCO3


each ml of 0.2 N HCl ≡ 1 x M.W. of NaHCO3 x 0.2

≡ .......... g NaHCO3 1 x 1000

Concentration:

Concn. of NaHCO3 = (E.P2 - 2E.P1) x F x 1000

= .......... g/L 10

Results:

E.P1 =

E.P2 =

Concn. of Na2CO3 =

Concn. of NaHCO3 =

sample standard

36

Illustrative diagrams for the titration of NaOH ≠ standard HCl Na2CO3 ≠ standard HCl NaHCO3 ≠ standard HCl

37

8.3 - 8.3 -

3.8 - 3.8 -

38

8.3 - 8.3 -

3.8 - 3.8 -

39

Summary

OH- CO3-- HCO3

-

E.Pph.ph.

(1st flask): all OH-

1 CO3

-- 2

No E.P.

E.PM.O.

(2nd flask): all OH- all CO3

-- all HCO3-

40

Exp.: Determination of Aspirin tablets (Back titration)

PPrriinncciippllee :: (Back or Residual Titration)

The method depends on boiling aspirin (acetylsalicylic acid) with

a known excess of standard alkali (NaOH) where aspirin is hydrolysed into

acetic and salicylic acids which react with sodium hydroxide giving sodium

acetate and sodium salicylate. Then the remaining unreacted part of

NaOH is back titrated against standard HCl using ph.ph. indicator.


NNootteess::

This method requires Blank Determination which is a separate

determination in which we repeat the same steps of the experiment but

without the sample and the E.P. in this case is called Blank Reading.


1) Transfer 1 Rivo tablet (equivalent to 320 mg acetylsalicylic acid) into

a clean flask.

2) Add 50 ml of 0.1 N NaOH.

3) Boil for 10 minutes inserting a funnel in the neck of the flask

to avoid evaporation of the solution.

4) Cool and add 10 drops ph.ph. indicator.



Aspirin (sample) known excess of st. NaOH

(50 ml) Sod. acetate Sod. salicylate

standard titrant

remaining unreacted

part

41



The equivalence factor (F) can be calculated on NaOH (1st stand.) or

on HCl (2nd stand.) but it is preferred to be calculated on the 1st standard.

Calculation of equivalence factor (F):

2 NaOH ≡ 1 Aspirin

1 NaOH ≡ 2

1 Aspirin

each ml of 0.1 N NaOH ≡ 2

1 x M.W. of Aspirin x 0.1

≡ .......... g Aspirin

1 x 1000


Concn. (Purity) of Aspirin = (Blank reading – E.P.) x F x 110000

= ........ gg%% 00..332200

Results:

E.P. =

Concn. (Purity) of Aspirin =

The Pharmacopoeial limits for the purity of aspirin:

According to BP 2007: from 95.0% to 105.0%.

According to USP 2007: from 90.0% to 110.0%.

1st standard

2

nd standard

Aspirin (sample)

sample

standard sample

standard

42

Exp.: Determination of Boric acid (H3BO3)


Boric acid (H3BO3) is a very weak monoprotic acid (pKa = 9.2) so it

can not be titrated against a standard alkali.

H3BO3 BO2- + H+ + H2O

But if a sufficient amount of Glycerol, Mannitol or similar organic

polyhydroxy compound is added to boric acid, it will be converted into

a relatively stronger monoprotic acid that can be directly titrated

against standard NaOH using ph.ph. indicator.

[Glyceroboric complex] H + NaOH [Glyceroboric complex] Na + H2O

Notes:

1) The reaction between glycerol and boric acid may be 2:1 (as shown in

the above equation) or 1:1 (as shown in the following equation):

2) M.O. indicator is unsuitable for the titration of glyceroboric acid

against NaOH because the pH of glyceroboric acid is about 4.2

metaborate ion

(very weak acid) { pH near 7 }

(relatively stronger acid) { pH ≈ 4.2 }

43

(i.e. the titration curve will start at pH 4.2 and go upward) so the

vertical part of the titration curve will be above the pH-range of M.O.

3) If the titration is done directly on boric acid without adding glycerol,

the end point will appear very rapidly due to the hydrolysis of the

produced sodium metaborate leading to the formation of NaOH again

giving pink color with ph.ph.

H3BO3 + NaOH NaBO2 + 2H2O


1) Transfer 10 ml of the sample into a clean conical flask.

2) Add 5 ml neutral glycerol


4) Titrate ≠ 0.1 N NaOH. {Color change at E.P. from colorless to pink }


Boric acid + 2 glycerol [Glyceroboric complex] H + 3H2O

[Glyceroboric complex] H + NaOH [Glyceroboric complex] Na + H2O

1 Boric acid ≡ 1 glyceroboric acid ≡ 1 NaOH

1 NaOH ≡ 1 Boric acid


each ml of 0.1 N NaOH ≡ 1 x M.W. of boric acid x 0.1

≡ ........ g boric à

1 x 1000

Concentration:

Concn. of boric acid = E.P x F x 1000

= .......... g/L

10

Sod. metaborate

sample standard

titrant

44

Exp.: Determination of Borax (Na2B4O7)

((BBoorraaxx == SSooddiiuumm BBoorraattee == SSooddiiuumm TTeettrraabboorraattee))


# Borax is hydrolyzed in water as follows:

Na2B4O7 + 7 H2O 2 NaOH + 4 H3BO3

The liberated NaOH can be titrated against 0.1 N HCl using M.O. indicator

without interference of boric acid as it is a very weak acid.




3) Titrate ≠ 0.1 N HCl. {Color change at E.P: from Yellow to Orange }


Na2B4O7 + 7 H2O 2 NaOH + 4 H3BO3

2 NaOH + 2 HCl 2 NaCl + 2 H2O

1 Borax ≡ 2 NaOH ≡ 2 HCl

1 HCl ≡ 2

1 Borax



1 x M.W. of borax x 0.1

≡ .......... g borax

1 x 1000

Concentration:

Concn. of borax = E.P x F x 1000

= .......... g/L 10

Tit. ≠

0.1 N HCl

add

glycerol

Tit. ≠

0.1 N NaOH

sample standard

titrant

45

Exp.: Determination of HCl & CH3COOH mixture



a) First step: (E.PM.O.) for HCl only

Titration of 10 ml of the mixture ≠ standard NaOH using M.O. indicator.

E.PM.O. ≡ HCl only

b) Second step: (E.Pph.ph.) for Total (HCl & CH3COOH)

Titration of another 10 ml of the mixture ≠ standard NaOH using ph.ph. ind.

E.Pph.ph. ≡ HCl + CH3COOH

Reaction equations:

HCl + NaOH → NaCl + H2O

CH3COOH + NaOH → CH3COONa + H2O

Notes:

1) During the determination of HCl only (1st E.P. using M.O.): NaOH neutralizes

HCl only without interference of acetic acid because the hydrogen

ion (H+) of HCl suppresses the ionization of acetic acid by common-ion

effect as shown in the following equations:

HCl H+ + Cl-

CH3COOH H+ + CH3COO-

and when the neutralization of HCl is completed, the pH of the solution

reaches the pH-range of M.O. and so M.O. changes its color.

While during the detn. of total HCl & acetic acid (2nd E.P. using ph.ph.):

NaOH neutralizes HCl first without any change in the color of ph.ph.

and then the titration is completed to make neutralization of acetic

46

acid also. At this point, the pH reaches the pH-range of ph.ph. and so

ph.ph. changes its color.

Question: (Complete the following)

In the detn. of (HCl/CH3COOH) mix. ≠ NaOH, E.PM.O. is lower than E.Pph.ph.

while in the detn. of Na2CO3 ≠ HCl, E.PM.O. is higher than E.Pph.ph. provided

that the 2 end points are done on 2 separate flasks.

2) The principle of this determination can be applied for the

determination of many (Strong acid/Weak acid) mixtures where:

E.PM.O. ≡ Strong acid

E.Pph.ph. ≡ Total (Strong acid + Weak acid)

Examples of strong acids: HCl, H2SO4

Examples of weak acids: Acetic acid (CH3COOH), Butyric acid (C3H7 COOH),

Phthalic acid ( ).

Note: In case of H2SO4 or Phthalic acid a = 2

1 (when calculating F).

47


A M.O. reading (E.P1) :




{ Color change at E.P.: from Red to Orange }

B ph.ph. reading (E.P2) :




{ Color change at E.P.: from Colorless to first Pink persistent for 30 sec.}


E.P1 (M.O. reading) ≡ HCl only

E.P2 (ph.ph. reading) ≡ HCl + CH3COOH

(E.P2 - E.P1) ≡ CH3COOH only

For HCl :

HCl + NaOH NaCl + H20

1 NaOH ≡ 1 HCl


each ml of 0.1 N NaOH ≡ 1 x M.W. of HCl x 0.1

≡ .......... g HCl 1 x 1000

Concentration:

Concn. of HCl = E.P1 x F x 1000

= .......... g/L 10

For CH3COOH :

sample standard

48

CH3COOH + NaOH CH3COONa + H20

1 NaOH ≡ 1 CH3COOH


each ml of 0.1 N NaOH ≡ 1 x M.W. of CH3COOH x 0.1

≡ ..... g CH3COOH 1 x 1000

Concentration:

Concn. of CH3COOH = (E.P2 - E.P1) x F x 1000

= .......... g/L 10

Results:

E.P1 =

E.P2 =

Concn. of HCl =

Concn. of CH3COOH =

sample standard

49

CChhaapptteerr IIIIII

PPrreecciippiittaattiioonn ttiittrraattiioonn

50

Precipitation Titration

Introduction :

Definition of Precipitation Titration:

It is a type of volumetric analysis that depends on the formation of

a precipitate.

Precipitation Titration

= Precipitimetric Titration

= Precipitate-formation Titration.

The most important precipitating agent used in precipitation titration

is Silver Nitrate (AgNO3).

Titrimetric methods based upon using silver nitrate (AgNO3) are

called Argentometric titration.

Argentometric titrations are mainly used for determination of halides

(Cl-, Br- & I-). It is also used for the determination of SCN-, CN- and

some compounds that form insoluble products (ppt) with AgNO3.

Types of Precipitation Titration :

Precipitation titrations are classified into 3 main types according to

the method of detection of end point:

1. Volhard's method depends on the formation of

Colored complex ions (Colored solution).

2. Fajan's method depends on the formation of

Colored adsorption compound.

3. Mohr's method depends on the formation of

Colored secondary precipitate.

51

MMoohhrr''ss MMeetthhoodd ..

It is a type of precipitation titration that depends on the formation of

a colored secondary ppt at the end point.

Main features of Mohr's method:

Sample: Cl- , Br- (Not for I- or SCN-).

Type of titration: Direct titration.

Standard Solution: Standard AgNO3.

pH: Neutral or slightly alkaline. (pH ≈ 6.5 -9).

Indicator: 1 ml of 5% Pot. Chromate (K2CrO4)

solution.

Color at E.P.: First permanent darkening of the yellow

chromate color [due to the formation of

Ag2CrO4 (brick red ppt) at the end point

leading to darkening of the solution].

52

Exp.: Determination of NaCl by Mohr's method


It depends on the titration of NaCl sample with standard AgNO3 using

potassium chromate (K2CrO4) as indicator.

Titration reaction:

Cl- + Ag+ AgCl

End point reaction:

2 Ag+ + CrO4-- Ag2CrO4

In this experiment, Cl- sample is titrated with Ag+ in presence of CrO4--

indicator where AgCl is precipitated as a white ppt before Ag2CrO4 THEN

at the end point (i.e. when Cl- is completely reacted with Ag+), the first

slight excess of Ag+ will react with CrO4-- indicator giving Ag2CrO4 (brick

red ppt) leading to darkening of the yellow color of the solution.

NNootteess::

3) Effect of pH: Mohr's method should be done in neutral or slightly

alkaline medium (pH ≈ 6.5 -9) because:

At pH 9: Ag+ will be precipitated as AgOH (brown to black ppt)

leading to Consumption of the titrant & Masking of the E.P. color.

Ag+ + OH- AgOH (brown to black ppt)

While at pH 6.5: The chromate ion (CrO4--) changes into acid chromate

(HCrO4-) then to dichromate (Cr2O7

--). Both HCrO4- & Cr2O7

-- form soluble

salts with Ag+ & so no colored ppt will be formed at the E.P.

2 CrO4-- + 2 H+ 2 HCrO4

- Cr2O7

-- + H2O

sample titrant white ppt

titrant indicator brick red ppt

chromate acid chromate dichromate

53

4) Ag2CrO4 is more soluble than AgCl so that No Ag2CrO4 will be

precipitated until all Cl- ions have been precipitated as AgCl provided

that the CrO4-- concentration should be adjusted to make Ag2CrO4

formed only at the end point and so prevent error in the end point.

1 ml of 5% K2CrO4 solution is suitable.

High concentration of K2CrO4 gives too soon (early) E.P.

because Ag2CrO4 will be rapidly precipitated before E.P.

Low concentration of K2CrO4 gives too late E.P.

because CrO4-- will be insufficient and so a large amount of

Ag+ (titrant) will be needed to precipitate Ag2CrO4 and so the E.P.

comes too late.

5) I- & SCN- can not be determined by Mohr's method because the

formed AgI & AgSCN strongly adsorb CrO4-- on their surfaces and so

the ppt formed at the end point will be an adsorption compound which

is less colored and so less sharp E.P. will be obtained.

6) BLANK EXPERIMENT should be done because: An additional excess

of AgNO3 (titrant) will be added after the correct E.P. to form

enough Ag2CrO4 to be seen over the heavy white ppt of AgCl & the

yellow color of K2CrO4 indicator leading to error (increase) in the end

point. This error can be corrected by performing blank determination

to know the volume of the additional excess of AgNO3 (titrant) that

causes error.

In blank determination, all steps of the experiment are performed in

absence of sample but using 10 ml dist. water instead of it and also

using talc or CaCO3 powder to imitate (تحاكي) the white AgCl ppt.

i.e. 10 ml dist. water + 1 ml 5% K2CrO4 + talc or CaCO3 powder

titate st. AgNO3 till the 1st darkening of the yellow chromate color.

54

The volume consumed of AgNO3 (titrant) is called Indicator Blank

and it should be subtracted from the observed E.P. in the original

experiment.

{ i.e. the calculation of concentration involves (E.P. - Ind. Blank)}.



5) Add 1 ml of 5% K2CrO4 indicator (by graduated pipette).

6) Titrate ≠ 0.01 N AgNO3

{ Color at E.P.: 1st permanent darkening of the yellow chromate color }

Before starting tit.: Cl- sample + CrO4-- ind. Clear bright

yellow solution (colorless) (yellow color)

During tit. ≠ AgNO3: AgCl + Cl- sample + CrO4-- ind. Turbid bright

yellow solution (white ppt) (colorless) (yellow color)

At the end point: AgCl + CrO4-- ind. + Ag2CrO4 Turbid dark

yellow solution (white ppt) (yellow color) (brick red ppt) few


Calculations in precipitation titration differ from that in acid-base titration in

the calculation of equivalence factor where (b) in precipitation titration equals

the number of univalent cation atoms present in or equivalent to ONE molecule

of the standard. In other words, (b) equals the total positive charge on the

cation radical of ONE molecule of the standard.

ex.: For AgNO3 b = 1, For NH4SCN b = 1, For Hg(NO3)2 b = 2

For this experiment:

Cl- + Ag+ AgCl

+1 +1 +2

55

1 AgNO3 ≡ 1 NaCl


each ml of 0.01 N AgNO3 ≡ 1 x M.W. of NaCl x 0.01

≡ .......... g NaCl 1 x 1000

Concentration:

Concn. of NaCl = (E.P.- Ind. blank) x F x 1000

= .......... g/L 10

Results:

E.P. =

Concn. =

sample standard

56

VVoollhhaarrdd''ss MMeetthhoodd ..

It is a type of precipitation titration that depends on the formation of

colored complex ions (colored solution) at the end point.

Main features of Volhard's method:

Sample: Cl- , Br-, I- & SCN-.

Type of titration: Back (Residual) titration.

Standard Solutions: 1st standard (25 ml): AgNO3.

2nd standard (titrant): NH4SCN or KSCN.

pH: Acidic medium (HNO3).

Indicator: Ferric salts such as:

1) Ferric alum = Ferric ammonium Sulfate

{ FeNH4(SO4)2.12H2O }.

OR 2) Ferric nitrate { Fe(NO3)3 }.

Color at E.P.: Faint orange or faint red color.

57

Introduction on Solubility Product (Ksp) :

♦ Ksp is calculated only for sparingly soluble salts (ex.: AgCl) and it is

constant for a certain salt at a specified temperature.

(ex.: Ksp of AgCl = 1.8x10-10, AgBr

♦ Meaning of Ksp:

For AgCl (for example): Ksp = [Ag+][Cl-]……at saturated solution (or at equilibrium).

♦ Ksp values for a certain salt can be known from reference tables.

♦ Significance of Ksp:

For the ppt of AgCl (for example) to be formed in a solution:

[Ag+][Cl-] must be Ksp of AgCl.

For different sparingly soluble salts:

Solubility product (Ksp) ≡ Solubility ………. & vice versa.

If different precipitates can be formed:

Ksp ≡ the salt is precipitated first (i.e: the salt is more insoluble or less soluble).

≡ more stable ppt.

N.B.: This rule is not always obeyed in case of non-symmetrical

salts ( i.e. salts in which the valence of the cation radicle the valence

of the anion radicle) such as Ag2CrO4 , PbI2 ......etc.

{although Ksp of Ag2CrO4 is less than Ksp of AgCl but Ag2CrO4 is more

soluble than AgCl…………refer to Mohr's method}.

58

Important example:

Ksp of AgI AgBr AgSCN AgCl

So:

If a mix. of I-, Br-, SCN-, Cl- (of near concentrations) is treated with Ag+:

If we have:

+ Ag+

The order of precipitation will be:

1- AgI

2- AgBr

3- AgSCN

4- AgCl

AgCl ppt

+ SCN-

AgSCN ppt

+ Cl-

AgBr ppt

+ SCN-

No reaction

AgI ppt

+ SCN-

No reaction

59

Exp.: Determination of NaCl by Volhard's method

PPrriinncciippllee :: {{BBaacckk ((RReessiidduuaall)) ttiittrraattiioonn}}

It depends upon the precipitation of Cl- as AgCl by the addition of

a known excess of standard AgNO3 THEN the remaining unreacted AgNO3 is

back titrated against standard NH4SCN using ferric alum indicator.

1 Cl- + Ag+ AgCl

2 Titration reaction:

Ag+ + SCN- AgSCN

3 End point reaction:

SCN- + Fe3+ [Fe(SCN)]2+

NNootteess::

1) Since Ksp of AgSCN AgCl so AgSCN is more stable (more insoluble)

than AgCl so after complete reaction between NH4SCN titrant & the

excess unused AgNO3, the SCN- ions will attack AgCl ppt and replace Cl-:

AgCl (solid) + SCN- AgSCN (solid) + Cl-

To overcome this problem: The ppted AgCl must be removed or coated before

titrating the excess unused Ag+ with NH4SCN & that can be achieved by:

1- Filtration of AgCl ppt before titration.

OR 2- Coating AgCl ppt by the addition of an organic solvent as

nitrobenzene or amyl alcohol which forms a water-proof layer

around AgCl ppt protecting it from reaction with NH4SCN.


In the detn. of NaCl by Volhard's method, if AgCl is not removed,

the observed E.P. will be higher than the true and the calculated Cl

- content

(concn.) will be lower than the true... because it's a back titration.

(sample) (25 ml st. AgNO3) white ppt

(remaining unused excess) (NH4SCN titrant) white ppt

(1st dr. excess after complete

reactn with the unused Ag+)

(indicator) faint orange

or faint red color

NH4SCN titrant

remaining unused AgNO3

+ Fe3+ indicator

60

2) Strong shaking near the E.P. is required to prevent adsorption of Ag+ ions

on the freshly ppted AgSCN which leads to occurrence of early E.P.


In the detn. of NaCl by Volhard's method, if the titration is done without

vigorous shaking at the E.P., the observed E.P. will be lower than the true and

the calculated concn. will be higher than the true.

3) The titration is performed only in acidic medium (HNO3) because:

a. In alkaline medium, Ag+ & Fe3+ ions will be ppted as AgOH &

Fe(OH)3.

b. HNO3 prevent precipitation of other Ag+ salts like carbonate

and sulfide because these salts are soluble in HNO3.

c. HNO3 helps the coagulation of AgCl ppt & that facilitates its

filtration or coating.

BUT too high concentration of HNO3 should be avoided because that

destroys the colored ferric thiocyanate complex formed at the E.P.

RReeqquuiirreedd GGllaasssswwaarree::

110000--mmll mmeeaassuurriinngg ((vvoolluummeettrriicc)) ffllaasskk .. well-closed

SSttooppppeerreedd ffllaasskk.. well-closed

CCoonniiccaall ffllaasskk..

BBuurreettttee && bbuurreettttee hhoollddeerr..

1100--mmll bbuullbb ppiippeettttee..

2255--mmll bbuullbb ppiippeettttee..

1100--mmll ggrraadduuaatteedd ppiippeettttee..

1100--mmll mmeeaassuurriinngg ccyylliinnddeerr..

FFiilltteerr ppaappeerrss ((WWhhaattmmaannnn nnoo..11))..

FFuunnnneell..

BBeeaakkeerr..

DDrrooppppeerr..

CCaallccuullaattoorr..

61


!!!!!! BBEEFFOORREE SSTTAARRTTIINNGG TTHHEE DDEETTEERRMMIINNAATTIIOONN,,

WWAASSHH AALLLL TTHHEE GGLLAASSSSWWAARREE WWIITTHH DDIISSTTIILLLLEEDD WWAATTEERR..

(because tap water contains Cl -).

11 Transfer accurately 10 ml of the sample into a 100-ml measuring flask.

22 Add 2 ml conc. HNO3.

33 Add 25 ml of 0.05 N AgNO3.

44 Shake well for few minutes to coagulate the precipitated AgCl.

55 Complete to the mark with distilled water.

66 Mix well.

77 Filter through a dry filter paper, rejecting the first few milliliters of

the filtrate.

88 Transfer 25 ml of the filtrate into a clean glass stoppered flask.

99 Add 1 ml ferric alum indicator.

1100 Titrate with 0.01 N NH4SCN with strong shaking.

Color at E.P.: Faint Orange Color not disappear on strong shaking.

62


In this exp., the equivalence factor (F) can be calculated on 0.05 N AgNO3

(1st standard "25 ml") OR on 0.01 N NH4SCN (2nd standard "titrant") but it is

preferred to be calculated on the 1st standard.

If (F) is calculated on the 1st standard (0.05 N AgNO3):


1 AgNO3 ≡ 1 NaCl

each ml of 0.05 N AgNO3 ≡ 1 x M.W. of NaCl x 0.05

≡ .......... g NaCl 1 x 1000


Concn. = [25 – 4 E.P. (

05.0

01.0 )] x F x 1000 = .......... g/L

10

If (F) is calculated on the 2nd standard (0.01 N NH4SCN):


1 NH4SCN ≡ 1 NaCl

each ml of 0.01 N NH4SCN ≡ 1 x M.W. of NaCl x 0.01

≡ .......... g NaCl 1 x 1000


Concn. = [25 (

01.0

05.0 ) – 4 E.P.] x F x 1000 = .......... g/L

10

Results:

E.P. =

Concn. =

sample standard

0.05 N 0.01 N 0.05 N

Dilution factor

0.01 N

Dilution factor

0.01 N

sample standard

0.05 N

63

Exp.: Determination of KBr by Volhard's method

PPrriinncciippllee :: {{BBaacckk ((RReessiidduuaall)) ttiittrraattiioonn}}

It depends upon the precipitation of Br- as AgBr by the addition of

a known excess of standard AgNO3 THEN the remaining unreacted AgNO3 is

back titrated against standard NH4SCN using ferric alum indicator.

1 Br- + Ag+ AgBr

2 Titration reaction:

Ag+ + SCN- AgSCN

3 End point reaction:

SCN- + Fe3+ [Fe(SCN)]2+

BBrroommiiddee iiss ddeetteerrmmiinneedd bbyy VVoollhhaarrdd''ss mmeetthhoodd iinn tthhee ssaammee wwaayy aass

cchhlloorriiddee bbuutt wwiitthhoouutt rreemmoovviinngg tthhee pppptteedd AAggBBrr bbeeccaauussee iitt iiss mmoorree

iinnssoolluubbllee ((lloowweerr KKsspp)) tthhaann AAggSSCCNN ssoo SSCCNN-- iioonn wwiillll nnoott aattttaacckk

AAggBBrr pppptt aafftteerr ccoommpplleettee rreeaaccttiioonn wwiitthh tthhee uunnuusseedd AAgg++..

PPrroocceedduurree ffoorr tthhee ddeettnn.. ooff KKBBrr bbyy VVoollhhaarrdd''ss mmeetthhoodd ::

!!!!!! BBEEFFOORREE SSTTAARRTTIINNGG TTHHEE DDEETTEERRMMIINNAATTIIOONN,,

WWAASSHH AALLLL TTHHEE GGLLAASSSSWWAARREE WWIITTHH DDIISSTTIILLLLEEDD WWAATTEERR..

(because tap water contains Cl – which interferes with the determination)

11 Transfer accurately 10 ml of the sample into a clean glass stoppered flask.

22 Add 2 ml conc. HNO3.

33 Add 25 ml of 0.05 N AgNO3.

(sample) (25 ml st. AgNO3) yellowish

white ppt

(remaining unused excess) (NH4SCN titrant) white ppt

(1st dr. excess after complete

reactn with the unused Ag+)

(indicator) faint orange

or faint red color

NH4SCN titrant

remaining unused AgNO3

+ Fe3+ indicator

64

44 Shake well for few minutes to coagulate the precipitated AgBr.

55 Add 1 ml ferric alum indicator.

66 Titrate with 0.01 N NH4SCN with strong shaking.

Color at E.P.: Faint Orange Color not disappear on strong shaking.


In this exp., the equivalence factor (F) can be calculated on 0.05 N AgNO3

(1st standard "25 ml") OR on 0.01 N NH4SCN (2nd standard "titrant") but it is

preferred to be calculated on the 1st standard.

If (F) is calculated on the 1st standard (0.05 N AgNO3):


1 AgNO3 ≡ 1 KBr

each ml of 0.05 N AgNO3 ≡ 1 x M.W. of KBr x 0.05

≡ .......... g KBr 1 x 1000


Concn. = [25 – E.P. (

05.0

01.0 )] x F x 1000 = .......... g/L

10

If (F) is calculated on the 2nd standard (0.01 N NH4SCN):


1 NH4SCN ≡ 1 KBr

each ml of 0.01 N NH4SCN ≡ 1 x M.W. of KBr x 0.01

≡ .......... g KBr 1 x 1000


Concn. = [25 (

01.0

05.0 ) – E.P.] x F x 1000 = .......... g/L

10

sample standard

0.05 N 0.01 N 0.05 N

0.01 N 0.01 N

sample standard

0.05 N

65

Exp.: Determination of NaCl by Fajan's method

PPrriinncciippllee ((TThheeoorryy oorr MMeecchhaanniissmm ooff aaccttiioonn ooff aaddssoorrppttiioonn iinnddiiccaattoorrss)) ::

It depends on the titration of NaCl sample with standard AgNO3 using

fluorescein indicator. At the E.P., the formed AgCl ppt becomes pink.

Titration reaction: NaCl + AgNO3 ind. nfluorescei AgCl + NaNO3

Before starting titration: the ions present in the solution are Na+, Cl - & In-.

Stage of

titration: A During titration

(before E.P.)

B 1st drop excess of AgNO3

immediately after the E.P.

Ions present in solution:

Cl- + Na+ + In- + NO3-

Ag+ + NO3- + Na+ + In-

Diagram:

Color of

the ppt:

White ppt in

green fluorescence

Pink ppt

in faint colored solution

Explan

ation:

At start of tit.,

(Cl-) ions present in

excess so the 1ry

adsorption layer will

be (Cl-) ions [own

ions of ppt] and

the 2ry adsorption

layer will be the

oppositely charged

(Na+) ions [counter-

ions].

Immediately after the E.P., AgNO3 is

present in excess so (Ag+) ions form the

1ry ads. Layer [own ions of ppt] {Note that all

(Cl -) ions from sample are already precipitated}.

The 2ry adsorption layer of the opposite

charge will be (NO3-) BUT since

fluorescein ions carry (-ve) charge also &

it is more strongly adsorbed than (NO3-),

so fluorescein replaces (NO3-) and the 2ry

layer becomes (In-) & the ppt becomes

pink due to deformation of the electronic

system of the indicator anion.

sample titrant

AgCl

Cl- Cl-

Cl- Cl-

Cl- Cl-

Na+

Na+

Na+

Na+

Na+

Na+

sample from sample & ind.

ind. from sample & ind.

ind. titrant 1st drop xss after E.P.

titrant

AgCl

Ag+ Ag+

Ag+ Ag+

Ag+ Ag+

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

AgCl

Ag+ Ag+

Ag+ Ag+

Ag+ Ag+

In-

In-

In-

In-

In-

In-

spontaneous

titrant

(Ag+ is precipitated as AgCl )

66

NNootteess::

1) Eosin indicator can not be used for the determination of (Cl-) because

eosin is too strongly adsorbed to the surface of the ppt and may

replace (Cl-) ions in the 1ry adsorption layer before the E.P. giving

early E.P. [stage A in the diagram]

2) Effect of pH: In Fajan's method, the pH should be adjusted to keep

the indicator dissociated.

For fluorescein indicator:

Fluorescein (In-Na+) is a sod. salt of a very weak acid and it's hydrolyzed

in water to give the parent acid (InH) which is very weak and so its

dissociation is suppressed in acidic medium due to common ion effect.

InH H+ + In-

H+

So fluorescein is used in neutral or slightly alkaline medium (pH=7-10).

[Note that strong alkaline pH is prohibited in argentometric tit. because

Ag+ will be precipitated as AgOH (brown to black ppt) leading to

Consumption of the titrant & Masking of the E.P. color].

For eosin indicator:

Eosin (In-Na+) is a sod. salt of a more stronger acid and so its

dissociation is not affected in acidic medium. Therefore, eosin can be

used at pH down to 2 ( i.e. pH 2).


1100 mmll ssaammppllee ddiilluuttee ttoo 5500 mmll wwiitthh ddiisstt.. wwaatteerr aadddd 1155 ddrrooppss

fflluuoorreesscceeiinn iinnddiiccaattoorr TTiittrraattee ≠ 0.05 N AgNO3.

At the E.P., the formed AgCl ppt becomes pin

67

CChhaapptteerr IIVV

CCoommpplleexxoommeettrriicc ttiittrraattiioonn

68

Complexometric titration

(Complex-formation titration)

Definition of Complex-formation titration :

It is a type of titrations that depends on the formation of soluble complexes.

Introduction :

♦ Complex = Coordination compound.

♦ Complex consists of an electron donor bonded to an electron acceptor

by a coordination bond.

♦ Example of Complexes:

Silver ammine complex " [Ag(NH3)2]+ " where NH3 is the donor (ligand)

while Ag+ is the acceptor (metal).

[H3N: → Ag+ ← :NH3]

Donor .

(= Ligand)

(= Complexing agent)

can donate lone pair

of electrons such as

N, O & S. .. .. ..

coordination bond Acceptor .

(= Metal)

can accept lone pair of

electrons in its vacant

orbitals ...... ...... .......

69

EEDDTTAA TTiittrraattiioonnss ..

It is the most common type of complex-formation titrations.

The main complexing agent (= ligand = donor) in these titrations is EDTA.

Basic components in any EDTA titration:

1) Metal (acceptor) sample .

2) EDTA (donor) the main complexing agent.

3) Indicator (donor).

4) Buffer (to adjust pH of the medium).

1 Metal :

A very large number of metals can be determined complexometrically

such as Mg2+, Ca2+, Zn2+, Al3+………………etc.

2 EDTA : { EDTA = Ethylenediaminetetraacetic acid }

It is used as disodium salt

because the parent acid is insoluble in water while the disodium salt is soluble.

EDTA is a hexadentate ligand [ i.e. it can bind to metal ion with

6 coordination bonds because it contains 6 ligand (donor) atoms as

shown in the structure].

N.B. Generally, multidentate ligands (i.e. bi-, tri-, hexa- …… etc) are

called "Chelating agents".

CH2 N

CH2 CO

O H

CH2 C O H

CH2 N

CH2 C O H

CH2 C O H

O

O

O

Na+

Na+

70

EDTA can form stable 1:1 complex with many divalent, trivalent and

tetravalent metal ions (i.e. 1 mole of EDTA ≡ 1 mole of metal ion) so

EDTA is always prepared as Molar solutions.

The parent acid of EDTA is represented as H4Y while its disodium salt

is represented as H2Y2-.

3 Indicator :

The indicators used in complexometric titrations are metal-sensitive

indicators so these indicators are called "Metal indicators".

The color of the free indicator must differs from the color of

the [metal-indicator] complex.

Also, these indicators are pH-sensitive indicators BUT during

titration, we adjust pH by using a suitable buffer and so the indicators

will be affected by the metal only.

Examples of metal indicators:

1 EBT indicator (Eriochrome Black T) 2 Murexide indicator

It is a salt of an acid. It is a salt of an acid.

Its parent acid is represented as

(H3In)

Its parent acid is represented as

(H5In)

EBT ind. is represented as follows:

H2In- HIn2- In3-

(pH 7) (pH 7-11) (pH 11)

Red

Blue

orange

Murexide is represented as follows:

H4In- H3In2- H2In3-

(pH 9) (pH 9-11) (pH 11)

Reddish

Violet Violet

Bluish

Violet

EBT is used at pH 10 (adjusted by

Ammonia Buffer) so the color of

the free EBT will be Blue.

Murexide is most commonly used at

pH 12 (adjusted by 8% NaOH) so

the color of free murexide will be

Bluish Violet.

Also, murexide is sometimes used

at pH 10 (amm. buffer) where the

color of free murexide is Violet.

The color of [metal-EBT] complex is

usually Wine-red color.

The color of [metal-murexide] complex

differs according to the metal & pH.

71

Ex. of metals determined using EBT:

Mg2+, Zn2+, Hg2+

Ex. of metals determined using murexide:

Ca2+ (at pH 12) & Ni2+ (at pH 10).

Question: (Choose the correct answer)

EBT and murexide are (metal sensitive - pH sensitive - both)

4 Buffer :

Role of buffer in complexometric titrations:

1) Consume the released protons from the reaction between metal and

EDTA and so make the reaction go forward.

Metal + EDTA (metal-EDTA) complex + 2H+

2) Keep pH constant and so keep the correct color of the indicator.

Examples of buffers used:

1) Ammonia buffer (pH = 10).

2) 8% NaOH (pH = 12).

Methods of EDTA titrations 1. Direct titrations:

Direct titration of metal sample ≠ standard EDTA.

2. Back (Residual) titrations:

Metal sample + known excess of standard EDTA

Back titration of the remaining unreacted EDTA ≠ standard metal solution.

3. Replacement (Substitution) titrations:

Metal sample + (Mg-EDTA) solution (metal-EDTA) + equivalent amount of Mg2+

Titration ≠ standard EDTA.

4. Alkalimetric titrations:

Metal sample + EDTA solution (metal-EDTA) complex + 2H+

Titration ≠ standard NaOH

5. Miscellaneous methods:

72

Ex.: Determination of Ag+ sample.

Examples of Complexometric titrations

Introduction :

General Principles for Complexometric Titrations:

The complex is formed between Metal & EDTA or between Metal & Indicator

{ Not between EDTA & Indicator because both of them are

electron- donors (ligands) }.

(Metal-EDTA) complex must be more stable than (Metal-Indicator) complex.

The change in color at the end point is due to change of:

[complexed ind. into free ind.]…………………………………………in direct titration

OR

[free ind. into complexed ind.]………………………………………….in back titration.

Sketch Diagrams for

Direct Complexometric Titrations EDTA

These symbols will be used in the following diagrams:

= Magnesium ion.

= Calcium ion.

= Disodium salt of

ethylenediaminetetraacetic

acid.

= Free EBT indicator.

= Free Murexide indicator.

= Complexed EBT indicator.

= Complexed Murexide ind. EDTA

Ca2+

EBT

Murexide

EBT

Mg2+

Mg2+

Murexide

Ca2+

Exp.: Complexometric Determination of MgSO4.7H2O


Direct titration of Mg2+ with standard EDTA at pH = 10 (amm. buffer)

using EBT indicator.

Equations: (Please, Refer to previous sketch diagrams)

Before starting titration:

Mg2+ + HIn2- MgIn- + H+

During titration EDTA:

Mg2+ + H2Y2- MgY2- + 2H+

Just before & at the E.P.:

MgIn- + H2Y2- MgY2- + H+ + HIn2-

(This reaction occurs because Mg-EDTA complex is more stable than Mg-EBT complex)



2) Add 2 ml of Amm. buffer.

3) Add few specks of EBT indicator.

4) Titrate 0.01 M EDTA.

{ Color change at E.P.: from Wine Red to Pure Blue }

7755

sample EBT ind.

EDTA (titrant)

Free EBT (blue)

complexed EBT (wine red)

Complexed

EBT ind.

Free

EBT ind.


Calculations in complexometric titrations differ from that in

other types of titration in the calculation of equivalence factor where

( b ) in complexometric titrations is Absent because we always use Molar

concentrations in complexometric titrations.


Mg2+ + H2Y2- MgY2- + 2H+

1 EDTA ≡ 1 MgSO4.7H2O

each ml of 0.01 M EDTA ≡ 1 x M.W. of MgSO4 .7H2O x 0.01

≡..g MgSO4.7H2O 1000

Concentration:


= .......... g/L 10

76

sample standard

standard sample

Exp.: Complexometric Determination of CaCl2.6H2O


Direct titration of Ca2+ with standard EDTA at pH = 12 (8% NaOH)

using murexide indicator.

Equations: (Please, Refer to previous sketch diagrams)

Before starting titration:

Ca2+ + H2In3- CaH2In-

During titration EDTA:

Ca2+ + H2Y2- CaY2- + 2H+

Just before & at the E.P.:

CaH2In- + H2Y2- CaY2- + 2H+ + H2In3-

(This reaction occurs because Ca-EDTA complex is more stable than Ca-murexide complex)



2) Add 2 ml of 8% NaOH.

3) Add few specks of murexide indicator.


{ Color change at E.P.: from Rose Red (Pink) to Bluish Violet }



Ca2+ + H2Y2- CaY2- + 2H+

77

sample murexide ind.

EDTA (titrant)

Free murexide (bluish violet)

complexed murexide (rose red)

standard (EDTA)

sample

Complexed

murexide ind.

Free

murexide ind.

1 EDTA ≡ 1 CaCl2.6H2O

each ml of 0.01 M EDTA ≡ 1 x M.W. of CaCl2.6H2O x 0.01

≡..g CaCl2.6H2O 1000

Concentration:


= .......... g/L 10

Exp.: Complexometric Determination of

Potash alum { KAl(SO4)2.12H2O }

PPrriinncciippllee :: ((bbaacckk ttiittrraattiioonn))

Al3+ salts can not be directly titrated EDTA because (Al-EDTA) complex

is formed slowly so Al3+ salts are best determined by back titration.

Equation: Al3+ + H2Y2- AlY- + 2H+

then the remaining unreacted EDTA is back titrated standard MgSO4

using EBT indicator & amm. buffer.

{ Color change at E.P.: from Blue to Wine Red }



2) Add 25 ml of 0.05 M EDTA.

3) Add 2 ml of amm. buffer.

4) Add few specks of EBT indicator. 78

sample standard

sample known excess of st. EDTA

(Al-EDTA) complex

Complexed

EBT ind.

Free

EBT ind.

5) Titrate 0.05 M MgSO4

{ Color change at E.P.: from Blue to Wine Red }


Equivalence factor (F): 1 EDTA ≡ 1 Al3+

each ml of 0.05 M EDTA ≡ 1 x M.W. of KAl(SO4)2.12H2O x 0.05 ≡………...g

KAl(SO4)2.12H2O 1000

Concentration:

Concn. = (25 - E.P.) x F x 1000

= .......... g/L 10

Results:

E.P. =

Concn. =

Exp.: Complexometric determination of CuSO4.6H2O

PPrriinncciippllee::

By titration ≠ standard EDTA using PAN indicator & Acetate

buffer.

{PAN = Pyridylazonaphthol}

Color change: from Pink (complexed PAN) to Yellowish Green (free PAN).

N.B. PAN indicator allows selective titration of Cu2+ in presence

of Ba2+, Ca2+, Mg2+ & Mn2+.

PPrroocceedduurree::

sample standard

Complexed

EBT ind.

Free

EBT ind.


2) Add 2 ml of acetate buffer.

3) Add 5 drops of PAN indicator.


{ Color change at E.P.: from Pink to Yellowish Green }


Cu2+ + H2Y2- CuY2- + 2H+

1 EDTA ≡ 1 CuSO4.5H2O


each ml of 0.01 M EDTA ≡

1 x M.W. of CuSO4 .5H2O x 0.01 ≡..g CuSO4.5H2O

1000

Concentration:

Concn. of CuSO4 .5H2O = E.P1 x F x 1000

= .......... g/L 10

Complexed

PAN ind.

Free

PAN ind.

sample standard

EDTA

Exp.: Complexometric Determination of (Ca2+/Mg2+) Mixture

(CaCl2.6H2O & MgSO4.7H2O)

PPrriinncciippllee :: ((ddiirreecctt ttiittrraattiioonn))

c) First step: ((DDeetteerrmmiinnaattiioonn ooff CCaa22++oonnllyy))..

By titrating 10 ml of the mixture ≠ standard EDTA using

8% NaOH & murexide indicator.

E.P1 ≡ CCaa22++ only.

d) Second step: ((DDeetteerrmmiinnaattiioonn ooff ttoottaall CCaa22++ && MMgg22++))..

By titrating another 10 ml of the mixture ≠ standard EDTA using

amm. buffer & EBT indicator.

E.P2 ≡ TToottaall (Ca2+ & Mg2+).

Notes:

1) When we use 8% NaOH (pH=12) & murexide, we determine Ca2+ ONLY

because: Mg2+ does not form any complex with EDTA at pH=12 due to

the precipitation of Mg2+ as Mg(OH)2 which is more stable than

(Mg-EDTA) complex. IN OTHER WORDS, we can say that the selectivity

of EDTA was increased to determine Ca2+ only without interference of Mg2+.

2) Ca2+ alone can not be directly determined ≠ EDTA using EBT ind.

because: EBT gives poor E.P. with Ca2+ ( i.e. the color change at E.P. is

not sharp) and that is because the binding between Ca2+ & EBT is very weak.

3) Although EBT can not be used in direct complexometric

determination of Ca2+ alone (as mentioned before), BUT it can be used

for determination of total Ca2+ & Mg2+…………………………HOW??!!

That is because:

Before starting titration: {sample (Ca2+& Mg2+) + amm. buffer + EBT }

EBT will be complexed with Mg2+ only because (Mg-EBT) complex is

more stable than (Ca-EBT) complex which is very weak (as mentioned before).

i.e. We have { free Ca2+ + free Mg2+ + (Mg-EBT) } in the flask.

During titration, the following reactions occur:

1) EDTA + free Ca2+ (Ca-EDTA) complex.

2) EDTA + free Mg2+ (Mg-EDTA) complex.

3) EDTA + (Mg-EBT) (Mg-EDTA) + free EBT.

i.e. EDTA reacts first with free Ca2+ then with free Mg2+ because

(Ca-EDTA) complex is more stable than (Mg-EDTA) complex (compare with EBT).

And finally EDTA reacts with (Mg-EBT) complex and so the color

change at E.P. will be due to Mg-reaction and that gives sharp E.P.

representing (or equivalent to) the total.


A Murexide reading (E.P1) : (≡ Ca2+ only)


2) Add 2 ml of 8% NaOH.

3) Add few specks of murexide indicator.


{ Color change at E.P.: from Rose Red (Pink) to Bluish Violet }

B EBT reading (E.P2) : (≡ Ca2+ + Mg2+)


2) Add 2 ml of Amm. buffer.

3) Add few specks of EBT indicator.


{ Color change at E.P.: from Wine Red to Pure Blue }

8822

E.P. blue color

Complexed

murexide ind.

Free

murexide ind.

Complexed

EBT ind.

Free

EBT ind.


E.P1 {murexide reading} ≡ Ca2+

E.P2 {EBT reading} ≡ Ca2+ + Mg2+

(E.P2 - E.P1) ≡ Mg2+

For Ca2+ :

Ca2+ + H2Y2- CaY2- + 2H+

1 EDTA ≡ 1 CaCl2.6H2O


each ml of 0.05 M EDTA ≡ 1 x M.W. of CaCl2.6H2O x 0.05

≡..g CaCl2.6H2O 1000

Concentration:

Concn. of CaCl2.6H2O = E.P1 x F x 1000

= .......... g/L 10

For Mg2+ :

Mg2+ + H2Y2- MgY2- + 2H+

1 EDTA ≡ 1 MgSO4.7H2O


each ml of 0.05 M EDTA ≡ 1 x M.W. of MgSO4 .7H2O x 0.05

≡..g MgSO4.7H2O 1000

Concentration:

Concn. of MgSO4 .7H2O = (E.P2 - E.P1) x F x 1000

= .......... g/L 10

Results:

E.P1 =

E.P2 =

Concn. of CaCl2.6H2O =

Concn. of MgSO4 .7H2O =

sample standard

EDTA (standard)

sample standard

EDTA (standard)

IInnssttrruuccttiioonnss ffoorr uussiinngg

tthhee GGllaasssswwaarree

الصيدلية التحليلية الكيمياء لعملي المطلوبة األدوات

استعمالھا طـريقة و

)تدريجه غير دقيق( :)Flask ml Conical 250 (دورق المخروطيال )١ .}طل بالماء فقـيغس{

)تدريجھا دقيق جدا( :) Pipette"Transfer Pipet orBulb"(ذات الفقاعة الماصة )٢ ml and 25 ml pipets 10: مطلوب منھا ماصتين

.}مأل بهغسل بالماء ثم بالمحلول الذي ست ت {

)ذ بالترتيبـتنف (طريقة االستخدام بالتفصيل .لماصة جيدا بالماءاغسل ا .١ . بمنديلجفف الماصة جيدا من الخارج .٢ .لماصة جيدا بالمحلول الذي ستمأل بهاغسل ا .٣ .جفف أصابعك و جفف فوھة الماصة جيدا .٤ . مللي١٠عالمة الـ) أعلى منأي(لما بعد امأل الماصة .٥ . بمنديلامسح الماصة جيدا من الخارج .٦ـ .٧ ة ال ى عالم ول عل ي ١٠اضبط سطح المحل سبابة(بتخفيف ضغط اصبعك ( ملل دريجيا، ) ال ت

).يساعد في ذلك أن تدير الماصة بيدك اليسرى و :ملحوظات ھامة على ھذه الخطوة

ابةالماصة يكون باصبع السبفوھة مسك. السفلي لسطح المحلولحد يكون لل١٠ضبط المحلول على عالمة الـ . في نفس مستوى العين١٠عند ضبط المحلول يجب أن تكون عالمة الـ . ة سكھا ألن إمال اء م يا أثن عھا رأس ون وض ى أن يك ا عل افظ دائم ة ح أل الماص د م بع

.الماصة قد تسبب دخول فقاعات ھواء داخلھا ١٠ة بعد ضبط المحلول على عالمة الـاحذر المشي بالماص. ١٠احذر مسح الماصة بمنديل بعد ضبط المحلول على عالمة الـ.

.تأكد من عدم وجود أي فقاعة ھواء في الماصة بعد مألھا، وإال تعيد مألھا مرة أخرى .٨ :فرغ محتويات الماصة داخل الفالسك كالتالي .٩ و الماصة رأسية° ٤٥اجعل الفالسك مائال بزاوية. مازالت داخل الفالسك ثانية على األقل و الماصة ١٥اترك الماصة تفرغ محتوياتھا ثم انتظر. فرغ آخر كمية في الماصة عن طريق لمس مقدمة الماصة برفق للجزء الجاف من الفالسك. ر) م في نھاية الماصةـ س٠.١تشغل تقريبا مسافة (اعلم أن آخر نقطة بالماصة محسوبة غي

. لذلك إنزالھا غير مطلوبفي حجم الماصة

)تدريجھا دقيق جدا( :) Burette"Buret" (السحاحة )٣ ml50 OR ml 25 : مطلوب سحاحة واحدة فقط

.}مأل بهتغسل بالماء ثم بالمحلول الذي ست {

)ذ بالترتيبـتنف (طريقة االستخدام بالتفصيل .لسحاحة جيدا بالماءاغسل ا .١ .احة جيدا بالمحلول الذي ستمأل بهلسحاغسل ا .٢ .امأل السحاحة بالمحلول ثم قم بتفريغ فقاعات الھواء من الجزء السفلي للسحاحة .٣ .عالمة الصفر) أعلى منأي(امأل السحاحة لما بعد .٤ .جفف السحاحة جيدا من الخارج .٥ون لل .٦ ك يك ة أن ذل ع مالحظ صفر م ة ال ى عالم ول عل داضبط سطح المحل سطحح سفلي ل ال

ا، وينطبق المحلول مع مراعاة أن تكون عالمة الصفر في نفس مستوى العين عند النظر إليھ ).E.P(على طريقة قراءة الـ ذلك أيضا

. بالطريقة الموضحة)Titration(ـ أثناء عملية اليجب مراعاة مسك السحاحة .٧

:)Pipette "Graduated Pipet ml 10"( مل ١٠رجة صة مدما )٤ ) بالماصة ذات الفقاعةنة تدريجھا غير دقيق مقار(

:)Measuring Cylinder ml 10( مل ١٠ مخبار مدرج )٥ )غير دقيقتدريجه (

):ml Beaker250 ( مل ٢٥٠بيكر )٦ )تدريجه غير دقيق(

.) FunnelGlass(زجاجيقمع )٧

.)Buret Holder(حامل سحاحة )٨

٩( White Plate.

ت عامةت عامةملحوظاملحوظا رجاع أي محلول إلى زجاجات المعمل بعد انتھائك من التجربةإاحذر أن تقوم ب.

ا بعد انتھاء المعمل نظيفة تمام كا على جميع أدواتاحرص على الحفاظ دائم.

Date post:	19-Jul-2018
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