1. Precipitation Titrations
K G. Baheti
Y. B. Chavan College of Pharmacy Aurangabad

2. Precipitation reaction
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate. 
Drop some ordinary table salt into a glass of water and watch it "disappear". We refer to this as dissolution, and we explain it as a process in which the sodium and chlorine units break away from the crystal surface, get surrounded by H2O molecules & and become hydrated ions.
NaCl(s) → Na+(aq)+ Cl–(aq)

3. NaCl(s) Na+(aq)+ Cl–(aq)
But if you keep adding salt, there will come a point at which it no longer seems to dissolve. If this condition persists, we say that the salt has reached its solubility limit, and the solution is saturated in NaCl. The situation is now described by
NaCl(s)  Na+(aq)+ Cl–(aq)
in which the solid and its ions are in equilibrium.

4. Expressing solubility
Solubilities are most fundamentally expressed in molar (mol L–1 of solution) or molal (mol kg–1 of water) units.
Temperature need to mention when solubility determines as temp affect the solubility of comps.
As per India pharmacopoeia Solubility designations "soluble", "insoluble", "slightly soluble", and "highly soluble" are used..

5. Precipitation reaction
The use of solubility rules require an understanding of the way that ions react.
Most precipitation reactions are single replacement reactions or double replacement reactions.
A double replacement reaction occurs when two ionic reactants dissociate and bond with the respective anion or cation from the other reactant.
The ions replace each other based on their charges as either a cation or an anion.
Double replacement reactions

6. Precipitation reaction
A double replacement reaction is specifically classified as a precipitation reaction when the chemical equation in question occurs in aqueous solution and one of the of the products formed is insoluble. An example of a precipitation reaction is given below:
CdSO4(aq)+K2S(aq)→CdS(s)+K2SO4(aq)
Both reactants are aqueous and one product is solid. Because the reactants are ionic and aqueous, they dissociate and are therefore soluble.

7. Theory of precipitation
Solution process and solubility product:
Solubility depend upon the breaking up of solute solute interaction by solute solvent interactions. When solvent overcome the solute forces of crystal then crystal get soluble.
Like dissolve like
During precipitation the opposite condition as mentioned above is desired. Intermolecular forces between molecules are high and solute forces replaces the solute solvent forces.
Solubility product for the reaction
AB A B-
K sp = [A+ ] + [B-]
Substance will precipitate out when the product of ionic concentration exceeds the Ksp value i.e. in above equation of Ksp, the BA will precipitate out when the product of out of [A+ ] [B-] exceed Ksp

8. Precipitation titration
Titrations with precipitating agents are useful for determining certain analyte. Example: Cl– can be determined when titrated with AgNO3
Conditions for precipitation titration :
Precipitate must be practically insoluble
Precipitation reaction must be rapid
Precipitation reaction must be quantitative
No interference by adsorption effect(co-precipitation)
Able to detect equivalent point during titration

9. Factors affecting solubility
Common ion effect
Effect of pH
Effect of temperature
Effect of solvent

10. Calculations of solubility product
Steps for calculations are given below
Write equation
Express solubility product of electrolyte
Calculate molar solubility
Calculate conc of each ion
Substitute values and calculate Ksp
Example: Calculate Ksp of MgCO3 if 1lit of its saturated solution contain 0.533gms of MgCO3 at 20oC
Write equation:
2. Express solubility product of electrolyte
3. Calculate molar solubility: wt of MgCO3 is Hence molar solubility is /84.32 = (6.32 X 10-3 mole/litre)

11. Calculations of solubility product
4. Calculate conc of each ion:
Each mole of magnesium carbonate dissociate and form 1 gm ion of magnesium and 1 gm ion of carbonate. Hence both ion have same concentration equal to molar solubility
[Mg++ ] = [CO3- - ] = 6.32 x10-3
5. Substitute values and calculate Ksp
Ksp = 6.32 x10-3 X x10-3
Ksp + 4 X10-6

12. Detection of end point in precipitation titration
1. Formation of coloured precipitate
2. Formation of soluble coloured compound
3. Use of adsorption indicator or Fajan’s Method
4. Turbidity method (Gay Lussac method)
Formation of coloured precipitate
Chlorides are present in all types of water resources at a varying concentration depending on the geo-chemical conditions in the form of CaCl2, MgCl2 and NaCl.
Chlorides are introduced into the water resources from the discharge of effluents from chemical industries, sewage disposal and seawater intrusion in coastal region.
The concentration of chloride ions more than 250 ppm is not desirable for drinking purpose. The total chloride ions can be determined by argentometric method. (Mohr’s Method)

13. Detection of end point in precipitation titration
In this method, first the analyte react with the titrant after the analyte is reacted completely the next drop if titrant react with indicator and formed small quantity of colored precipitate which indicate end point of titration (Mohr’s method)
Example:
Assay of NaCl with silver nitrate with dilute potassium chromate solution as indicator.
Ksp(AgCl) = 1.2 x = [Ag+] [Cl-]
Ksp(Ag2CrO4) =1.7 x = [Ag+2] [CrO4-2]
We expect that the salt with smaller Ksp should precipitate first, this is true if both salt dissociate to yield same number of ions. But in this case the chloride ions are in excess than that of chromate ion and concentration of chromate ion very dilute i.e M, hence the chloride precipitate first and then chromate will precipitate as colured compound

14. Detection of end point in precipitation titration
Reactions in the Mohr’s Method are given below
In this method Cl‒ ion solution is directly titrated against AgNO3 using potassium chromate (K2CrO4 ) as the indicator.
AgNO Cl‒ AgCl ↓ NO3‒
(in water) (White precipitate)
At the end point, when all the chloride ions are removed. The yellow colour of chromate changes into reddish brown due to the following reaction.
2AgNO3 + K2CrO Ag2CrO4 ↓ KNO3
(yellow) (Reddish brown)

15. Detection of end point in precipitation titration
Conditions for Mohr’s method
Very dilute solution of pot chromate should be used
The titration should be carried out neutral or slight alkaline condition i.e. pH 6.5 to 9. In acidic condition, hydrogen chromate will be formed and in highly alkaline condition silver hydroxide(Ksp =2.3 X 10-8)
Solubility product of silver chromate increase with rise in temperature.
Ammonium salt, iodide salts and thocyanate salt can not be done by Mohr’s method because
In the presence of ammonium salt, the ammonia do have effect of solubility of silver salt due to increase in pH
AgI and AgSCN adsorb chromate strongly hence false , indistinct end point results

16. Detection of end point in precipitation titration
Assay of sodium chloride by Mohr’s Method:
Preparation and standardization of 0.1 N Silver nitrate solution
Preparation of potassium chromate indicator solution
Application : Determination of NaCl, KCl, Sodium chloride injection

17. Detection of end point in precipitation titration
Formation of soluble Precipitate (Determination of chloride by Volhard Method)
This is an indirect method for chloride determination where an excess amount of standard Ag+ is added to the chloride solution containing Fe3+ as an indicator. The excess Ag+ is then titrated with standard SCN- solution until a reddish brown color is obtained which results from the reaction:
NaCl AgNO3  AgCl NaNO3 + excess Ag+
Excess Ag SCN-  AgSCN ↓
Fe3+ (Yellow) + SCN- = Fe(SCN)2+ (Reddish brown)
The boiling of solution for 10mins s essential to coagulate the precipitate of silver chloride. Nitrobenzene is added in this method which prevent the interactions between silver chloride and ammonium thiocyanate by forming coat over the silver chloride ppt. for

18. Volhard’s method
Assay of Ammonium chloride and sodium chloride by Volhard Method
Preparation and standardization of 0.1 N Silver nitrate solution
Preparation and standardization of 0.1 N ammonium thiocyanate solution

19. Differentiate between Mohr’s method and Volhard method
Direct titration of halide with silver nitrate
Indirect titration or back titration
Indicator – Pot Chromate
Indicator Ferric ammonium sulphate
End point- red precipitate of silver chromate
End point- red soluble complex of ferric thiocyanate
Condition for titration : Neutral to alkaline (pH 6.5 – 9.0)
Condition for titration : Acidic solution
Titration of Iodide and cyanate is not possible
Can be used for determination of chloride, bromide and iodide
As solubility of silver chromate increases with rising temp, titration are carried out at RT
As the color of ferric thiocynate complex fades above 25oC, the titration are carried below 20oC

20. Use of adsorption indicator or K Fajan’s method
In this method indicator adsorb on the surface of precipitate at equivalent point with color change
Examples: Acidic dye Fluorescin, eosin or basic dyes rhodamine.
When Nacl solution is titrated with silver nitrate, the silver chloride precipitate will adsorb chloride ions which are initially in excess. Thus the chloride ion forms the primary adsorbed layer, which in turn will secondary adsorbed layer of oppositely charged Na+ ions

21. Use of adsorption indicator or K Fajan’s method
• The precipitation titration in which silver ions is titrated with halide or thiocyanate ions in presence of adsorption indicator is called fajan’s method. The indicator, which is a dye, exists in solution as the ionized form, usually an anion.
• The method is generally used for the quantitative analysis of halide ions or thiocyanate ions .
• In the titration of Cl– ions with AgNO3 in presence of adsorption indicator here AgNO3 is kept in burette and the Cl– ion solution with indicator is taken in titration flask
a) Before the equivalence point :before the eqv.point ,colloidal particles of AgCl are negatively charged due to the adsorption of Cl– from the solution . The adsorbed Cl– from the primary layer which attract the positively charged Na+ ions from the solution to form a more loosely held secondary layer as shown in figure on previous slide
b) After the eqv. point ,excess of silver ions Ag+ displace the Cl– ions and form the primary layer which attract the negatively charged nitrate ions NO3 – .
c) At the end point ,anion of indicator in (weak organic acid or base )replace the negatively charged ion NO3– from the second layer and give the intense color . This intense color gives the end point of titration

22. Use of adsorption indicator or K Fajan’s method
Structures of
adsorption indicators

23. Use of adsorption indicator or K Fajan’s method
Assay of sodium chloride by K Fajan’s method

24. Turbidity Method : Gay Lussac Method
This is the turbidity procedure for determination of Ag+ with Cl- or vice versa.
It was introduced by Gay Lussac in 1832.
In this method standard solution of sodium chloride is treated with solution of silver nitrate or vice versa.
This method is based of formation of turbidity due to precipitation. After eq. point, the precipitation reaction ceases and addition of extra drop will not result in turbidity.
Example.
Determination of silver nitrate with standard sodium chloride solution, is carried out in the presence of free nitric acid and small qty of barium nitrate (it help in coagulation) .
Procedure: Weight 0.4 gm of silver nitrate in well stopper the bottle, add 100ml of water, a few drops of conc. Nitric acid and small crystal of barium nitrate. Titrate it with standard 0.1M sodium chloride solution, by adding 20ml at once. Seal the bottle with stopper and shake vigorously until the precipitate of silver chloride has coagulated and settled, leaving a clear solution. Still the silver ions are in excess. Continue to add sodium chloride solution, 01ml at one time and shake after each addition till no turbidity is produced. Note the volume of sodium chloride consumed. It is pilot reading. Repeat the titration by adding 01ml less than the pilot reading and continue adding 0.02ml after that. Note the end pilot where no turbidity is produced.
Nephelo-turbidity method can also be used.

25. Thank you : Any questions??????