TITRATION OF TRIPROTIC ACID WITH STRONG BASE:

TRIPROTIC Vs NaOH: [H₃PO₄ Vs NaOH]
ILLUSTRATIVE EXAMPLE: Phosphoric acid (H₃PO₄), is a Triprotic acid with Ka₁= 10^-3 , Ka₂=10^-8 and Ka₃= 10^-12 .Consider the titration of 0.10M 100 ml H₃PO4 with 0.1M NaOH Solution and answers the following questions.

(A) Write out the reactions associated with Ka₁, Ka₂ and Ka₃.

(B) Calculate the pH after the following titration points:

(1) 100 ml of 0.1M H₃PO₄ + 0.0 ml of 0.1 ml NaOH
(2) 100 ml of 0.1M H₃PO₄ + 50 ml of 0.1 ml NaOH
(3) 100 ml of 0.1M H₃PO₄ + 100 ml of 0.1 ml NaOH
(4) 100 ml of 0.1M H₃PO₄ + 150 ml of 0.1 ml NaOH
(5) 100 ml of 0.1M H₃PO₄ + 200 ml of 0.1 ml NaOH
(6) 100 ml of 0.1M H₃PO₄ + 250 ml of 0.1 ml NaOH
(7) 100 ml of 0.1M H₃PO₄ + 300 ml of 0.1 ml NaOH 

(C) Sketch the titration curve for this titration.

(D) What weak acid and what conjugate base make the best phosphate Buffer at pH ~7.0?

(A) Write out the reactions associated with Ka₁, Ka₂ and Ka₃.

 SOLUTION:

Step-1    

Step-2  

 Step-3    

(B)  Calculate the pH after the following titration points;

(1) 100 ml of 0.1M H₃PO₄ + 0.0 ml of 0.1 ml NaOH 

SOLUTION:

By approximation we know that if

(2) 100 ml of 0.1M H₃PO₄ + 50 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100= 10 and base 0.1x50 = 5.0

Here H₃PO₄/NaH₂PO₄ remain same in solution and which are act acidic buffer

(3) 100 ml of 0.1M H₃PO₄ + 100 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x100 = 10

Here only NaH₂PO₄ remain same in solution and which are under go amphoteric hydrolysis. PH of amphoteric salts is independent of concentration of salt.

(4) 100 ml of 0.1M H₃PO₄ + 150 ml of 0.1 ml NaOH

SOLUTION:

Given millimoles (M x V) of acid= 0.1x100 = 10 and base 0.1x150 = 15

Here only NaOH (5 millimoles) and NaH₂PO₄ both are remain present in solution hence which are further go titration.

Here NaH₂PO₄ and Na₂HPO₄ remain same in solution and which are act acidic buffer

(5) 100 ml of 0.1M H₃PO₄ + 200 ml of 0.1 ml NaOH

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x 200 = 20

Here only NaOH (10 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here only Na₂HPO₄ remain same in solution and which are under go amphoteric hydrolysis. pH of amphoteric salts is independent of concentration of salt.

(6) 100 ml of 0.1M H₃PO₄ + 250 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x 250 = 25

Here NaOH (15 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here Na₂HPO₄ (10 millimoles) and (5.0millimoles) both remain same in solution and which are further go titration.

Here Na₂HPO₄ and Na₃PO₄ remain same in solution and which are act acidic buffer

(7) 100 ml of 0.1M H₃PO₄ + 300 ml of 0.1 ml NaOH 

SOLUTION:
Given millimoles (M x V) of acid= 0.1x100 = 10 and base 0.1x 300 = 30

Here NaOH (20 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here NaOH (10 millimoles) and Na₂HPO₄ (10 millimoles) both remain same in solution and which are further go titration.

Finally 10 millimoles (Molarity=10/400) N₃PO₄ is formed which undergo polyvalent salt hydrolysis 

HYDROLYSIS OF ANION (PO₄^-3):

Step wise illustration of hydrolysis of poly basic acids and polyacidic base is given as -

Na+ ion do not under goes hydrolysis while PO₄^-3 undergoes step wise hydrolysis

Experimentally we know that   Ka₁>>Ka₂>>Ka_3, hence x >>y>>Z   so y and z can be neglected with respect to x   it mean total OH^-  ions count from only first step OH- ions coming from 2^nd and 3^rd  hydrolysis is ignored 

Special Case (1):  

Special Case (2) 

This is the quadratic equation solve by following formulae

So approximation not valid hence follows 2^nd case

(C) Sketch the titration curve for this titration:

SOLUTION:    

S.N.	Given condition	comments	PH
1	100 ml of 0.1M H3PO4 + 0.0 ml of 0.1 ml NaOH	H3PO4 Polyprotic acid	02.02
2	100 ml of 0.1M H3PO4 + 50 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka1 (1st half E-Point)	03.00
3	100 ml of 0.1M H3PO4 + 100 ml of 0.1 ml NaOH	Amphoteric salt Hydrolysis PH=1/2(Pka1+Pka2)	05.50
4	100 ml of 0.1M H3PO4 + 150 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka2 (2st half E-Point)	08.00
5	100 ml of 0.1M H3PO4 + 200 ml of 0.1 ml NaOH	Amphoteric salt Hydrolysis PH=1/2(Pka2+Pka3)	10.00
6	100 ml of 0.1M H3PO4 + 250 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka3 (3st half E-Point)	12.00
7	100 ml of 0.1M H3PO4 + 300 ml of 0.1 ml NaOH	Poly anionic salt Hydrolysis 3rd Equivalent point	12.07

 Graphical representation of titration:

FACTER'S AFFECTING ON SOLUBILITY

TOPIC COVER:
(1) Effect of temperature on Solubility
(2) Effect of common ions on Solubility
(3) Effect of simultaneous Solubility
(4) Effect of solvent on Solubility
(5) Effect of pH on Solubility
(i) Effect of pH on Solubility of metal hydroxide
(ii) Effect of pH on Solubility of salt of weak acid
(iii) Effect of pH on Solubility of salt of strong acid
(6) Effect of buffer solution on Solubility
(7) Effect of complex  formation  on Solubility

(1) Effect of temperature on Solubility:
In general most cases Solubility increases on temperature . however we must follow two cases .
(1) In endothermic reaction Solubility Increases on increasing temperature.
(2) In exothermic reaction Solubility decrease on increasing temperature .

(2) Effect of common ions on Solubility:
(1) Solubility of a salt decrease in the presence of common ion.
(2) Higher the concentration of common ion smaller Solubility.
ILLUSTRATIVE EXAMPLE (1): Calculate the Solubility of AgCl in .. ( Kal AgCl is 10-10)
(1) In pure water
(2) In 0.1M NaCl
(3) In 0.01 M NaCl
(4) In 10-5 M NaCl
(5) In 0.05 M HgCl2
(6) In 0.01 M AgCl
(7) In 0.1M KNO3
SOLUTION :

ILLUSTRATIVE EXAMPLE (2): Arrange the following in increasing order of their Solubility.
(a) AgCl           Ksp = 2×10-10
(b) BaSO4       Ksp = 4×10-8
(C) CaF2.         Ksp = 1.08×10-10
(D) Hg2I2.       Ksp = 9×10-17
SOLUTION:

(3) Effect of simultaneous Solubility:
ILLUSTRATIVE EXAMPLE (1):
Calculate simultaneous Solubility of AgSCN and AgBr in water . ( Ksp AgSCN=10-12 Ksp AgBr =5×10-13)

ILLUSTRATIVE EXAMPLE (2):
CaCO3 and BaSO4 have Solubility products value 1.0×10-8 and 5.0× 10-9 respectively, if water is shaken up with both solids till equilibrium is reached calculate the concentration of CO3-2 ion is ?

ILLUSTRATIVE EXAMPLE (3):
The Ksp value of CaCO3 and CaC2O4 in water are 4.7 × 10-9 and 1.3 ×10-9, respectively , at  25°c .If mixture of two is washed with water , what is Ca2+ ion concentration in water ?
(Ans- 7.707×10-5 )
ILLUSTRATIVE EXAMPLE (4):

(4) Effect of solvent on Solubility:
Solubility of solute in solvent purely depends on nature of both solute and solvent , a polar solute dissolved in polar solvent and non polar solute dissolved in non polar solvent . A polar solute has low solubility or insoluble in a non polar solvent . For this reason if you want to decrease the Solubility of an inorganic salt (polar salt ) in water you mixed the water with an organic solvent (non polar )

PRIDICTING OF PRECIPITATION:
We know that:
IP= Ionic product  and Ksp= Solubility product
CASE(1): If IP < Ksp  then Solution is unsaturated
CASE(1):If IP > Ksp  then Solution is Oversaturated or ppt formation occurs.
CASE(1): If IP =Ksp  then Solution is saturated is No more solute dissolved.

ILLUSTRATIVE EXAMPLE (1):
A 200 ml  of 1.3×10-3 M AgNO3 is mixed with 100 ml of 4.5×10-5 M Na2S Solution will precipitatation occurs ?
(Ksp = 1.6×10-19) .

ILLUSTRATIVE EXAMPLE (2):
50ml of 6.9×10-3M CaCl2 mixed with 30 ml of  0.04 M NaF2. Will precipitatation of CaF2 occurs ?
( Ksp for CaF2= 4.0×10-11)
ILLUSTRATIVE EXAMPLE (3):
How much solid Pb(NO3)2 must be added to 1.0 L of 0.0010 M NaSO4 Solution for precipitatation of PbSO4 (Ksp=1.6×10-8) to form.
(assume no change in volume when the solid is added).

ILLUSTRATIVE EXAMPLE (4):
PbCl2 has Ksp=1.6×10-5 , If equal volume of 0.030 M PB(NO3)3 and 0.030M KCl are mixed , will precipitatation occurs ?
ILLUSTRATIVE EXAMPLE (5):

ILLUSTRATIVE EXAMPLE (6):

(5) Effect of pH on Solubility:
Many weak soluble ionic compound have Solubility which depends upon  the pH of the Solution for example metal hydroxide and  salt of weak acids.

(i) Effect of pH on Solubility of metal hydroxide
ILLUSTRATIVE EXAMPLE (1): Zince hydroxide ( Zn(OH)2 ) has Ksp 4.5×10-17 in pure water calculate it's molar Solubility and pH of resulting solution .
(Ans- S=2.2×10-6 M and pH = 8.6434 )
ILLUSTRATIVE EXAMPLE (2): At what pH the  Zince hydroxide will start precipitate (pHs) and at what pH precipitation is completed (pHc) from the Solution containing 0.1M Zn+2 ? ( Given Ksp of Zn(OH)2 is 4.5×10-17)
(Ans- pHs = 6.33 and pHc = 8.33 )

ILLUSTRATIVE EXAMPLE (3):  A solution containing 0.1 M Ca+2 and 0.02 M Mg+2 , is it possible to seperate one of these ions by precipitatating it as  hydroxide while keeping the other in Solution ?.
(Given Ksp of Ca (OH)2 is 5.5×10-6 and Ksp of Mg(OH)2 is 5.0×10-12)

(ii) Effect of pH on Solubility of salt of weak acid:
For salt of weak acids (eg Sulphides , Carbonate , Oxalates , and Phosphates ) the smaller the value of Ksp the lower the pH at which the salt precipitate (pH ---Ksp) exactly same as metal hydroxide .
It is noted that the salt having smaller Ksp , precipitate in more acidic medium and  other hand  the salts having Higher Ksp , precipitate in less acidic medium.
For example the precipitatation  of Ca +2 at low pH , CO3-2 will be turned to HCO3-1 Or  may be to H2CO3 , below at pH=< 8( see the Diagram)  while at pH >=13  all the carbonic acid species are present as CO3-2 , therefore CaCO3 will precipitate in basic medium and will dissolve in acidic medium .

ILLUSTRATIVE EXAMPLE (1):
Which one will precipitate in more acidic medium CaCO3 (Ksp=4.8×10-9) or MgSO4 (Ksp=1.0×10-5) ?

ILLUSTRATIVE EXAMPLE (2):
A solution containing 0.1 M Ti+ and 0.05 M Cd+2 . Is it possible to separate these two ions by precipitatating one of as Sulphides ?
( Ksp(CdS)=2.0×10-28 , Ksp(TiS)=2.0×10-22)

(iii) Effect of pH on Solubility of salt of strong acid :
Note that the pH has no effect on Solubility of  strong acids salts  eg Cl- Br- and SO4-2 etc because the concentration of of these conjugated base in the same either in acidic or basic medium . However metal ions can be separated by these anions  according to their Ksp value as the hydroxide or salts of weak acids.

ILLUSTRATIVE EXAMPLE (1):
Ca(OH)2 has Ksp = 7.9 ×10-6 , what is the pH of Solution made by equilibrating solid Ca(OH)2 with water ?
ILLUSTRATIVE EXAMPLE (2):
Cu(OH)2 has Ksp =1.6×10-19 calculate the
(1) What is the pH of a saturated solution of Cu(OH)2 ?
(2) What  is the maximum Cu+2 concentration possible in a  Neutral Solution ? (pH=7)
(3) What is maximum pH of Solution in which concentration of  Cu+2 is 0.50 M.

(6) Effect of buffer solution on Solubility:
Whenever a salt dissolved in a buffer solution it's pH remain same .
ILLUSTRATIVE EXAMPLE (1):
The Solubility of Pb(OH)2 in water =6.70×10-6. calculate it's Solubility in a buffer solution of pH=8 .   [ IIT-1999]
ILLUSTRATIVE EXAMPLE (2):
Calculate Solubility of AgCN in a buffer of pH=3 . ( Given Ksp=1.2×10-15 and Ka HCN is 4.8×10-10 ).
ILLUSTRATIVE EXAMPLE (3):
Calculate Solubility of AgCN in a buffer of pH=3 . ( Given Ksp=1.2×10-16 and Ka HCN is 4.8×10-10 ).
ILLUSTRATIVE EXAMPLE (4):
Calculate the molarity Solubility of Cu(OH)2 [Ksp=2.2×10-20] in
(1) Distilled water
(2) pH =13.0 NaOH (aq)
(3) pH= 4.0 buffer
ILLUSTRATIVE EXAMPLE (5):
ILLUSTRATIVE EXAMPLE (4):
Calculate Solubility of AgCN in a buffer of pH=3 . ( Given Ksp=8×10-10 and Ka HCN is 9×10-10 ).

(7) Effect of complex  formation  on Solubility:
The Solubility of many salt can be increased by addition of a species that can form complex ion with one of the ions (usually the cation ) formed when poorly soluble salt dissolved.
ILLUSTRATIVE EXAMPLE (1):
Find Solubility (s) of AgCl in 'C'M NH3 (aq) (given , Ksp of AgCl and Kf [Ag(NH3)2]+.
SOLUTION:
[Kf =complex formation constant=K solubility constant=1/K(insolubility)=1/K(dissociation)]

ILLUSTRATIVE EXAMPLE (2):
Calculate Solubility of  AgCl(s) in 2.7 M NH3(aq) solution . ( Ksp AgCl=10-10 and Kf[Ag(NH3)2]=1.6×10+7).
SOLUTION:
ILLUSTRATIVE EXAMPLE (3):
Calculate the Solubility of AgCN in 0.01M KCN Solution assuming
(1) No Complex formation
(2) Complex formation
(KspAgCN=8×10-8 and Kf[Ag(CAN)2]=4×10+7)

ILLUSTRATIVE EXAMPLE (4):
Find the Solubility of MX(s) ( Ksp=1.6×10-10) in 0.05M NaCN(aq).
(1/K dissociation [M(CN2)]=2.5×10-8 )

ILLUSTRATIVE EXAMPLE (5):
What is Molar solubility of AgBr in 0.1M NaS2O3 ? ( Given Ksp AgBr = 5×10-13 and Kf [Ag(S2O3)2]-3 =5×10+13)
ILLUSTRATIVE EXAMPLE (7):
Calculate the Molar Solubility of AgBr in 1.0 M NH3 at 25° ( Ksp AgBr =5×10-13 and Kf [Ag(NH3)2]+ =1.7×10+7)
ILLUSTRATIVE EXAMPLE (6):

ILLUSTRATIVE EXAMPLE (6):

BUFFER SOLUTION

Topics cover :
(1) DEFINITION
(2)TYPE OF BUFFER SOLUTION
(3) pH OF ACIDIC BUFFER
(4) pH OF BASIC BUFFER
(5) DILUTION OF BUFFER
(6) BUFFER CAPACITY OR BUFFER  INDEX
(7) IDEAL BUFFER SOLUTION

(1) DEFINITION:
The aqueous electrolyte solution which resist the any change in pH even after addition of small amount of strong acid or strong base  called  buffer Solution.

(2)TYPE OF BUFFER SOLUTION:
(1) Simple Buffer Solution or Neutral buffer
(2) Mixed Buffer Solution

(3) pH OF ACIDIC BUFFER:

(4) pH OF BASIC BUFFER:

(5) DILUTION OF BUFFER:

(6) BUFFER CAPACITY OR BUFFER  INDEX:

(7) IDEAL BUFFER SOLUTION:

ILLUSTRATIVE EXAMPLE (1):
Pkb(NH4OH) is 5 and a buffer solution contains 0.1M NH4OH and 0.1M (NH4)SO4 calculate pH of this buffer solution ?
(Ans-5.3 )

ILLUSTRATIVE EXAMPLE (2):
Pls of HX is 4.7 , (1) find the pH of Solution having 0.5 M HX and KX 0.25M ? (2) pH of this solution if it is diluted 10000 times ?
(Ans- (1) 4.4 (2) )

ILLUSTRATIVE EXAMPLE (3):
Calculate pH of acidic buffer mixture containing 1.0 M HA (Ka=1.5x10-1) and 0.1M NaH .
(Ans- 0.824)

ILLUSTRATIVE EXAMPLE (4):
Calculate the mass of NH3 and NH4Cl required to prepared a buffer solution of pH  =9.0 when total  concentration of buffering reagents is 0.6 mole /litre .( Pkb for NH3 is 4.7, log2 is 0.3010)
(Ans- a=0.2 mole ,b=0.4 mole)

ILLUSTRATIVE EXAMPLE (5):
One liter buffer solution is prepared by mixing of 1.0 mole HCOOH (formic acid) and 1 mole HCOONa . (Given Pka HCOOH =4.0) then calculate
(i) pH of buffer
(ii) pH of Solution if 1/3 mole of HCl is added
(iii) pH of Solution if 1/3 mole of NaCl is added
(iv) pH of Solution if it diluted to 10 liter
(v) pH of Solution if it diluted to 1000 liter
(Ans- i-4.0 , ii- 3.7 , iii-4.3010 , vi- 4.0 v- 4.07)

ILLUSTRATIVE EXAMPLE (6):
Calculate the pH of an aqueous solution originally containing 0.4 M acetic acid and 0.2 M sodium acetate (ka CH3COOH= 1.8x10-5 ).
(Ans- 4.4)

ILLUSTRATIVE EXAMPLE (7):
Calculate pH of Solution Originally having 0.2 M (NH4)SO4 and 0.4 M NH4OH (given Kb NH4OH =1.8x10-5)
(Ans- 9.26)
ILLUSTRATIVE EXAMPLE (8):
In 100 ml of 0.4M C6H6COOH Solution,0.1M C6H6COONa is added , calculate pH of resulting solution (given Ka C6H6COOH is 4x10-5)
(Ans-4.1)

ILLUSTRATIVE EXAMPLE (9):
A solution contains 0.2 mole acetic acid and 0.10 mole of sodium acetate ,made up to 10 liter volume , calculate the pH of Solution ( given Ka CH3COOH is 1.8x10-5)
(Ans-4.44)
ILLUSTRATIVE EXAMPLE (10):
What mass of , in gram ,of NaNO2 must added to 700 ml of 0.165 M HNO2 to produce a Solution with pH of 3.50 ? ( Ka HNO2 is 6.0x10-4)
(Ans- 15.1gm)

ILLUSTRATIVE EXAMPLE (11):
In 500 ml of a buffer solution containing 0.8 M CH3COOH and 0.6 M CH3COONa , 0.2 M HCl is added. Calculate the pH of Solution before and after adding HCl. (Ka CH3COOH is 1.8x10-5) .
(Ans- i - 4.62  ii- 3.96)

ILLUSTRATIVE EXAMPLE (12):
A mixture of 0.2 mole RNH2 and 0.4 mole RNH3Cl is  mixed .the volume of Solution prepared is 10 liter  (given Kb RNH2 is 10-5) calculate.
(i) pH of resulting solution
(ii) pH of Solution if diluted to 1000 litres
(iii) pH of Solution if 200 ml buffer is mixed with 2 milimoles of H+
(iv) pH of Solution if 200 ml buffer is mixed with 2 milimoles of OH-
(Ans- i-  5.30 , ii- 5.31 , ii- 8.3 , iv -9.0)

BRONSTED LOWERY ACID-BASE CONCEPT

According to Bronsted theory the species which donate protons (H+) in any medium is consider as acid and the species which accept proton is consider as base.

Acid and base characters are realised in the presence of each other.
For example

CONJUGATE ACID-BASE PAIRS
(1) conjugate pair is acid-base pair differing in single proton (H⁺)
(2) conjugate acid is written by adding  H⁺ and conjugate is written by removing H⁺ .

(3) Strong acid has weak conjugate base and vice versa. Similarly strong base has weak conjugate and vice versa.

(4) Equilibrium always moves from strong acid to weak acid and strong base to weak base.

(5) Conjugate acid - base pair differ by only one proton. Reaction will always proceed from strong acid to weak acid or from strong base to weak base.

MERITES OF BRONSTED CONCEPT:
(1) the role of solvent clearly defined.
(2) the acidic and basic  character may be observed in non aqueous medium also .
(3) the acidic ,basic or Amphoteric nature of most of the substance may be defined.
(4) the acid having greater tendency to donate protons are stronger acid and base  having greater tendency to accept protons are stronger base .
(5)In conjugate pair ,if one is strong then other must be weak .
The weak acid or base are normally determined by comparing the the stability of different acid or base
DEMERITES OF BRONSTED CONCEPT:
(1) Proton is a nuclear particle hence reaction should not explained in term of proton.
(2) the neutralized process becomes multiples step process.
(3) Most of the Amphoteric solvent become Amphoteric.

AMPHOTERIC SPECIES (Amphiprotic):
The species which have a tendency to donate proton as well as accept proton (H⁺) such species are known as Amphoteric species.
For example H₂O,NH₃ HS^- ,HPO₃^- ,HC₂OO₄^- , H₂O₄ etc

ILLUSTRATIVE EXAMPLES:
(1)The conjugate base of HCO₃ is –
  (A) H₂CO₃        (B) CO₂             (C) H₂O      (D) CO₃^-   
(2) The conjugate acid of HSO₃^- is -
  (A) SO₃^2-          (B) SO₄^2-                  (C) H₂SO₄    (D) H₂SO₃

(Ans: 1-D 2-D)

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