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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):

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