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):

Amphoteric salts hydrolysis:

Example of amphoteric salts  NaHS, NaHCO₃, Na₂HPO₄, NaH₂PO₄

(A) HCO3- act as conjugate acid as well as conjugate base:

Both reaction will support each other extent of hydrolysis and extent of dissociation is same.

(B) Here H₂PO₄^- and HPO₄^-2 are amphoteric anions. The pH of amphoteric salts anions is independent of concentration of salts.

Here HPO₄^-2 is conjugate base of H₂PO₄^- and H₃PO₄ is conjugate acid of H₂PO₄^-Similarly PO₄^-3 is conjugate base of HPO₄^-2 and H₂PO₄^-1 is conjugate  acid of HPO₄^-2

When these salts are dissolved in water [H₃O+] concentration can be determined as;

ILLUSTRATIVE EXAMPLE: Calculate pH of solution of

(1) 100 ml 0.1M H₃PO₄ + 100 ml 0.1M NaOH.

(2) 100 ml 0.1M H₃PO₄ + 200 ml 0.1M NaOH.

(3) 100 ml 0.1M H₃PO₄ + 300 ml 0.1M NaOH.

(4) 100 ml 0.1M H₃PO₄ + 400 ml 0.1M NaOH.

                                                     

[3] CATIONIC AS WELL AS ANIONIC HYDROLYSIS:

Take a salt (CH3COONH4) of the weak acid (CH3COOH) and the weak base (NH4OH) . and dissolve in water, therefore, the salt completely dissociates as given below.

The ions get hydrolysed according to the reaction.

Such salts undergoes hydrolysis because ,the aqueous solution contains unionised acid as well as  base molecules .

The nature of aqueous solution of such salt depends on the equilibrium constant for cationic or anionic hydrolysis.

Multiplying and dividing by H⁺ & OH^– and rearranging,

There is an important issue that needs clarification before we move on further. In this case,

 we can see that both the ions (i.e., cation and anion) get hydrolyzed to produce a weak acid and a weak base (hence, we can’t predict whether the solution is acidic, basic or neutral). We have considered the degree of hydrolysis of both the ions to be the same. Now we present an explanation as to why this is incorrect and then state reasons for the validity of this assumption

n.

 Actually the hydrolysis reaction given earlier, 

Now, we calculate the pH of the solution as:

If the reaction for hydrolysis is in equilibrium then all the reversible processes occurring in water must be in equilibrium .

The H+ or OH- ions may be calculated from the dissociation constant of acid or base , here calculation of H+ from acid is given as below .

We know that at 25° Pkw of water is 14 .

Hence 

             pH = 7+ 1/2[Pka~pkb]

If Kh1<Kh2 then ka>kb. and  pKa <pKb

as results Solution become acidic 

If Kh1>Kh2 then ka<kB and  pKa>pKb

as results  solution become  basic 

ILLUSTRATIVE EXAMPLE (1): calculate the pH of 0.2 M NH4CN Solution. ( Given Ka HCN is 3x10-10 and kb NH4OH is 2.0x10-5)

(Ans-pH 9.5 )

ILLUSTRATIVE EXAMPLE (2):

Calculate the DOD and pH of 0.2M NaCN Solution (Given Ka of HCN is 2.0x10-10)

(Ans- DOD = √2×10-10 and pH =11.5)

ILLUSTRATIVE EXAMPLE (3):

Calculate the DOD (h) and pH of 0.2 M C6CH5NH3Cl Solution (Given Ka C6CH5NH3Cl is =4.0×10-8)

(Ans- DOD =√20×10-4  and pH is 6.6)

ILLUSTRATIVE EXAMPLE (4):

ILLUSTRATIVE EXAMPLE (5):

[2] CATIONIC SALT HYDROLYSIS:

(2) CATIONIC HYDROLYSIS OR ACIDIC SALTS HYDROLYSIS:

          (Salt of a Weak Base and a Strong Acid)

      Let the acid be HCl and the base be NH₄OH. Therefore the salt would be NH₄Cl.

      NH₄Cl completely dissociates into NH₄⁺and Cl^– ions.

      HCl being a strong acid dissociates completely to give H⁺ ions and Cl^– ions.

In this hydrolysis, NH₄OH and H⁺ are being produced. This implies that the solution is acidic

To calculate pH,

Multiplying and dividing by OH^– and rearranging,

Now, substituting the concentrations,

ILLUSTRATIVE EXAMPLE (1):

ILLUSTRATIVE EXAMPLE (2):

ILLUSTRATIVE EXAMPLE (3):

ILLUSTRATIVE EXAMPLE (4):

ILLUSTRATIVE EXAMPLE (5):

     

[1] ANIONIC SALT HYDROLYSIS:

When a salt is dissolved in a solvent, it first dissociates into its constituent ions. This process is called dissolution. Now, if these ions chemically react with water, the process is called hydrolysis.

Salt hydrolysis is may be consider as the reverse of process of neutralization

We can also say that “combination of any of the ion furnished by the salt with water molecules is called hydrolysis “

Cationic hydrolysis will make the solution acidic but anionic hydrolysis will make the solution basic

(1) NEUTRAL SALTS: (Salts of strong acids and strong bases)

  A salts formed by complete neutralization of strong acid and strong base are called neutral salt such salts will not undergoes hydrolysis so aqueous solution of such salts be must neutral.

The salts that undergo hydrolysis after dissolution are
(2) ALKALINE SALTS:(Salts of weak acids and strong bases)
(3) ACIDIC SALTS:(Salts of weak bases and strong acids)
(4)  Salts of weak acids and weak bases

(1) ANIONIC HYDROLYSIS OR ALKALINE SALTS HYDROLYSIS:

                  (Salt of a Weak Acid and Strong Base) 

      Let us take a certain amount of weak acid (CH₃COOH) and add to it the same amount of a strong base (NaOH). They will react to produce CH₃COONa. 

      CH₃COONa being a strong electrolyte, completely dissociates into its constituent ions.

      Now, the ions produced would react with H₂O. This process is called hydrolysis

We know that NaOH is a strong base and therefore it would be completely dissociated to give Na⁺ and OH^– ions.

      Canceling Na⁺ on both the sides,

We can note here that ions coming from strong bases do not get hydrolysed. We should note here that the solution will be basic. This is because the amount of CH₃COOH produced and OH^– produced are equal. But CH₃COOH will not completely dissociate to give H⁺ ions. Therefore [OH^–] ions will be greater than [H⁺] ions.

      Since the reaction is at equilibrium,

This equilibrium constant K_c is given a new symbol, K_h

If we multiply and divide the above equation by [H⁺] of the solution, then

CASE (1): If a is very much less than 1, then 1-a= 1 this approximation valid when C/Kh is greater than 100”

CASE (2): If C/Kh is lower than 100 than calculate h by formation of quadratic equation.

ILLUSTRATIVE EXAMPLE (1): A 0.0258 M solution of the sodium salt, NaH of the weak monoprotic acid, HA has a pH of 9.65. Calculate Ka of the acid AH.

ILLUSTRATIVE EXAMPLE (2): What is the pH of 0.10 M CH₃COONa solution. Hydrolysis constant of sodium acetate is 5.6 × 10^-10   ?

SOLUTION: Hydrolysis of the salt may be represented as

ILLUSTRATIVE EXAMPLE (3): Calculate pH of 1.0 x 10^-3 M Sodium phenolate (Na⁺O^-C₆H₅ ) Ka for C₆H₅OH is 1.0 x10^{-10 .}

SOLUTION:  (Ans-  pH=10.43)

(2) CATIONIC HYDROLYSIS    OR  ACIDIC SALTS HYDROLYSIS:

         (Salt of a Weak Base and a Strong Acid)    .......