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)

                                                                 @@@

ARRHENIUS ACID-BASE CONCEPT

CHARACTERS OF ACIDS:
(1) acids convert blue litmus to red and methyl orange  indicator to  red .
(2) Sour in taste.
(3) It liberate hydrogen gas with active metals.
(4) Acids neutralised the effect of base 
(5) acids increases the conduction of water.
CHARACTERS OF BASES:
(1) bases convert the red litmus to blue and methyl orange indicator to yellow.
(2) Phenylphthlene indicator (white) to pink.
(3) Bitter in taste and soapy in touch.
(4) Bases neutralised the effect of acid.
(5) They increases conductance of water.
ACID BASE CONCEPT:
(1) ARRHENIUS ACID-BASE CONCEPT
(2) BRONSTED LOWERY ACID-BASE CONCEPT 
(3) LEWIS ACID-BASE CONCEPT

(1) ARRHENIUS ACID-BASE CONCEPT
the substance which produces H+ in aqueous solution is consider as acid and the substance which give produces OH- in aqueous solution is consider as base
Arrhenius theory depend upon dissociation of water.

Type of Arrhenius acids:

Type of Arrhenius Bases :

         

Feature of Arrhenius theory:

(1) OH¯ ion is present also in hydrated form of H₃O₂¯, H₇O₄¯, H₅O₃¯

      H⁺ ion is present also in hydrated form of H₃O⁺, H₅O₂⁺, H₇O₃+, H₉O₄⁺

(2) Neutralisation reaction can be easily explain by the Arrhenius theory .acids furnishing H⁺ ion  in water to large extent are strong acid.

Strength of acid or base:

(3) If Ka increases then concentration of H⁺ increases hence acidic strength is increases.

Similarly the base furnishing OH^- ions to the large extent are strong base

Kb is dissociation constant for bases, if Kb is increases OH- increases, hence basic strength of base

(4)The term strong is used only for those acids or bases or bases which dissolved almost completely in water.

LIMITATIONS OF ARRHENIUS THEORY

(1) This theory explain nature of a substance only aqueous medium . It cannot be applied for non aqueous solution.
(2) It could not explain formation of hydronium ions like H₃O- , H₅O₂- , and H₇O₃- .
(3) the nature of aqueous solution of AlCl₃, CuSO₄ ,BF₃, B(OH)₃ etc are acidic and aqueous solution of NH₃ ,NaCO₃ ,RNH₂ R2NH,  R₃N , _C2H5N  etc are basic in nature cannot be explain by Arrhenius Concept.
(4) there are many  Amphoteric hydroxide Zn(OH)₂  Al(OH)₃ ,Pb(OH)₂ , which cannot be  explain  by Arrhenius Concept.
(5) Arrhenius explain only when H⁺ is released it cannot explain when H+ is taken.

                                                 

LEWIS ACID-BASE CONCEPT:

According to Lewis acid base theory- acids are electron pair accepter (electron deficient) and bases are electron pair donor and the combination of Lewis acid and Lewis base occur through coordinate bond formation.
For example

Others examples of Lewis acid-base neutralization.

Lewis acids:

(1) Cations are act as Lewis acids example, H⁺ , Ag⁺ , Cu⁺² , Na⁺ , Fe⁺² , Hg⁺²  etc.
(2) Electron deficient compounds are act as Lewis acids for examples BF₃ ,ACl₃ ,BCl₃ ,FeCl₃ etc 
(3) The compounds  which central atom with available vacant d-orbital  can add additional pairs of electrons even though it already has an octet or more of electrons act as Lewis acids for examples SiCl₄,PCl₅, SOF₄, SbCl₃, SbF₅ ,SnCl₄, IF₇, SF₄ etc 
(4) Compounds containing multiple bonds for example CO₂ , SO₂, SO₃ ,CS₂ etc.

Lewis bases:

(1) Anions are act as Lewis base H^- ,CH₃^- ,NH^2- , OH^- , X^- etc. but all the anions are not Lewis base for example (PCl₅)^- 
(2) The Neutral molecules having lone pairs of electrons act as Lewis base for examples NH₃ ,R-NH₂, R-NH-R, R₃-N, H₂O , R-OH ,R-O-R , PH₃, R-PH₂ , R₂PH ,R₃P , H₂S , R-SH , R-S-R etc
(3) Compounds with non polar multiples bond  are also act as Lewis base, example Alkenes(C₂H₄) , Alkynes(C₂H₂ ) Dienes, polyenes, benzene, Polynuclear homoaromatic compounds. (Ligand in coordination compounds)

Chemical Reactions according Lewis acid-base concept:

We can now write equations for many reactions that involve a Lewis acid reacting with a Lewis base. The following are a few examples:

Acid-base behavior according to the Lewis theory has many of the same aspects as does acid-base theory according to the Brønsted–Lowry theory.

(1)There is no acid without a base. An electron pair must be donated to one species (the acid) by another (the base).

(2) An acid (or base) reacts to displace a weaker acid (or base) from a compound.

(3)The interaction of a Lewis acid with a Lewis base is a type of neutralization reaction because the acidic and basic characters of the reactants are removed.

An acid-base reaction takes place readily between BF₃ and NH₃ ,

                                         BF₃  +NH₃ → H₃N:BF₃

However, when the product of this reaction is brought into contact with BCl₃, a reaction takes place:

                                  H₃N:BF₃ +BCl₃ → H₃N:BCl₃  + BF₃

In this reaction, a Lewis acid, BF₃ , has been replaced by essentially a stronger one BCl₃ .  And replacement of Lewis  acid by other Lewis acid follow Second order Nucleophilic substitution (S_N²).

Merits of Lewis concept:
(1) The acid/base nature of substance may also defined without any solvent.
(2) It is highly useful to explain coordination compounds in which central metal atom into behave as Lewis acid and ligands behave as Lewis base.
(3) The acid having greater tendency to accept lone pair bare stronger acid and base have greater tendency to donate lone pair are stronger base.

Demerites of Lewis concept: 
(1) It is extremely generalized Concept , almost all the compounds either become acid or base  by this Concept.

Related Questions:

Why is BCl3 a strongerLewis acid than BF3 ?

Arrange the silicon halides into decreasing order of Lewis acids Character? SiF3, SiCl3, SiBr3, SiI3

Silianol (SiH3OH) is more acidic than methanol (CH3OH) why?

Why thionyl chloride behaves as Lewis acid as well as Lewis base?.

What is Lewis acid character order of Boron halides, arrange in increasing order?

What are the order of extent back bonding, Lewis acid character and nucleophilicity of (BF3, BCl3, BBr3, BI3)boron trihalides?

Arrange the silicon halides into decreasing order of Lewis acids Character? SiF4, SiCl4, SiBr4, SiI4

Levelling effect and relative strength of acids:

When strong acids like HClO₄, HCl, HBr, HI, HNO₃ etc are dissolved in water they are equally ionised (100%) it means they are equally strong in water this is called levelling effect of water.
Similarly for strong bases also , NaOH, KOH  Ca(OH)₂ etc are in water behave the same. hence relative strength of strong acids or base can not be compare in water .
(1) Water does no show levelling effect for weak acids or base because they are ionised upto different extent in water.
(2) All the acids are stronger than H₃O+ ion, consider as strong acids and weak than H₃O+ ion are consider as weak acids.
(3) All the bases stronger than OH^- are consider as strong base and weaker than OH^-, then they are consider  as weak base.

Definition: If more than one acids or bases are showing same acidic or basic strength in same solvent it is called labelling effect and the solvent is called labelling solvent for example:

(1) HClO₄ and HI equally dessociation in aqueous medium (99.99%)  it means they have equal strength.

(2) HCl, HNO₃  and H₂SO₄ are equally strong in water because their strength are “levelled” to solvent species H₃O+. only by putting them into a more acidic solvent do they weak acids. With determinate Pka values which differetiate their strengths. Thus in glacial ethanoic acid (acetic acid ) as solvent, the order of  acidic strength is as: H2SO₄ > HCl > HNO₃ .

(3) Similarly HF and HCl are equally dessociation in NaOH (100 %).
If their strength differ in same solvent ,it is called differential effect and the solvent is called differential solvent.

(4) For example HClO₄ dessociate 99.99 % in acetic acid while HI dissociate 99.8% in acetic acid, it means acidic strength of HClO₄ is more than HI in acetic acid.

(5) Many of inorganic oxoacid are strong acid (with more negative Pka value) inaqueous solution. But, as we have seen, use of solvent with a lower proton affinity than water like acetic acid, makes it possible to differentiate between the strength of these acids and measure Pka value.

H₂CO₃ < H₃PO₄ < H₂SO₄ < HClO₄ (Increasing acidic sterngth) these acidic strength can  also explain if the folmulae of the these oxy acids are written as base or OH formate or like that OC(OH)₂ < OP(OH)₃< O₂S(OH)₂< O₃Cl(OH) then it is clear that the acidic strength incrases as the number of oxygen atoms not involve in O-H bonding increases.

The reason behind levelling and diffential  effect is ability of solvent to donate or accept protons .
For acids, acidic solvents are differential and basic solvent are levelling.

Related Questions:

What are "pyro" oxy acids?

What are "Ortho" or "Meta" oxyacids?

What is Meta Boric Acids?

What are the structural difference between oxides ( P4O6 and P4O10) of phosphorous?

What are common structural features of oxides (P4O6 and P4O5) of phosphorous?

What is "calgon" ? Give structure and its uses?

Structure of Oxy acids of Phosphorous:

What is structure of (HPO3) metaphosphoric acid?

What is metaphophoric (HPO3)?

SOLUBILITY(S) AND SOLUBILITY PRODUCT (Ksp)

Topics covered

(1)SOLUBILITY AND SOLUBILITY PRODUCT

(2) FACTER'S AFFECTING SOLUBILITY

(1) Effect of temperature on Solubility

(2) Effect of common ions on Solubility

(3) Effect of simultaneous on 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 of salt 

(7) Effect of complex formation on Solubility

SOLUBILITY(S):

Solubility of any substance (electrolyte) in solvent represent its amount present in the given amount of solution in order to make the solution saturated at a constant temperature.

MOLAR SOLUBILITY:

Number of mole of a solute dissolved in one litre saturated solution at constant temperature is called molar solubility.

Unit of solubility(s)=mole/litre

If solubility is equal or more than 0.1 then salt is soluble.

If solubility is equal or less than 0.1 then salt is partially soluble or sparingly soluble.

The solubility of substance in water may be classified as.

(1)Molecular Solubility: define for molecular solid, example glucose ,urea  and other covalent soluble solid.

(2)Ionic Solubility: define for electrolyte

SOLUBILITY PRODUCT (Ksp):

It is the product of the ionic concentration of the ions of binary solid electrolyte in saturated state at constant temperature.

(1) Let solubility of a compound A_x B_y be s moles L^–1 it means that if more than s moles are dissolved in solvent (one litre )  only s moles will be soluble, rest will be insoluble, following equilibrium is established.

(2) According to law of mass action -

                                                    K_SP is called solubility product.

(3) At a certain temperature solubility product of a compound is constant, it means that ions are formed in the manner that product of their concentration is always a constant. However, it becomes clear that if one of ions (A⁺ or B^–) is added from outside, it would tend to increase K_SP because [A⁺] or [B^–] has increased, so that extra ions will react with other ions to convert into insoluble part and it precipitates.

(4) 

Case-1 If 

It means it is a saturated solution.

Case-2 If  

                                       
It means more solute can dissolved.

Case-3

                                      
It means solution is supersaturated i.e. precipitation will start to occurs.

RELATIONSHIP BETWEEN SOLUBILITY AND SOLUBILITY PRODUCT:

The equilibrium for a saturated solution of a salt A_xB_y may be expressed as.

Let the solubility of the salt A_xB_y in water at a particular temperature be ‘s’ moles per litre then

NOTE-  (i) Ksp is the equilibrium constant which depends upon temperature only.

(ii)In general while calculating molar solubility we neglect hydrolysis of ions.

EXAMPLE(1): The solubility of CaCO₃  is 0.03 g/L calculate Ksp ?

SOLUTION: solubility in term of mole /liter is 0.03/100 mole/liter (M wt of CaCO₃ is 100 u)

we know that  

Ksp =(s)²   

         = (0.03/100)²

       =9 x 10^-8

(1)AB type of salts:

(2)AB2 or A2B type of salts:

(3)AB3 or A3B type of salt:

(4)A2B3 type of salts:

EXAMPLE(2):  The solubility product expression for La₂(CO₃)₃ is Ksp =?

SOLUTION:First, write the equilibrium expression. The subscript following each atom in the solid is the number of ions of that atom in solution. The total charges must add to zero:

Given the information above and the equilibrium expression:

EXAMPLE(2): What will be the solubility in moles/litre if solubility product of F₃B₂ is 2 × 10^–30.

SOLUTION:

(2) FACTER'S AFFECTING SOLUBILITY

(1) Effect of temperature on Solubility: In General,most case Solubility increases on increasing temperature.However we must follow two case 

(1) in endothermic reactions

IONISATION AND IONIC PRODUCT OF WATER (Kw)

Water is a polar ,protic inorganic compound that is at room temperature a tasteless and odourless liquid nearly colourless with hint of blue. It covers about 71.4% of earth. Water is universal solvent due to its ability to dissolved many substance .The IUPAC name of is Water  and  Oxidane.

Pure water is a very weak electrolyte -

On applying the law of mass action at equilibrium,

              Then

Since, ionization takes places to a very small extent, so the concentration of unionized water molecule is regarded as constant. Thus the product of K[H2O] gives another constant Kw.

                       So    

The product of concentration of H+ and OH. ion in water at a particular temperature is known as ionic  product of water.

IMPORTANT POINT ABOUT WATER:

(1) Mass of 1 litre of water = 997 gm.

(2) Molar concentration of water = 55.5 gm-mole / litre.

(3) Number of water molecule in 1 litre of water = 55.5×6.023×10²³ = 3.34×10²⁵.

(4) Concentration of H⁺ ion in one litre of neutral water = 10^-7 moles / litre.

(5) Concentration of OH^- ion in one litre of neutral water = 10^-7 moles / litre.

(6) Number of H+ ion in one litre of neutral water = 6.023 x 10¹⁶.

(6) Number of OH.- ion in one litre of neutral water = 6.023 x 10¹⁶.  

IONIZATION CONSTANT (Ki) OR DESSOCIATION CONSTANT (Kd) OF WATER:

At 25^c  Kw=1x10^-14  and conc^. of water 1000/18=55.55 mole /liter

Ki or Kd are also called acid or base constant of water.

Hence Pka and Pkb of water is equal (Pka=Pkb)

DISSOCIATION CONSTANT OF WATER (DOD):

Pure water is weak electrolyte and dissociates as

EFFECT OF TEMPERATURE ON IONIC PRODUCT OF WATER ( Kw):

(1) Ionisation constant of water is endothermic process, so on increasing temperature Keq increase. Kw increases with increase in temperature.

(2) During self ionisation of water dissociation concentration of water is remain constant=55.55 mole /liter.

(3) Kw is a thermodynamics equilibrium constant for water it depend upon only temperature, if temperature is increases then value of Kw is increases. and temperature is decreases then Kw decreases

(4) The numerical value of Kw increases considerably with temperature from 0.11 × 10^–14 at 0°C to 50 × 10^–14 at 100°C. It is 1.0 × 10^–14 at 25°C which we will use frequently the variation of ionic product of water with temperature is given by

On the basis of above equation we calculate the value of kw at different temperature which are given below .

We know ionic product of water at 25 ^c is

The pH scale was marked from 0 to 14 with central point at 7 at 25ºC, taking water as solvent.

If the temperature and the solvent are changed, the pH range of the scale will also change. For example

 at 25 ºC (Kw = 10^–14) Neutral point, pH= 7

 at 80 ºC (Kw = 10^–13) Neutral point, pH = 6.5

ILLUSTRATIVE EXAMPLE (1): H₂O has its pH 6.5 predict the nature of solution when

(a) Kw = 10^–14 at 25 ºC

(b) Kw = 10^–15 at 70 ºC

(c) Kw = 10^–12 at 90 ºC

SOLUTION:

(a) acidic  (b) acidic (c) basic

ILLUSTRATIVE EXAMPLE (2): The ionization constant for water is 1x10^-13.6 at 37ºC.What will be H₃O⁺ and OH^-Concentration at that temperature.

(1) 3.75 x10^-8          (2) 1.75 x 10^-8                      (3) 1.58 x 10-7            (4) 1.85 x 10^-8

SOLUTION:

ILLUSTRATIVE EXAMPLE (3): Calculate the ionic product of water at 25^c if Kw is 1.0 X10^-14 m² at 25^c

(Ans - 5.5x10-14)

ILLUSTRATIVE EXAMPLE (4): At 25 ^c the degree of dissociation of water is 1.8 x 10^-9 calculate the dissociation constant and ionic product of water.

(Ans-3.24x10^-18)

ILLUSTRATIVE EXAMPLE (5): At 30 ^c the self ionization constant of liquid ammonia (NH₃) is 10^-30.the density of NH3 at -30 ^c is 0.85 gm /liter. 

(1) Calculate the ionic product of liquid ammonia?

(2) Calculate % extent of self ionization of NH₃ molecules?

(3) How many NH₄⁺ ions are present in 5ml of liquid NH₃?

(1- 2.5x10-27 m², 2- 10^-13% 3- 25x10^-6x6.022x10²³ )

ILLUSTRATIVE EXAMPLE (6):  At 25^c the ionic product of heavy water (D₂O) is 1.44x 10^-15 m² Calculate it's dissociation constant and degree of dissociation (density of D₂O=1.02) gm/ml)

(Ans-7.44x10^-10)                                               

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