HETEROGENEOUS EQUILIBRIUM (NERNST’S DISTRIBUTION LAW)
HENRY’S LAW The mass of gas (m) dissolved by unit volume of a given liquid is proportional to the pressure (P) of the gas at constant temperature. ∝ = K1 (Constant) Limitations The pressure should be low and the temperature should be high Gas should behave like an ideal gas Gas should not undergo compound formation with the solvent Gas should be noninteracting with solvent Gas should not undergo association or dissociation in the solvent Gas remain in same molecular formula.
NERNST’S DISTRIBUTION LAW At constant temperature when different quantities of a solute are allowed to distribute between two immiscible solvents in contact with each other then at equilibrium the ratio of the concentration of the solute in two layers is constant. Shake almost 2 hours Let us consider solid I2 is added to water and CCl4 in contact with each other
NERNST’S DISTRIBUTION LAW At constant temperature when different quantities of a solute are allowed to distribute between two immiscible solvents in contact with each other then at equilibrium the ratio of the concentration of the solute in two layers is constant. Let us consider solid I2 is added to water and CCl4 in contact with each other Now according to Nernst Distribution Law I2 CCl4 I2 H2O = Constant (at constant temperature) KD Distribution coefficient or partition coefficient between the two solvents at the given temperature of the solute This is the mathematical form of the Nernst’s distribution law. = It is an equilibrium constant, its variation with temperature will be governed by Van’t Hoff equation
NERNST’S DISTRIBUTION LAW Condition of Validity of This Law Same molecular species should be present in both solvents Solution should be ideal hence dilute Two solvents should be immiscible
NERNST’S DISTRIBUTION LAW DERIVATION Solute X Let us take two immiscible solvents A and B in contact with each other Let solute ‘x’ is added, which is dissolved in both the solvents At equilibrium at a given temperature
NERNST’S DISTRIBUTION LAW DERIVATION Let us take two immiscible solvents A and B in contact with each other Let solute ‘x’ is added, which is dissolved in both the solvents At equilibrium at a given temperature x(A) = x(B) x(A) x(B) ǁ
NERNST’S DISTRIBUTION LAW FOR A SOLUTE WITH DISSOCIATION Solute X Let us take two immiscible solvents A and B in contact with each other Let solute ‘x’ is added, which is dissolved in both the solvents X undergoes partial dissociation in solvent B to the species Y and Z X ⇌ Y + Z CA Concentration of solute in solvent A CB Let Total concentration of solute in solvent B Degree of dissociation of solute ‘x’ in solvent B
NERNST’S DISTRIBUTION LAW FOR A SOLUTE WITH DISSOCIATION X ⇌ Y + Z X ⇌ Y + Z The equilibrium concentration of the species in solvent B (1-)CB CB CB The chemical potential of the undissociated species in the two solvents would be same x(A) x(B) ǁ x(B) = x(A) Ration of concentration of the undissociated species in the two solvents would be constant. 1 − CB = KD () 1 − A CA 1 − B CB = KD () Where A and B are the degree of dissociation of the solute in solvent A and B respectively.
NERNST’S DISTRIBUTION LAW FOR A SOLUTE WITH ASSOCIATION Solute X Let us take two immiscible solvents A and B in contact with each other Let solute ‘x’ is added, which is dissolved in both the solvents X undergoes association in solvent B to the species Xn CA Concentration of solute in solvent A CB Let Total concentration of solute in solvent B Degree of association of solute ‘x’ in solvent B nX ⇌ Xn
NERNST’S DISTRIBUTION LAW FOR A SOLUTE WITH ASSOCIATION nX ⇌ Xn nX ⇌ Xn (1-)CB n CB The equilibrium constant K = [Xn] [X] n = n CB [ 1− CB] n [ 1 − CB] n = CB nK 1 − CB = CB nK 1 Distribution law is given by = = 1 − CB x(A) x(B) ǁ
NERNST’S DISTRIBUTION LAW FOR A SOLUTE WITH ASSOCIATION nX ⇌ Xn x(A) x(B) ǁ = 1 − CB = CB nK 1 = 1 × 1 K and are the constant and dependent on temperature 1 Constant at constant T ′ = 1
NERNST’S DISTRIBUTION LAW FOR A SOLUTE WITH ASSOCIATION nX ⇌ Xn x(A) x(B) ǁ ′ = 1 ′ = For n-mer formation in solvent B only If the solute associates in both the solvents, such as n1 -mer in solvent A and n2 -mer in solvent B, then distribution law will be ′ = 1 2
APPLICATION OF DISTRIBUTION LAW Solvent extraction. Determination of equilibrium constant. Determination of hydrolysis constant and degree of hydrolysis.Determination of molecular complexity of solute.
SOLVENT EXTRACTION Extraction of a solute from a solution by another immiscible solvent. A solute is generally extracted from an aqueous solution with the help of suitable organic solvents such as benzene, ether, chloroform etc. These are called the extracting solvent. The extracting solvent should meet the following requirements It should be completely immiscible with solution containing impurities. The solute should be more soluble in the extracting solvent. The extracting solvent should be more volatile.
SOLVENT EXTRACTION Aqueous solution containing solute (or impurity) Extracting Solvent Separating funnel inverted and shaken several times with stopcock.
SOLVENT EXTRACTION Aqueous Layer Organic Layer
SOLVENT EXTRACTION Both the aqueous and organic layers are separated.
The extraction can be done in two ways SOLVENT EXTRACTION By using the entire volume of the extracting solvent in one lot. By using the extracting solvent in fractional quantities and repeating the process of extraction several times.
Show that multistep extraction is more advantageous than single step extraction. SOLVENT EXTRACTION V0 volume of the solution containing w g solute (or impurity) V1 volume of extracting Solvent Let w1 g of the solute remain unextracted in the original solvent after first stage of extraction.
SOLVENT EXTRACTION Let w1 g of the solute remain unextracted in the original solvent after first stage of extraction. Concentration of the solute in the original solution is equal to w1 V0 Concentration of the solute in the extracting solvent is equal to (w−w1) V1 KD = w1 V0 (w−w1) V1 KD (w-w1 )V0 = w1V1 KDV0w – KDw1V0 – w1V1 = 0 W1 = KDV0 KDV0+ V1 × W --------------------(i)
In the second extraction again the same volume V1 of the solvent is used. SOLVENT EXTRACTION let w2 g of the solute remain unextracted in the original solvent after second stage of extraction. Concentration of the solute in the original solution is equal to w2 V0 Concentration of the solute in the extracting solvent is equal to (w1−w2) V1 KD = w2 V0 (w1−w2) V1 KD (w1 − w2)V0 = w2V1 KDV0 w1 – KDw2V0 – w2V1 = 0 W2 = KDV0 KDV0+ V1 × w1
SOLVENT EXTRACTION W2 = KDV0 KDV0+ V1 × w1 = KDV0 KDV0+ V1 × KDV0 KDV0+ V1 × W [Putting value of W1 from equation (i)] W2 = ( KDV0 KDV0+ V1 ) 2 × W Similarly, for 3rd step of extraction W3 = ( KDV0 KDV0+ V1 ) 3 × W Similarly, for n-th step of extraction Wn = ( KDV0 KDV0+ V1 ) n × W
SOLVENT EXTRACTION = W (1 + V1 KDV0 ) n = W (1 + Y) n Taking Y = V1 KDV0 = W (1 + nY +−−−− −) −−−−− −() = KDV0 KDV0 + V1 × For single step, Volume of extracting solvent used = V1 For n no of steps, total volume of extracting solvent used = nV1
SOLVENT EXTRACTION If all the volume n1V1 of the extraction solvent is used in one lot, the amount of solute unextracted (W′ ) is given by W′ = KDV0 KDV0 + nV1 × W = W 1 + V1 KDV0 W′ = W 1 + Y = V1 KDV0 −−−−− −() By equation (iii) (ii), we get, W′ Wn = (1 + nY +−−−− −) (1 + nY) W′>Wn Hence multistep extraction will be better than the single step extraction.
DETERMINATION OF EQUILIBRIUM CONSTANT Let us consider the following reaction in aqueous medium, KI + I2 ⇌ KI3 The equilibrium constant KC = [KI3]eq [KI]eq[I2]eq ---------(i) [KI3]eq [KI]eq [I2]eq Concentration of KI3 in aqueous medium at equilibrium Concentration of KI in aqueous medium at equilibrium Concentration of free iodine in aqueous medium at equilibrium
DETERMINATION OF EQUILIBRIUM CONSTANT 30 ml I2 solution in Organic Solvent 100 ml H2O 30 ml I2 solution in Organic Solvent 100 ml KI solution in H2O Bottle - I Bottle - II Additional Reaction KI + I2 ⇌ KI3 Shake the bottle for almost 1.5 hours and attain equilibrium
DETERMINATION OF EQUILIBRIUM CONSTANT Bottle - I Bottle - II Additional Reaction KI + I2 ⇌ KI3 We can determine the KD value of I2 between Organic and aqueous medium At constant temperature, Bottle-I and Bottle-II have same KD value. With the help of KD value, Concentration of different species can be determined. Hence, KC can be determined
DETERMINATION OF EQUILIBRIUM CONSTANT Bottle - I Aqueous Layer Titration Approx. 10 ml (N/10) KI solution 25 ml aqueous layer 100 ml Conical flask Titrate with (N/100) Na2 S2O3 solution with starch as indicator [I2]aq is determined from titre vale
DETERMINATION OF EQUILIBRIUM CONSTANT Bottle - I Organic Layer Titration Approx. 10 ml (N/10) KI solution 2 ml Organic layer 100 ml Conical flask Titrate with (N/10) Na2 S2O3 solution with starch as indicator [I2]Org is determined from titre vale
DETERMINATION OF EQUILIBRIUM CONSTANT Bottle - I , = . 2 . 2 = [I2]Org [I2]aq value can be calculated At constant temperature, this is also the value for Bottle-II Thus, value of Bottle-II is known
Bottle - II DETERMINATION OF EQUILIBRIUM CONSTANT Organic Layer Titration Approx. 10 ml (N/10) KI solution 2 ml Organic layer 100 ml Conical flask Titrate with (N/10) Na2 S2O3 solution with starch as indicator [I2]Org is determined from titre vale (Let, C1 )
Bottle - II DETERMINATION OF EQUILIBRIUM CONSTANT Aqueous Layer Titration 100 ml Conical flask Titrate with (N/100) Na2 S2O3 solution with starch as indicator 25 ml aqueous layer Total concentration of iodine in aqueous layer in moles/lit. (Concentration of free I2 and concentration of KI3 ) (Let, C2 )
DETERMINATION OF EQUILIBRIUM CONSTANT Calculations . 2 . 2 . = Value known from Bottle-I at constant temperature . 2 . = C1 KD [I2]eq = C1 KD Concentration of KI3 at equilibrium = Total concentration of Iodine in aqueous medium - Concentration of free Iodine in aqueous medium [KI3]eq = C2 − C1 KD C2 C1 KD
DETERMINATION OF EQUILIBRIUM CONSTANT Calculations Let, C = initial concentration of KI solution Concentration of KI at equilibrium = initial concentration of KI solution - Concentration of KI3 at equilibrium C C2 − C1 KD [KI]eq= C − C2 − C1 KD
DETERMINATION OF EQUILIBRIUM CONSTANT Calculations KC = [KI3]eq [KI]eq[I2]eq From equation (i), we have, [KI3]eq = C2 − C1 KD [KI]eq= C − C2 − C1 KD [I2]eq = C1 KD KC = C2 − C1 KD C − C2 − C1 KD × C1 KD N.B: All concentrations terms should be in moles/lit unit.
DETERMINATION OF PARTITION COEFFICIENT OF BENZOIC ACID BETWEEN BENZENE AND WATER C6H5 − CO2H In Benzene n-mer Formation (C6H5 − CO2H)n In Water C6H5 − CO2H Same molecular species (Solute) (Solvent) (Solvent) CA Concentration of benzoic acid in water CB Total Concentration of benzoic acid in benzene Degree of association of benzoic acid in benzene Number of benzoic acid molecules that combine to form associated molecule (C6H5 − CO2H)n n
DETERMINATION OF PARTITION COEFFICIENT OF BENZOIC ACID BETWEEN BENZENE AND WATER n(C6H5 − CO2H) ⇌ (C6H5 − CO2H)n (1-)CB n CB The equilibrium constant K = [(C6H5−CO2H)n] [C6H5−CO2H] n = n CB [ 1 − CB] n [ 1 − CB] n = CB nK 1 − CB = CB nK 1 −−−− −()
DETERMINATION OF PARTITION COEFFICIENT OF BENZOIC ACID BETWEEN BENZENE AND WATER According to distribution law KD = [C6H5 − CO2H]Bz [C6H5 − CO2H]H2O = 1 − CB CA = CB nK 1 n CA With the help of equation (i) KD = (CB) 1 n CA × nk 1 n K and Dependent on temperature Constant at constant temperature nk 1 n Constant at constant temperature
DETERMINATION OF PARTITION COEFFICIENT OF BENZOIC ACID BETWEEN BENZENE AND WATER KD = (CB) 1 n CA × nk 1 n KD ′ = (CB) 1 n CA = Constant logKD ′ = 1 n logCB − logCA logCA = 1 n logCB − logKD ′ A plot of logCA vs logCB will give a straight line with slope 1 n and intercept −logKD ′ from which both ‘n’ and KD can be determined.
DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD The composition of a complex formed by mixing of ammonia and cupric sulfate can be determined by partition method. When ammonia is shaken with CCl4 and CuSO4 solution, it distributed itself in the two liquids in such a way that the ratio of concentrations of free ammonia in the two phases obeys the partition law KD = [NH3]aq [NH3]org Knowing the initial concentration of Cu+2 ions, the coordination number of Cu+2 ion can be calculated
DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD 30 ml Organic Solvent (CHCl3 ) 30 ml 1(M) aq. Solution of NH3 25 ml 1(M) aqueous Solution of NH3 + 5 ml 0.5(M) CuSO4 Sol. Bottle - I Bottle - II Additional Reaction Cu+2 + nNH3 Shake the bottle for almost 1.5 hours and attain equilibrium 30 ml Organic Solvent (CHCl3 ) ⇌ [Cu(NH3)n ] +2
DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD Bottle - I Aqueous Layer Titration 100 ml Conical flask Titrate with (N/2) H2 SO4 solution until Yellow colour becomes Red [NH3]aq is determined from titre vale 5 ml aqueous Layer 10-15 ml distilled water 2-3 drops Methyl Orange Ind.
DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD Bottle - I Organic Layer Titration 100 ml Conical flask Titrate with (N/50) H2 SO4 solution until Yellow colour becomes Red [NH3]Org is determined from titre vale 5 ml Organic Layer 10-15 ml distilled water 2-3 drops Methyl Orange Ind.
, = . 3 . 3 = [NH3]aq [NH3] value can be calculated At constant temperature, this is also the value for Bottle-II Thus, value of Bottle-II is known DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD Bottle - I
DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD Bottle - II Organic Layer Titration 100 ml Conical flask Titrate with (N/50) H2 SO4 solution until Yellow colour becomes Red [NH3]Org is determined from titre vale (Let, C1 ) 5 ml Organic Layer 10-15 ml distilled water 2-3 drops Methyl Orange Ind.
DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD Bottle - II Aqueous Layer Titration 100 ml Conical flask Titrate with (N/2) H2 SO4 solution until Yellow colour becomes Red Total [NH3]aq is determined from titre vale (Let, C2 ) 5 ml aqueous Layer 10-15 ml distilled water 2-3 drops Methyl Orange Ind.
DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD Calculations Value known from Bottle-I at constant temperature . 3 = C1KD Concentration of ammonia bound as cupra-ammonium complex = Total concentration of ammonia in aqueous medium - Concentration of free ammonia in aqueous medium C2 C1KD . 3 . 3 = = C2 − C1KD
DETERMINATION OF CO-ORDINATION NUMBER OF Cu+2 ION BY PARTITION METHOD Calculations Co-ordination number of Cu+2 ion = Concentration of ammonia bound as cupra−ammonium complex initial concentration of Cu+2 ion in aq. medium Co−ordination number of Cu+2 ion = C2 − C1KD [Cu+2] Take nearest whole no.