lec22: Applications of Kohlrausch law

Transcription

1 lec22: Applications of Kohlrausch law 1-Calculation of molar conductivity of a weak electrolyte at infinite dilution It is not possible to determine the value of m for weak electrolytes since we cannot obtain the limiting value of the molar conductivity for a weak electrolyte. This is done indirectly by the molar ionic conductance for the individual ions of the weak electrolyte as follows: For e.g. molar conductance of acetic acid at infinite dilution can be calculated from the molar conductance at infinite dilution of hydrochloric acid, sodium acetate and sodium chloride as follows: 0 HCl = λh+ + λ Cl (1) 0 CH3COONa = λ0 Na+ + λ0 CH3COO (2) 0 NaCl = λ0 Na++ λ0 Cl (3) Add eq. 1 and 2 and subtract 3 we get: 0 HCl + 0 CH3COONa - 0 NaCl = λ0 H+ + λ0 Cl- + λ0 Na+ + λ0 CH3COO- - λ0 Na+ - λ0cl The limiting molar conductivities of KCl, KNO3, and AgNO3 are msm2mol -1, msm2mol -1 and msm2mol -1 respectively at 25 C. Calculated the limiting molar conductivity of AgCl at this temperature? 0 AgCl= 0 KCl + 0 AgNO 3-0 KNO 3 =( ) msm 2 mol -1 = msm 2 mol Calculation of Degree of Dissociation of Weak Electrolytes The degree of dissociation of weak electrolyte such as NH4OH, acetic acid can be determined by measuring the molar conductivity Λ, of the solution of the electrolyte at any given dilution. For e.g. the degree of dissociation (α) of a weak electrolyte at the concentration C mole per liter may be given by the following relation:

2 α = / 0 Where, Λ is the equivalent conductance of electrolyte at any concentration and Λ0 is the equivalent conductance of the same electrolyte at infinite dilution. Hence, measurement of Λ permits evaluation of α if Λ0 is known. Let us consider dissociation of acetic acid: CH 3 COOH+H 2 O==CH 3 COO - +H 3 O + (1- α)c αc αc Where C is the equivalent concentration.dissociation constant is represents as following: K C =[ H 3 O + ][ CH 3 COO - ]/[ CH 3 COOH ] K C =(αc)( αc)/ (1-α)C K C = α 2 C/1-α K C = ( / 0) 2 /1-( / 0) ΛC=K C Λ 2 0 /Λ-KCΛ 0 By plotting this equation,k C can be calculated. Equivalent conductance of N acetic acid solution is ohm -1 cm 2 and ohm -1 cm 2 of infinite diluted solution. Calculate the degree of dissociation. α=λ/λ 0 α=48.15 ohm -1 cm 2 /390.6 ohm -1 cm 2 α= K C = α 2 C/1- α K C =(0.1232)( )/( ) =1.781x Determination of Solubility of Sparingly Soluble Salts

3 Salts like AgCl, BaSO4, CaCO3, Ag2CrO4, PbSO4, PbS, Fe(OH)3 etc. are ordinarily regarded as sparingly soluble and have a very small but definite solubility in water. The solubility of such sparingly soluble salts is obtained by determining the specific conductivity (κ) of a saturated salt solution. AgCl Ag + + Cl - Assume concentration of solid is a constant. Concentration of Ag + = concentration of Cl - = C Ksp = [Ag + aq] [Cl - aq] Ksp = C 2 The measurement of the specific conductivity, κ of the saturated solution leads to a value of the concentration. i.e. C = κ / Λ 0 or S= κ / Λ 0 i.e C is the solubility(s).units of S is the same of that C,equi.cm 3 As, Ksp = C 2 Ksp = (κ / Λ 0 ) 2 Specific conductivity of saturated AgCl solution at 25 o C is 3.41x10-6 ohm -1 cm -1,while the specific conductance of water that used as solvent is 1.6x10-6 ohm -1 cm -1.Calculate the solubility of AgCl in.water at this temperature in units of mol L -1.Given that equivalent conductance of infinite diluted AgCl solution is equi ohm -1 cm 2. κ AgCl = κ solution - κ water =3.41x10-6 ohm -1 cm x10-6 ohm -1 cm -1 =1.81x10-6 ohm -1 cm -1 As AgCl is 1:1 electrolyte S = κ / Λ 0 =(1.81x10-6 ohm -1 cm -1 )/ (138.3 equi ohm -1 cm 2 ) =1.31x10-8 equi -1 cm -3

4 =1.31x10-5 equi -1 L Determination of ionic product of water With the help of specific conductivity of water, the ionic product of water can be determined. The ionization of water may be represented as, H2O H+ + OH- The product of the concentrations of H+ and OH ions expressed in mol/l is called ionic product of water and is represented by Kw. [H+] [OH-] = Kw The measured specific conductivity of the purest form of water is x 10-6 S cm -1. The molar conductance is given by Λm = Kv x 18 Thus, = x 10-6 x 18 = x 10-6 Scm 2 mol -1 The molar conductance of water at infinite dilution can be obtained by m (H2O) = Λ (H+) + Λ (OH-) Therefore, Λ (H2O) = = Scm 2 mol -1 and for water, α = Λm / Λ = x 10-3 x 18/ CH+= COH- = x 10-3 x 18/ (Concentration of water = 1000/18 = M CH+= COH- = α x x 10-9 = x = x 10-7 Kw = (1.003 x 10-7 ) = (1.006 x ) 6- Determination of Ionic Mobility of an Ion Hypothesis a solution contains C equi of 1:1 electrolyte. the applied voltage is E,volt.cm-1.The passing current is: I=E/R

5 If the electrolyte is strong=c + +C -.The cation cross U + distance in sec toward cathode and anions cross U - distance in sec toward anode.i.e C + U + equivalents of cations move to cathode and C - U - equivalents of anions move to anode. The total equivalent concentrations that move toward two electrodes is(c + U + )( C - U - ).The total current is represents as: L=F(C + U + + C - U - ) L=CF(U + +U - ) Λ=L/C=F(U + +U - ) Λ=F(U + +U - ) Λ= λ + + λ - F U + = λ + FU + =+ λ - Calculate the specific conductance of 0.1 NaCl solution at 25 0 C given that ionic mobilities of sodium and chloride ions are 42.6 x10-5 and 68x10-5 volt -1 cm 2 sec -1 respectively. C=0.1/1000= equi cm 3 L=CF(U + +U - ) =( equi cm 3 )(96500 C equi -1 )[42.6 x10-5 volt -1 cm 2 sec x10-5 volt -1 cm 2 sec -1 ] = ohm -1 cm Determination of hydrolysis constant of salts When aniline chloride is dissolved in water,it is hydrolysis as following: C 6 H 5 NH 3 Cl==== C 6 H 5 NH 3 + +Cl - C 6 H 5 NH 3 + +H 2 O==== C 6 H 5 NH 2 +H 3 O + (1- α)c αc αc Where is the hydrolysis degree and C is the salt equivalent concentration, thus:

6 Λs=(1- α) Λ C +αλ a +α Λ b Where S,C,a,and b are a salt,concentration,acid and base respectively. Where a is HCl and base is aniline. While equivalent conductance of weak base is too small,so it cam be negligible. Λs=(1- α) Λ C +αλ a α= (Λs- Λ C )/ Λ a - Λ C Λ C can be determined from measurements of equivalent conductance of aniline chloride dissolved in aniline. The aniline is a weak base,so it retarded the dissolution of salt and Λ C is somewhat equal of equivalent conductance of non decomposed salt.. while Λ a is determined from ionic conductance of its ions. The hydrolysis constant(k h ) of aniline chloride can be represents as following: K h =[ H 3 O + ][ C 6 H 5 NH 2 ]/ [C 6 H 5 NH + 3 ] K h =(αc) (αc)/(1-α)c K h =α 2 C/1-α K h can be used to calculation of decomposition constant of base(k b ) as following: K h =K w /K b Where K w is a ion product of water.