Electrochemistry

CONDUCTIVITY-TYPES-VARIATION WITH DILUTION-KOHLRAUSCH LAW - TRANSFERENCE NUMBER -DETERMINATION - IONIC MOBILITY - APPLICATION OF CONDUCTANCE MEASUREMENTS - CONDUCTOMENTRIC TITRATION

1. ELECTROCHEMISTRY Dr.P.GOVINDARAJ   Associate Professor & Head , Department of Chemistry SAIVA BHANU KSHATRIYA COLLEGE ARUPPUKOTTAI - 626101 Virudhunagar District, Tamil Nadu, India

2. ELECTROCHEMISTRY Definition It is a branch of physical chemistry which deals about G eneration of electricity by chemical reaction Occurring of chemical reaction by electricity Total reaction: 2Cl - + 2 Na + Cl 2 (g) + 2Na(s) Electricity Cell reaction : Zn(s ) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) At anode : Zn(s)→ Zn 2+ + 2e - At cathode : Cu 2+ + 2 e - → Cu(s) At anode : 2Cl - → Cl 2 (g) + 2e - At cathode : 2Na + + 2 e - → 2Na(s)

3. Definition A state of the system in which both forward and backward processes are balanced is called equilibrium Types of equilibrium There are four types of equilibrium Physical equilibrium Chemical equilibrium Phase equilibrium Ionic equilibrium (Acid – base equilibrium) EQUILIBRIUM

4. EQUILIBRIUM PCl 5 ⇌ PCl 3 + Cl 2 ICE ⇌ WATER CH 3 COOH + H 2 O ⇌ H 3 O + + CH 3 COO - acid base acid base PHASE EQUILIBRIUM IONIC EQUILIBRIUM CHEMICAL EQUILIBRIUM PHYSICAL EQUILIBRIUM

5. Definition Substances which allow electrical current to pass through them are known as conductors and this process is known as conduction Types of conductors There are two types of conductors 1.Metallic conductors 2.Electrolytic conductors CONDUCTORS

6. Metals which conduct electricity due to the mobility of outer shell electrons of atoms present Inside them are known as metallic conductors and this process is known as metallic conductors a nd this process is known as metallic conductance ( or) electronic conductance There is no chemical reaction in the metallic conductance Definition METALLIC CONDUCTORS

7. Solutions of chemical compounds or in their fused state containing opposite charged ions conduct electricity by the mobility of ions are known as Electrolytic conductors (or) electrolytes and this process is known as electrolytic conductance (or) ionic conductance The electrolytic conduction is accompanied by chemical reaction ELECTROLYTIC CONDUCTANCE Definition:

8. There are two types of electrolytes Strong electrolytes Weak electrolytes Strong electrolytes: Strong electrolytes are compounds which are completely dissociated (ionized) into positive and negative charged ions when dissolved in water. Examples: NaCl + H 2 O Na + + Cl - Al 2 (SO 4 ) 3 + H 2 O 2Al 3+ + 3   AgNO 3 + H 2 O Ag + +   The total number of positive ions and negative ions produced are equal to the formula of the electrolytes TYPES OF ELECTROLYTES

9. Weak electrolytes Weak electrolytes are compounds which are partially dissociated (ionized) into positive and negative charged ions when dissolved in water Examples: CH 3 COOH + H 2 O ⇌ CH 3 COO - + H + NH 4 OH + H 2 O + OH -   Equilibrium exists between the dissociated ions and the undissociated electrolyte TYPES OF ELECTROLYTES

10. Resistance (R) and specific resistance ( ρ ): The restriction with which the electrolytic solution oppose the flow of electricity is called as resistance of the Electrolyte The unit of resistance in SI system is ohm The resistance (R) is directly proportional to the length of the electrolytic conductor (l) and inversely proportional to the area of cross section (A) of the electrolytic conductor, i.e ., R  l/A , R = ρ l/A and ρ = R x A / l , Where ρ = specific resistance ELECTRICAL CONDUCTANCE QUANTITIES When l = 1m and A = 1 m 2 , then the above equation becomes ρ = R i.e ., specific resistance is defined as the resistance of a one meter cubic electrolytic Solution

11. ELECTRICAL CONDUCTANCE QUANTITIES Conductance (G) and Specific conductance (k ): The ease (power) with which electricity flows through a electrolytic solution is called as conductance of the solution The unit of conductance in SI system is Siemen Since resistance and Conductance are inverse relationship, the reciprocal of the resistance of the solution is also called as conductance of the solution The unit of conductance is represented as ohm -1 or mho The reciprocal of specific resistance is called as specific conductance (or) specific conductivity(k) of the electrolytic solution i.e., G = 1 / R i.e., k = 1 / ρ On substituting ρ = R x A / l in the above equation k = 1/ R x l / A -------(1)

12. ELECTRICAL CONDUCTANCE QUANTITIES The unit of specific resistance is k = x = ohm -1 m -1 (or) Sm -1 Since 1/ R is equal to conductance (G). The equation (1) becomes k = conductance x l/A When l = 1m and A = 1 m 2 , then the above equation becomes k = Conductance Thus, specific resistance (k) is also defined as the conductance of one meter cubic Electrolytic s olution  

13. EQUIVALENT AND MOLAR CONDUCTANCE Gram equivalent weight: Equivalent weight of chemical substance expressed in grams is called as equivalent weight For acid Equivalent weight = 2. For base Equivalent weight = 3. For reducing (or) oxidizing agent Equivalent weight =   Formula :

14. EQUIVALENT AND MOLAR CONDUCTANCE Examples: Equivalent weight of NaOH = ( NaOH + H 2 O → Na + + OH - ) = 40 / 1 Gram equivalent weight of NaOH = 40 grams Equivalent weight of Ca (OH) 2 = ( Ca (OH) 2 + H 2 O → Ca 2 + + 2OH - ) = 74 / 2 = 37 Gram equivalent weight of Ca (OH) 2 = 37 grams Equivalent weight of H 2 SO 4 = (H 2 SO 4 + H 2 O → 2H + + SO 4 2- ) = 98/2 = 49 Gram equivalent weight of H 2 SO 4 = 49 grams  

15. EQUIVALENT AND MOLAR CONDUCTANCE Equivalent weight of KmNO 4 = (2KMnO 4 +5H 2 SO 4 → K 2 SO 4 +2MnSO 4 +3H 2 O +5/2O 2 ) = 158/5 = 31.6 (i.e., Mn 7+ + 5e - → Mn 2+ ) Gram equivalent weight of KmnO 4 = 31.6grams  

16. Molecular weight of the chemical substance expressed in grams is called Gram Molecular Weight (or) One Mole Example: Gram Molecular Weight (or) 1 Mole of NaOH = 40 grams Gram Molecular Weight (or) 1 Mole of Ca (OH) 2 = 74 grams Gram Molecular Weight (or) 1 Mole of H 2 SO 4 = 98 grams Gram Molecular Weight (or) 1 Mole of KMnO 4 = 158 grams Note : 1 Mole of chemical substance contain 6.023 X 10 23 number of molecules . EQUIVALENT AND MOLAR CONDUCTANCE Gram Molecular Weight (or) 1 Mole of substance:

17. EQUIVALENT AND MOLAR CONDUCTANCE Equivalent conductance is defined as the conducting power of all the ions produced by one gram equivalent weight of an electrolyte in a given solution Mathematically , λ equ = k x V -------(1) where k = Specific Conductance of the solution V = Volume in cc containing 1 gram equivalent of the electrolyte If N is the normality then V = and equation (1) becomes λ equ = k x SI unit of λ equ is Sieman metre square per equivalent (or) Sm 2 equ -1   Equivalent Conductance( λ equ ):

18. EQUIVALENT AND MOLAR CONDUCTANCE Molar conductance is defined as the conducting power of all the ions produced by one gram molecular weight (1 mole) of an electrolyte in a given solution It is represented as λ m λ m = K x V -------(1) K ----- Specific conductance of the solution V ----- Volume in cc containing 1 mole of the electrolyte If M is the molarity of the solution then the equation (1) becomes λ m = K x 1000/M SI unit is Siemen meter square per mol (or) Sm 2 mol -1 Molar Conductance( λ m ) :

19. EQUIVALENT AND MOLAR CONDUCTANCE Measurement of conductance and cell constant: Cell constant : Cell constant  is of particular  cell  is determined as the ratio of the distance between the electrodes of the  cell  to the area of the electrodes .   cell constant  K = l/A = m -1 . k = conductance x l/A , λ equ = k x & λ m = K x 1000/M  

20. EFFECT OF DILUTION ON CONDUCTANCE Effect of dilution on conductance: On dilution specific conductance of the electrolytic solution decreases, but equivalent and molar conductance increases and reaches a maximum value Reason : On dilution, number of ions per unit volume decreases and hence specific conductance decreases On dilution, concentration of the electrolytic solution decreases and hence λ m & λ equi increases and reaches maximum value

21. EFFECT OF DILUTION ON CONDUCTANCE Graphical representation: The variation of molar conductance ( λ m ) and equivalent conductance ( λ equi )with square root of concentration of electrolytic solution for weak and strong electrolyte is graphical represented as

22. EFFECT OF DILUTION ON CONDUCTANCE For strong electrolyte, λ m (or) λ equi is high even at low dilution, this is due to that the strong electrolytes are completely dissociated into ions and gives maximum number of ions at low dilution Further increase in dilution, λ m (or) λ equi of strong electrolytes increase slowly due to increase in the mobility of ion. For weak electrolytes, λ m (or) λ equi is low value at low dilution, this is due to the less number of weak electrolytes dissociated at low dilution and produce less number of ions, so that λ m (or) λ equi are low value at low dilution On increasing the dilution, dissociation of the weak electrolytes gradually increased and the number of ions gradually increased, so that λ m (or) λ equi for weak electrolytes gradually increased and steeply increased at very high dilution (or) very low concentration

23. The molar conductivity of strong electrolytes at infinite dilution (λ ) can be determined by extending this straight line to zero concentration However the molar conductivity of weak electrolytes increases steeply at very low concentration and hence their molar conductivity of weak electrolytes at infinite dilution (λ ) cannot be determined by extrapolating the to zero concentration EFFECT OF DILUTION ON CONDUCTANCE

24. KOHLRAUSCH’S LAW At time infinite dilution, the molar conductance of an electrolyte can be expressed as the sum of the contributions from its individual ions i.e., λ  m = v + λ  + + v - λ  - where, v + and v - are the number of cations and anions per formula unit of electrolyte, λ  + and λ  - are the molar conductivities of the cation and anion at infinite dilution For example : The molar conductivities of HCl at infinite dilution can be expressed as λ  HC l = v H+ λ  H + + v Cl - λ  Cl – For HCl , v H+ = 1 and v Cl - = 1 λ  HCl = 1 x λ  H + + 1 x λ  Cl – λ  HCl = λ  H + + λ  Cl –

25. Kohlrausch law is used to calculate the molar conductance at infinite dilution for weak el e ct r o l y t es . Example: The molar conductance at infinite dilution for acetic acid is obtained from the values of 𝑚   for HCl, CH 3 COONa and NaCl by applying Kohlrauch law 𝐻𝐶𝑙   =   +   𝐻+ 𝐶𝑙− -------(1) -------(2) ------ - (3) 𝐶𝐻 3 𝐶𝑂𝑂𝑁𝑎   =   +   𝑁𝑎+ 𝐶𝐻3𝐶𝑂𝑂− 𝑁 𝑎𝐶 l   =   +   𝑁𝑎+ 𝐶𝑙− Equation (1) + (2) – (3) gives   𝐻𝐶𝑙 𝐶𝐻 3 𝐶𝑂𝑂𝑁𝑎 +   - 𝑁 𝑎𝐶𝑙   = 𝐻 +   +   𝐶𝐻 3 𝐶𝑂𝑂− 𝐶𝐻 3 𝐶𝑂𝑂𝐻 =   ------(4) 𝐻 𝐶 𝑙 𝐶𝐻 3 𝐶𝑂𝑂𝑁𝑎 Substituting the values of   ,   & 𝑁 𝑎 𝐶 𝑙   𝐶𝐻 3 𝐶𝑂𝑂𝐻 in equation (4) gives the value of   APPLICATION OF KOHLRAUSCH’S LAW

26. TRANSFERENCE NUMBER (OR) TRANSPORT NUMBER Definition: The fraction of total current carried by the cation (or) anion is termed as transport (transference) number Transport number (t + ) for cation is given by t + = 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒄𝒂𝒓𝒓𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒄𝒂𝒕𝒊𝒐𝒏 = 𝒒 𝒕 𝒐 𝒕 𝒂 𝒍 𝒄𝒖 𝒓 𝒓𝒆 𝒏 𝒕 𝒄 𝒂 𝒓 𝒓 𝒊𝒆 𝒅 𝒃𝒚 𝒃𝒐𝒕𝒉 𝒕 𝒉 𝒆 𝒊 𝒐𝒏 𝒔 𝑸 + Transport number (t - ) for anion is given by + t = 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒄𝒂𝒓𝒓𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒂𝒏𝒊𝒐𝒏 = 𝒒 𝒕𝒐𝒕 𝒂 𝒍 𝒄𝒖 𝒓 𝒓 𝒆𝒏 𝒕 𝒄 𝒂 𝒓𝒓𝒊𝒆𝒅 𝒃𝒚 𝒃𝒐𝒕𝒉 𝒕 𝒉 𝒆 𝒊𝒐 𝒏 𝒔 𝑸 − Where Q = q + + q -

27. Since the quantity of current carried by the particular ion is directly proportional to the mobility (speed) of ion (u), the equation (1) & (2) becomes + t = 𝑢 + 𝑢 + 𝑢 + − Where u + is the mobility (speed) of cations and it is the measure of current carried by the cations u - is the mobility (speed) of anions and it is the measure of current carried by the anions Adding equation (3) & (4) we get ------(3) and - t = 𝑢 − 𝑢 + + 𝑢 − -------(4) t + + t - = 𝒖 + + 𝒖 − 𝒖 + + 𝒖 − = 1 i.e., sum of the transport number of anion and cation of an electrolytic solution must be unity TRANSFERENCE NUMBER (OR) TRANSPORT NUMBER

28. Measurement of Transport number by Hittorf’s method: According to Hittorf , “The fall in concentration of ion around any electrode is directly proportional to speed of that ion moving away from it” i.e., Fall in concentration around anode  Speed of cation Fall in concentration around cathode  Speed of anion 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑐𝑎𝑡𝑖𝑜𝑛 (𝑢 + ) = 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑛𝑖𝑜𝑛 (𝑢 − ) 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛 𝑐𝑎𝑡ℎ 𝑜𝑑𝑒 We know that ----- - (1) + t = 𝑢 + + 𝑢 + 𝑢 − ------(2) TRANSFERENCE NUMBER (OR) TRANSPORT NUMBER

29. Substituting (2) in (1) we get + t = 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒 +𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑐𝑎𝑡ℎ𝑜𝑑𝑒 + t = 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒 𝑇𝑜𝑡𝑎𝑙 𝑓𝑎𝑙𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑏𝑜𝑡ℎ 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒 ---------(3) If the concentration are measured in terms of gram equivalent, then equation (3) becomes + t = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑜𝑠𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑛𝑜𝑑𝑖𝑐 𝑐𝑜𝑚𝑝𝑎𝑟𝑡𝑚𝑒𝑛𝑡 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑜𝑠𝑡 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑎𝑟𝑡𝑚𝑒𝑛𝑡 ---------(4) Since the total number of gram equivalent of ions lost from the anodic and cathodic compartment is equal to the number of gram equivalents of ions discharged (or) deposited on each electrode TRANSFERENCE NUMBER (OR) TRANSPORT NUMBER

30. So the equation (4) becomes + t = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑜𝑠𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑛𝑜𝑑𝑖𝑐 𝑐𝑜𝑚𝑝𝑎𝑟𝑡𝑚𝑒𝑛𝑡 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒 ---------(5) The number of gram equivalents deposited on each electrode must be equal to the number of gram equivalents of copper deposited in a copper coulometer, which is connected in series with the experimental solution, on passing same quantity of electricity So the equation (5) becomes + t = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑜𝑠𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑛𝑜𝑑𝑖𝑐 𝑐𝑜𝑚𝑝𝑎𝑟𝑡𝑚𝑒𝑛𝑡 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑢𝑙𝑜𝑚𝑒𝑡𝑒𝑟 --------- ( 6) By determining both numerator and denominator of equation (6) experimentally the transport number of cation (t + ) can be calculated and the transport number of anion (t + ) can be calculated as t - = 1 – t + TRANSFERENCE NUMBER (OR) TRANSPORT NUMBER

31. CONDUCTOMETRIC TITRATION Definition: Conductometric titration is the titration in which the end point is determined by measuring the electrical conductance of an electrolytic solution by conductometer Example: Acid base titration (HCl Vs NaOH) Principle and Experiment: The known volume of HCl (V 2 ) in the conductivity vessel and the standard NaOH (N 1 ) is taken in the burette The conductance of HCl is due to the presence of H + and Cl - ions As NaOH is added gradually, the H + ions are replaced by slow moving Na + ions as represented below H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) → Na + (aq) + Cl - (aq) + H 2 O (l)

32. On continuous addition of NaOH, the fast moving H + ions are gradually replaced by the slow moving Na + ions. So that the conductance will go on decreasing until the acid has been completely neutralized Further addition of NaOH after the end point, the conductance will begin to increase due to the introduction of fast moving OH - ions The end point is obtained graphically on plotting the conductance Vs the volume of NaOH added CONDUCTOMETRIC TITRATION

33. Using the end point V 1 , the strength of HCl is calculated using the formula 2 N = 𝑉 𝑁 1 1 𝑉 2 where V 1 is the volume of NaOH obtained from graph N 1 is the strength of NaOH V 2 is the volume of HCl VOLUME OF NaOH CONDUCTOMETRIC TITRATION

34. IONIC MOBILITY It is defined as the distance travelled by an ion per second under a potential gradient of 1 volt per metre. Mathematically, Ionic mobility = 𝑺𝒑𝒆𝒆𝒅 𝒐𝒇 𝒂𝒏 𝒊𝒐𝒏 𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝒈𝒓𝒂𝒅𝒊𝒆𝒏𝒕 Potential gradient is obtained by dividing the potential difference applied at the electrode with the distance between the electrodes Mathematically, Potential gradient= 𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒉𝒆 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒅𝒆𝒔

35. Lithium and sodium ions have comparatively low ionic mobility where as H + ions has high ionic mobility IONIC MOBILITY Explanation: Lithium and sodium ions have comparatively low ionic mobility . This is due to the hydration of ions as a result of higher charge density around these ions because of their smaller radii The higher charge density causes these ions to be more highly hydrated by ion – dipole interactions than the larger ions. Since a hydrated ion has to drag along a shell of water as it moves through the solution,its mobility is less than that of an anhydrated ion

36. Explanation: The high mobility of H + ions in hydroxylic solvents such as H 2 O can be explained by Grotthus type mechanism in which a proton moves rapidly from H 3 O + to a hydrogen – bonded water molecule and is transferred further along a series of hydrogen – bonded water molecules by a rearrangement of hydrogen bonds, as illustrated in the following diagram IONIC MOBILITY

37. APPLICATION OF CONDUCTANCE MEASUREMENTS 1. Determination of degree of dissociation (  ) of weak electrolytes The degree of dissociation of a weak electrolyte at any dilution can be calculated by the following relationship  =  m /  m  whe r e m   is the molar conductance at infinite dilution and obtained by Kohlrausch’s law  m is the molar conductance at particular concentration and obtained by conductance measurement using the following formula λ m = K x 1000/M where K is the specific conductance and is obtained by multiplying the conductance measured with cell constant (K = Conductance x cell constant )

38. 1. Determination of ionic product of water (K w ) Water is slightly dissociated as H 2 O ⇌ H + + OH - The product of the concentration of hydrogen and hydroxyl ions expressed in mol dm -3 known as ionic product of water (K w ) i.e., [H + ][OH - ] = K w ----------(1) The specific conductance (K) of the purest water at 25 C is determined experimentally by conductance measurement Since the concentration of H + ions and OH - ions are equal, the equation (1) becomes c x c = K w c 2 = K w APPLICATION OF CONDUCTANCE MEASUREMENTS

39. We know that the relationship between specific conductance (K), molar conductance (  m ) and concentration is  m = 𝐾 𝑐 ----------(3) c = 𝐾  m ----------(4) The specific conductance (K) of the purest water at 25 C is determined experimentally by conductance measurement The molar conductance at infinite dilution of water is obtained by Kohlrausch’s law 2 𝑚(𝐻 𝑂)   𝐻 + 𝑂𝐻 − =   +   ---------(5) APPLICATION OF CONDUCTANCE MEASUREMENTS

40. 2 𝑚 𝐻 2 𝑂 Assuming that  𝑚(𝐻 𝑂) differs very little from   and the concentration of H +  (or) OH - ion can be determined by substituting K and  𝑚 in equation (5) Then the ionic product of water is obtained by substituting the concentration of H + and OH - ions in the equation (2) K w = c 2 APPLICATION OF CONDUCTANCE MEASUREMENTS

41. D E T E R M IN A T I ON OF S O L U B LITY A N D SOL U BILITY PRO D U C T OF SPARINGLY SOLUBLE SALT i.e., Ag C l ⇌ Ag + ( aq ) + Cl - ( aq ) The solubility product of sparingly soluble salt (AgCl) is expressed as K sp = [Ag + ][Cl - ] ------------(1) where, [Ag + ] and [Cl - ] are the concentration of Ag + and Cl - ions in aqueous solution Let the solubility of AgCl is x mol m -3 i.e., the concentration of AgCl in the aqueous solution (c) = x mol m -3 The conductance of the AgCl in the aqueous solution and water used in the preparation of the solution are determined using conductivity meter When water is added to the sparingly soluble salt like AgCl, very minute quantity of salt will pass in solution and equilibrium with the unionized AgCl salt H 2 O

42. The difference multiple with cell constant gives the specific conductance (K) of the solution due to the dissolved salt (AgCl). Let the value be Z Sm -1 The m olar conduc t ance of t he Ag C l solu t ion is  m (AgCl) = K /C -----------(2) and subs t i t ute ‘x’ for the concent r at i on and Z f or 𝑚(𝐴𝑔𝐶𝑙) Assuming that  m (AgCl) =   specific conductance, equation (2) becomes 𝑚(𝐴𝑔𝐶𝑙)   = Z /x -----------(3) 𝑚 ( 𝐴𝑔 𝐶 𝑙 ) x = Z (S m -1 )/   (Sm 2 mol -1 ) • 𝑚 ( 𝐴𝑔 𝐶 𝑙 )   is obtained from Kohlrausch’s law 𝑚 ( 𝐴𝑔 𝐶 𝑙 ) 𝐴 𝑔 +    = =  +  −  𝐶 𝑙 D E T E R M IN A T I ON OF S O L U B LITY A N D SOL U BILITY PRO D U C T OF SPARINGLY SOLUBLE SALT

43. Substituting the calculated values Z and in equation (3), the solubility of AgCl ‘x’ molm -3 can be determined On dividing the solubility with 1000, gives the solubility (x) in terms of mol dm -3 i.e., concentration of AgCl in aqueous solution in x mol dm -3 On multiplying the solubility (x mol dm -3 ) with gram molecular weight of AgCl gives solubility of AgCl in terms of g dm -3 i.e., Solubility of AgCl = x mol dm -3 x 143.5 g mol -1 Solubility of AgCl = y g dm -3 The concentration if AgCl in aqueous solution is x mol dm -3 i.e., [Ag+] = x mol dm -3 ----------- ( 5) [Cl - ] = x mol dm -3 D E T E R M IN A T I ON OF S O L U B LITY A N D SOL U BILITY PRO D U C T OF SPARINGLY SOLUBLE SALT

44. Solubility product (K sp ) is obtained by substituting equation (5) in equation (1) K sp = (x mol dm -3 ) (x mol dm -3 ) K sp = x 2 mol 2 dm -6 D E T E R M IN A T I ON OF S O L U B LITY A N D SOL U BILITY PRO D U C T OF SPARINGLY SOLUBLE SALT

45. DETERMINATION OF HYDROLYSIS CONSTANT OF A SALT The relationship between hydrolysis constant of the salt and degree of hydrolysis of the salt is K h = cx 2 ------------(1) Where K h is Hydrolysis constant of the salt c is the concentration of the salt in moles per litre x is the degree of hydrolysis Example: The equilibrium existed in the hydrolysis of NH 4 Cl is NH 4 Cl + H 2 O ⇌ NH 4 OH + HCl (1-x) moles x moles x moles Where x is the degree of hydrolysis

46. The electrical conductance of an aqueous solution of N 4 Cl salt is due partly to the ions of the un hydrolysed salt and partly to the H+ and Cl- ions i.e., The molar conductance of this solution  m, as determined experimentally, will be the sum of the conductance of (1-x) moles of N 4 Cl and x moles of HCl + x  m(HCl)  m = (1-x)  ’ m ---------(2) where  m(HCl) is the molar conductance of HCl at infinite dilution  ’ m is the molar conductance of the unhydrolysed NH 4 Cl salt at the given concentration The value of  ’ m is obtained by adding slightly excess of NH 4 OH to the salt solution. This suppresses the hydrolysis of NH 4 Cl to such a large extent that the molar conductance of the solution can be taken as  ’ m . i.e., the conductance of the unhydrolysed salt DETERMINATION OF HYDROLYSIS CONSTANT OF A SALT

47. On rearranging the equation (2), we get ---------(3) where x is the degree of hydrolysis of NH 4 Cl  m is the molar conductance of the salt solution before adding excess of NH 4 OH  ' m is the molar conductance of the salt solution after adding excess of NH 4 OH are determined by conductance measurement  m and  ' m  is obtained by Kohlrausch’s law m(HCl) λ HCl = λ H + + λ Cl – DETERMINATION OF HYDROLYSIS CONSTANT OF A SALT

48. Substituting all the measured values in (3), we get the degree of hydrolysis (x) of NH 4 Cl salt Substituting the calculated degree of hydrolysis of NH 4 Cl salt (x) and the concentration of NH 4 Cl salt solution in equation (1) , the hydrolysis constant (K h ) can be calculated K h = cx 2 DETERMINATION OF HYDROLYSIS CONSTANT OF A SALT

49. THANK YOU