Expt. 7 - Conductivity of Electrolytes

Experiment 7CONDUCTIVITY OF ELECTROLYTESConductance, G, is the reciprocal of resistance, R. Resistance is expressed in ohms,Ω. The SI unit of conductance is the siemans, S, and 1 S = 1 Ω -1 . Although resistance is thequantity measured, conductance is more appropriate for the description of electrolytesolutions. This is because the conductance of a solution is the sum of the conductances of thesolvent and the solutes. The conductivity κ of an electrolyte is related to its conductancethrough the cell constant k cellκ = k cell x GThe conductivity has SI units S m -1 , and may be interpreted as the conductance of a cube ofsolution, 1 m on each edge. In this experiment, k cell is determined by implicitly in theconductivity meter by measuring the conductance of standard solutions.In the first approximation, the ions move independently during conduction, and theconductivity is expected to be proportional to concentration. An approximately concentrationindependent quantity, the molar conductivity Λ, can be defined in SI units by dividing theconductivity by the concentration in mol m -3 . In terms of the molarity c, the molar conductivityis given, in units of S m 2 mol -1 , byΛ = κ / 1000cIn fact, Λ decreases slowly with concentration, being approximately linear in thesquare root of concentration. At infinite dilution, the ions do conduct independently, and thelimiting molar conductivity, Λ o , obeys a rule of additivity,Λ o (AX) = Λ o (AY) + Λ o (BX) - Λ o (BY)where AX, AY, BX, and BY are strong uni-uni electrolytes. The above rule can be extendedto electrolytes like CaCl 2 in a manner that includes the stoichiometric coefficients.An alternative treatment of conductivity, found often in the literature, uses theequivalent conductivity, Λ e . For an electrolyte of the form (A Z+ ) ν+ (X Z- ) ν- , the molarconductivity, Λ is related to Λ e byΛ = (ν + Z + )Λ e = (ν - Z - )Λ eFor a weak electrolyte AB with molarity c, the concentrations of A + and B - are eachαc, where α is the degree of dissociation. The experimental value of κ allows the calculationof α. The apparent molar conductivity of AB, A app , is given by1

Λ app = κ / 1000cOnly part of the weak electrolyte is actually present as ions. The molar conductivity of theionized part is given by Λ = κ / (l000αc). Assuming that Λ may be approximated by Λ o , itfollows thatα = A app /Λ oThe additivity rule is applied to get Λ o from the proper combination of strong electrolytes,because it is not practical to extrapolate Λ to infinite dilution.Experimental ProcedureConductance is measured by immersing the cell in a beaker containing the solution ofinterest. Great care must be taken to wash the cell with the solution first. The followingsolutions are to be measured.1. The standard solutions.2. The distilled water used to prepare the solutions. Add 25 kΩ or 125 kΩ to the resistancebox. The resistance of the water will be quite large and not very reproducible. Repeatedrinsing will be necessary.3. The strong electrolytes, NaCl, CaCl 2 , Na 2 SO 4 , NaC 2 H 3 O 2 , and HCl, at 0.01 M, 0.004 M,and 0.001 M.4. The weak acid, HC 2 H 3 O 2 , at 0.05 M, 0.01 M, 0.005 M, 0.001 M, and 0.0005 M.5. A saturated solution of CaSO 4 (from the shelf) diluted 1:10. Do not shake this solution.Treatment of DataConvert all measured resistances to conductances and subtract the contribution fromwater to get the net conductance of the electrolytes. From the net conductance of the KClcalculate the cell constant, and from this, the molar conductivities of the remaining solutions,except the CaSO 4 .1. For the strong electrolytes, plot Λ versus √c and determine Λ o by extrapolation.2. For acetic acid, determine Λ o from the results for NaCl, Na 2 C 2 H 3 O 2 , and HC1. Calculateα and K a' the acid dissociation constant, for each concentration.2

3. For calcium sulfate, determine Λ o from the results for NaCl, CaCl 2 , and Na 2 SO 4 , andcalculate the concentration in moles per liter of the original saturated solution of calciumsulfate, assuming Λ = Λ o .Illustrative ProblemAt 25°C, K = 3.40 x 10 -3 S m -1 for 0.00100 M NH 4 OH. Values of Λ o are: NH 4 C1 =0.01497, NaOH = 0.02481, NaCl = 0.01265 S m 2 mol -1 . Find the dissociation constant ofammonium hydroxide.Answer: 1.80 x 10 -5 .3