Conductometric Titration
Essay Preview: Conductometric Titration
Report this essay
UNIVERSITY OF DAR ES SALAAMCOLLEGE OF NATURAL AND APPLIED SCIENCES (CONAS).[pic 1]CHEMISTRY DEPARTMENT.CH 341: CONDUCTOMETRIC TITRATION.Name of Candidates: GEOFFREY MACHELE 2015-04-03235 MDUDA EVANCE 2015-04-03237 Date of experiment: 6th December, 2017. Date of submission: 13th December, 2017. ABSTRACTThe aim of the experiment was to measure the conductance as a function of volume through titrimetric and volumetric methods. Titrations were carried out involving strong base-strong acid, strong base-weak acid, and precipitation titration. The endpoint or equivalence point was indicated by a sharp break in the conductance curve. The unknown concentrations of NaOH, CH3COOH and BaCl2 were 0.04 M, 0.1 M and 0.0905 M respectively.THEORYTitration is a method of determining the concentration of an unknown solution (the analyte) by reacting it completely with a standardized reagent that is a solution of known concentration (the titrant). The point at which all of the analyte is consumed is the equivalence point. It can be determined using a pH electrode or by use of a conductivity cell to measure the conductivity of the analyte.Conductometry is often applied to determine the total conductance of a solution or to analyse the end point of titration that include ions. Electrolytic conductivity is a measure of the ability of a solution to carry electric current, electric solutions conduct electric current by the migration of ions under the influence of electric field. (Barrow 1979). The conductivity of a solution is dependent on several factors, including the concentration of the solute, the degree of dissociation of the solute, the valence of the ion(s) present in the solution, the temperature, and the mobility of the ions in the solution.According to Ohm’s Law, the current strength flowing through a conductor is directly proportional to the potential difference (E) and inversely proportional to the resistance (R) of the conductor. I = E/R……………………………………………………………………… (1)The electric conductivity of a solution, K is given by the following equation;K = kL ohm-1 m-1…………………………………………………………… (2)[pic 2]Where k is the cell constant and L is the conductance.[pic 3]Equivalent conductivity of an electrolyte, ˄ is defined by the equation ˄ = × 10 -3 ohm-1 m2 equiv-1……………………………………… (3)[pic 4]Where C is the concentration given in equivalent per litre. The equivalent conductivity can be considered as the sum of ionic contribution ˄= ƛ+1 + ƛ-1………………………………………………………………… (4)
Where ƛ+1 and ƛ-1 are the ionic equivalent conductivities for the positive and negative ion respectively. The combination of equation 3 and 4 gives. K × 10-3 = c ƛ+1 + c ƛ-1………………………………….……………………. (5)A mixed solution may contain several different ions, and the electrolytic conductivity of the solution is equal to the sum of the contribution from the different ionic species. K × 10-3 = ∑ici ƛi……………………………………………………… (6)At infinity dilution the ionic equivalent conductivities are independent of each other. When a solution of NaOH is titrated with HCl, the electrolytic conductivity of the mixed solution at any stage of titration is given by the following equationK × 10-3= [Na+] ƛ(Na+) +[OH-] ƛ(OH-) +[H+] ƛ(H+) + [Cl-] ƛ(Cl-)…………………………. .(7)During titration the OH- and H+ are consumed by the neutralization reaction, before the equivalence point the concentration of H+ is negligibly small and OH- is gradually replaced by Cl- in the solution. After equivalence point the concentration of OH- is negligibly small and the excess HCl leads to an increase in the concentration of both H+ and Cl-, these concentrations change lead to change in the electrolytic conductivity of the solution.When small concentration of H+ is neglected the equation (07) can written as K × 10-3= ƛ(Na+) + ƛ(OH-) + ƛ(Cl-)………………………………………… .. (8)[pic 5][pic 6][pic 7]Where a is the original number of moles of NaOH, x is the number of moles of HCl at any stage of titration and V is the volume of the titration.Then according to equation (02) we have; L ×const = {ƛ(Na+) + ƛ(OH-) }- ƛ(OH-) – ƛ(Cl-)}……………………………….. (9)[pic 8][pic 9] Given,, {ƛ(Na+) = K1 and ƛ(OH-) = K2 [pic 10][pic 11]