Conductimetric Titration and Gravimetric Determination of a Precipitate Essay

rook This experiment demonstrated that by titrating barium hydroxide, Ba(OH)2 solution with a sulfuric acid, .1 M H2SO4 solution the forecast of equality can be obtained. Since they were ionic compounds, then the lowest conductivity reading was the point of equivalence be compositors case at that reading they were both at a non-ionic state since all their ions pitch been completely reacted. A first when the H2SO4 was added the conductivity was high, 17.8 umho, then as more(prenominal) H2SO4 was added it went to its lowest, 5.3 umho. The subsequent adding of more H2SO4 caused the conductivity to go again to a new peak, 10.3 umho, this was followed by another decrease in conductivity to 8.9 umho, from then on, as more H2SO4 was added the conductivity increased continuously until the end of the experiment. The tabulated resulted graph and the graph displayed on the pH sensor were preferably different, wherein by tabulation the lowest was 5.3 umho, while the pH sensor graph had its lowest way downstairs 5.3 umho. Therefore, in that respect was an error it could be that the solution was not properly mixed during titration. There was only lavish time for one trial. From calculation, the molarity of the Ba(OH)2 between 0.45-0.54 molarity when the conductivity was between 8.9 umho and 9.3 umho respectively. The molarity of the Ba(OH)2 should be the equivalent as the H2SO4 which was .1 M. IntroductionThe experiment was to demonstrate how to find the absorption of Ba(OH)2 needed to react with .1 M H2SO4 thereof conductimetric titration was used. The theory is that during titration as the solutions react the ions in both solutions cause the conductance of electricity. When the reaction stops, meaning that all the ions have been removed from the reactants then the conduction would be at the lowest point. That is the point of equivalence wherein the ratios of both solutions are the same. In this case both would be 0.1 molar. From then on, any more addition of t he .1 M H2SO4 would cause an increase of conductance because of the added ions. ResultThere was only enough time for one trial. The graph below shows the suppositional result which was different from the displayed result. Samplecalculation 4 x 106 /8.9 = .45 x 106 M Ba(OH)2.MaterialsLabpro or CBL 2 interfaceConductivity probeRing stand250mL beakerMagnetic stirrerStinning barFilter paper- picturesque dradeFilter funnel10 mL pipetPipet bulb and pumpBa(OH)2 solution.1 M H2SO4distilled water50mL buretBuret secure50 mL graduated cylinderUtility clampDiscussionEarlier in the experiment as the 0.1 M H2SO4 was being added the displayed graph showed a unseasonable result because the ionization data was collected too early before all the ions had been removed, thus there was a misleading result that the point of equivalent was reached. Later as more of the acid was added the sure point of equivalence was found, which was 8.9 umho. If there was enough time then the experiment could be red one in a more timely fashion. Could it be that the experiment was prearranged to give a faulty result just for a learning experience? ConclusionConductivty titration is another method that can be used to find the dousing of an unknown solution, albeit that the experimenter must be patient so that the ionization results can be had at the equimolarconcentration of both solutions. The error was evident because the acid was 0.1 M H2SO4, thus according to the readings, with 1 mL of 0.1 M H2SO4 and at 5.3 umho, that should have been the point of equivalence, meaning that at 1 mL of 0.1 M H2SO4 both substances would be equimolar, but that was not the case. Equimolarity was achieved at 4 mL of 0.1 M H2SO4 with the conductivity is 8.9 umho as displayed on the pH sensor graph.