# Conductimetric Titration and Gravimetric Determination of a Precipitate

Abstract: This experiment demonstrated that by titrating barium hydroxide, Ba(OH)2 solution with a sulfuric acid, .1 M H2SO4 solution the point of equivalence can be obtained. Since they were ionic compounds, then the lowest conductivity reading was the point of equivalence because at that reading they were both at a non-ionic state since all their ions have been completely reacted. A first when the H2SO4 was added the conductivity was high, 17.8 umho, then as more 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 quite different, wherein by tabulation the lowest was 5.3 umho, while the pH sensor graph had its lowest way below 5.3 umho. Therefore, there was an error; it could be that the solution was not properly mixed during titration. There was only enough 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 same as the H2SO4 which was .1 M. Introduction:

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The experiment was to demonstrate how to find the concentration of Ba(OH)2 needed to react with .1 M H2SO4; thus 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 the .1 M H2SO4 would cause an increase of conductance because of the added ions. Result:

There was only enough time for one trial. The graph below shows the theoretical result which was different from the displayed result. Sample
calculation: 4 x 106 /8.9 = .45 x 106 M Ba(OH)2.

Materials:

Labpro or CBL 2 interface
Conductivity probe
Ring stand
250mL beaker
Magnetic stirrer
Stinning bar
Filter funnel
10 mL pipet
Pipet bulb and pump
Ba(OH)2 solution
.1 M H2SO4
distilled water
50mL buret
Buret clamp