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CHEMISTRY 101L DATA SUMMARYEXPT.Conductometric Titration55Name name Lab Section 432IntroductionFigure 1. The conductivity minimum is 33µS at 4.47mL and this graph shows how from this minimum itwill increase from this point on. It will actually increase all the way to 164µS which is close to what itwas initially at the beginning. It’s noticeable how the conductivity is almost where it was at the beginning.Data Table 1 Volume and Conductivity (Gross)Volume of Buret Gross (mL) Corrected Volume Gross (mL) Conductivity Gross (µS) 7.1 mL 0 mL 166 µS7.61 mL 0.51 mL 152 µS8.18 mL 1.08 mL 134 µS8.52 mL 1.42 mL 122 µS9.02 mL 1.92 mL 98 µS9.66 mL 2.56 mL 72 µS10.18 mL 3.08 mL 49 µS10.65 mL 3.55 mL 37 µS11.09 mL 3.99 mL 34 µS11.57 mL 4.47 mL 33 µS12.08 mL 4.98 mL 56 µS12.69 mL 5.59 mL 94 µS13.23 mL 6.13 mL 122 µS14 mL 6.9 mL 164 µSFigure 2. The conductivity minimum on this graph is 17µS at 4.23mL and this graph shows how it changesand continues to increase from the minimum all the way to 205. Although the conductivity sensitive wasdecreasing rather slowly when it reached the minimum it started to increase rather rapidly. It’s noticeablein this trial how the conductivity at the end exceeded what it was at the start. Data table 2Volume and Conductivity (Sensitive) Volume of Buret Sensitive(mL) Corrected Volume Sensitive (mL) ConductivitySensitive(µS)14mL 0 mL 122 µS15.01mL 1.01 mL 107 µS16.01mL 2.01 mL 84 µS16.26mL 2.26 mL 73 µS16.5mL 2.5 mL 65 µS16.72mL 2.72 mL 56 µS16.95mL 2.95 mL 45 µS17.21mL 3.21 mL 36 µS17.36mL 3.36 mL 31 µS17.5mL 3.5 mL 28 µS17.62mL 3.62 mL 26 µS17.71mL 3.71 mL 23 µS17.8mL 3.8 mL 21 µS17.91mL 3.91 mL 19 µS18.06mL 4.06 mL 18 µS18.18mL 4.18 mL 17 µS18.23mL 4.23 mL 17 µS18.4mL 4.4 mL 25 µS19.9mL 5.9 mL 84 µS20.62mL 6.62 mL 115 µS21.52mL 7.52 mL 150 µS22.63mL 8.63 mL 205 µSCalculations Trial 2 X=mlx+b y=m2x+b2 m1x+b1=m2x+b2Y1= -28.141x+130.82 y2=41.963x-161.54 -28.141x+130.82=41.963x-161.54 -28.141x-41.963x=-161.54-130.82 X(-28.141-41.963)= -161.54-130.82 x=(-161.54-130.82)/(-28.141-41.963)= -292.36/-70.104=4.1704mL.01M H2SO4 1molsH2SO4/4.1704*10^-3L4.1704*10^-5molsH2SO4.005M Ba(OH)2=mols/.01L.00005molsBa(OH)2.00005/4.1704*10^-5=11H2SO4:1Ba(OH)2Trial 1X=mlx+b y=m2x+b2 m1x+b1=m2x+b2Y1=-33.452x+165.23 y2=54.62x-212.84-33.452x+165.23=54.62x-212.84 -33.452x-54.62x= -212.84-165.23X(-33.452-54.62)= -212.84-165.23 x=(-212.84-165.23)/(-33.452-54.62)= (-378.07/-88.072)=4.2927mL.01mH2SO4 1molsH2SO4/4.2927*10^-3L4.2927*10^-5molsH2SO4.005M Ba(OH)2=mols/.01L.00005molsBa(OH)2.00005/4.2927 *10^-5=11H2SO4:1Ba(OH)2Balanced EquationBa(OH)2aq+H2SO4 l Ba(SO4)s+2H2OlConclusion: The conductivity went down and then goes back up because due to the ion concentration from the acid-base reaction changing, then this also changes the conductivity. During an acid-base titration the H+ and OH- ions react to form H20 which is neutral. We are adding more sulfuric acid to the barium hydroxide solution and therefore changing the conductivity. It is initially an ion rich solution and continues to dilute more to water with a precipitate. Since there are no more ions to conduct electricity then the conductivity will drop to that of water. The results show how the conductivity in both the gross and sensitive trail goes down and then goes back up. The reasoning is explained in the first paragraph and the minimum for the gross trial is 33µS while theminimum for the sensitive trial is 17µS. What is noticeable is how in the first trial the last data we got almost reached what the conductivity was initially since the last point was 164 µS and the first point was 166 µS. Actually the first pint was higher. However in the second trial the last data which was 205µS is higher than the first point which was 122 µS by a lot. Although we tried to be as accurate as possible a possible error that occurred is that we added solution too soon once. When we got the equations for both trials then we were able to figure out how to balance the equation given at the beginning of the lab. The information we got for our data table was accurate enough to where we were able to see how to balance the equation. Since we found out that the ratio for H2SO4 to Ba(OH)2 is 1:1 then we knew we just had to balance and make both sides have the same amount of each element. We didn’t have too much variances from the expected results because the data went down and went back up as planned and we were able to figure out how to balance the equation. So this experiment did show how conductivity works with titration when it comes to acid-base reactions and the balanced equation confirms the conservation of

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