CHEM WET LAB 2.pdf - Measurement of an Equilibrium Constant...

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Unformatted text preview: Measurement of an Equilibrium Constant - Aqueous Layer Concentration of standard 0.010}? [-1 Ema-M OK ”325203 solution Temperature of the Equilibrium 20.5 C WWI-m OK Mixture First Determination Aqueous Layer Final Buret 20.01 mL WWI-m 0" Reading Aqueous Layer Initial Buret 10.01 mL Emu-m 0|: Reading Volume of Aqueous Aliquot 10.00 mL 10.00 ml nm- OK ”325203 Solution Final Buret 33.99 mL mm .nm- OK Reading ”325203 Solution Initial Buret 1.01 mL WWI-m OK Reading Volume of Thiosulfate Titer 32.98 mL 32.98 ml EMF Mm- OK Show a complete sample Volumemazszofl = 32.98 mL = 0.03298 L OK calculation of how you calculated - = the number of molfi of Concentratl00{Na25203) 0.010}? M thiosulfate used in the first "(N325203) = C(NEZSZOB) X “"325203) titration of the aqueous phase = (0.010?3 H) x (0.03298 L) = 3.533354 x 10'4 mol 0325203 "(Nizszoal = ”(5203) = 3.533754 x 10“ mol 5203 Total number of moles of 3.54e—4 m0l 3.539E—4 mol EWF win- —.50 thiosulfate used in the first titration of the aqueous phase For every mole of reducible iodine 2 mol 2 mol mw- OK how many moles of thiosulfate are needed? Show a complete sample n([2} = n(Na25203) x (1 mol {2:012 m0l Na25203} OK calculation of how you calculated _ -4 thetotal number of moles of — 3.538354 3: 10 mol "325203 3:31 mol [2} reducible iodine in the aqueous {2 ITICll "325203) phase = 1.3693}? x 10'4 mol I2 Total Number of Moles of 1.??e—4 mol 0.00013? mol EME- r- OK Reducible Iodine in the Aqueous Phase Show a complete sample “(12): 13593;? x 10-4 m0l I2 OK calculation of how you calculated the total concentration of reducible iodine in the aqueous phase VHZ} = 10.00 mL = 0.01000 L C(12) = n([2) + Vflz) = 1.?693E'II' x 10'4 mol [2 + 0.01000 L = 0.01?693?? H [2 Total Concentration of Reducible 0.01??? H 0.01?? H WWI-m OK Iodine in the Aqueous Phase Second Determination Aqueous Layer Final Buret 30.02 mL EW- I 0" Reading Aqueous Layer Initial Buret 20.01 mL 0" Reading Volume of Aliquot 10.01 mL 10.01 ml OK N02520:; Solution Final Buret 32.89 mL 0" Reading NaZSZDE Solution Initial Buret 0.01 mL 0" Reading Volume of Thiosulfate Titer 32.88 mL 32.38 ml 0K Total Concentration of Reducible 0.01?6 H 0.01?52 H -.50 Iodine in the Aqueous Phase Third Determination Aqueous Layer Final Buret 40.02 mL mum-m 0K Reading Aqueous Layer Initial Buret 30.02 mL mum-m 0K Reading Volume of Aliquot 10.00 mL 10.00 ml mm-Im 0K Nazszflg Solution Final Buret 34.93 mL WWI-m 0K Reading Nazszog Solution Initial Buret 2.01 mL mm-m OK Reading Volume of Tnioeultete Titer 32.91 mL 32.92 ml mm-Im 0K Total Concentration of Reducible 0.01??? M 0.01?55 [-1 Emu“ —_50 Iodine in the Aqueous Phase Average of the Best Two Concentrations of Reducible Iodine in the Aqueous Layer which are the 'best two'_ total 111e best two total reducible iodine concentrations are of OK redumble '0de concentratwns? the first (0.01T69 H:- and third (0.01T66 H:- determinations. Average of the Best Two 0.01??? H 0.01??? H “mum 0K Concentrati o ns Calculate the difference between 3.?69—5 [-1 0.0000 H mm” n“- -.50 the best two concentrations Measurement of an Equilibrium Constant — CH2C|2 Layer First Determination CHZCIZ Layer Final Buret Reading 24.45 mL 0K CHZCIZ Layer Initial Buret Reading 1?.45 mL 0K Volume of CHECIE Aliquot 100 mL 100 ml OK l-lazszo3 Solution Final Buret 37.22 mL OK Reading ”325203 Solution Initial Buret 0.52 mL OK Reading Volume of Thiosulfate Titer 35.?0 mL 35.?0 ml WWI-m or: _ Show a complete sample Volumemazszog) = 36.?0 mL = 0.036?'0 L 0K calculation of how you calculated Concentration(N325203) = 0.010}? M the number of moles of thiosull‘ate used in the first "(N325203) = C("325233) >( “(Nazszosl' titration of the CHzclz phase = (0.01038 H) >( (0.036?“ L} = 3.93?91 x 10'4 mol [4325203 n(Na25203) = ”(5203) = 3.93791 x 10“ mol 52o3 Total number of moles of 3.9494 mol 3.930E—4 mol WWI-fl or: thiosulfate used in the first titration of the CHZCIZ phase For every mole of iodine how 2 mol 2 mol OK many moles of thiosull‘ate are needed? Show a complete sample n([2) = n(N325203} x (1 mol 12),!(2 mol "325203) OK calculation of how you calculated _ -4 the total number of moles of _ 3'93?91 X 1B ITIOI "325203 X L ITIOI I2} iodine in the CHZCIE phase {2 mol N325203) = 1.955955 x 15'4 mol 12 Total Number of Moles of Iodine 1.9? 52—4 mol 0.00019?‘I m5l 0K in the CHzClz Phase _ Show a complete sample “(1:2): 1353955 :1: 10-4 mol [2 OK calculation of how you calculated the concentration of iodine in the _ _ CH2CI2 phase VH2) — ?.00 mL — 0.00?00 L Caz} = “(Izl + W12} = 1.968955 x 10'4 mol [2 + 0.00?00 L = 0.02812?920 H12 Concentration Iodine in the 0.0281 H 0.0281 H Fe.- CHzclz We mm- 0|: Second Determination CHzClz Layer Final Buret Reading 31.49 mL OK CH2C|2 Layer Initial Buret Reading 24.45 mL OK Volume of Aliquot 3-".04 mL 104 ml OK NazSZDE Solution Final Buret 41.02 mL OK Reading NazSZDE Solution Initial Buret 5.01 mL OK Reading Volume of Thiosulfate Titer 36.01 mL 36.01 ml OK Total Concentration of Iodine in 0.0274 H 0.02?4 H WWI-m 0" the CHzClz Phase Third Determination CHZCIZ Layer Final Buret Reading 38.49 mL OK CHECIE Layer Initial Buret Reading 31.49 mL OK Volume of Aliquot 7.00 mL 3.00 ml OK NazSEOE Solution Final Buret 42.42 mL OK Reading ”325203 Solution Initial Buret 6.01 mL OK Reading Volume of Thiosulfate Titer 35.41 mL 35.41 ml mm-Im OK Total Concentration of Iodine in 0.02?9 H 0.02?9 H mum-IE 0" the CHzClz Phase Average of the Best Two Concentrations of Iodine in the CH2CI2 Layer which are the 'best two' iodine in The best two total reducible iodine concentrations are of OK CH2”: We" CONCEPVBPWS? the first (0.0281 M) and third (5.52:9 M) determinations. Average of the Best Two 5.5255 H 5.5255 H mm-Im OK Concentrations Calculate the difference between 2.22e4 0.0002 M mam-“n -.50 the best two concentrations Concentrations of Iodine Species Show a complete sample K = [[2]CHZC| + [[2]H20 = 1?!) OK calculation of how you calculated the concentration of I: in water [[21H20 = 0.028016?96 H + 175 = 1.591053 x 10'4 H Concentration OH; in water 1.59e—4 H 0.000159 H mm: mm OK Show a complete sample I - = I _ I OK calculation of how you calculated [ 3 ha” [TOT] [ ZJHZO _4 the concentration of 13' in water = g'giggiggg: : ' 1591863 >( ll] H Concentration of 13' in water U.Ul?5 H l].l]1?5 H mmF OK Show a complete sample [1—] = [I 20!] M _ [I -] OK calculation of how you calculated {3“} _ I 3 {5“} the concentration of I' in water ;g-igg4';4_f§iglb:515806 M Concentration of I' in water 0.132 H 0.183 M mm-mm OK I lat' my; “W'mfagf'g K: = [Ia—km] + (01(an x [[21H20) 0" ca cu l0l"l o ow you ca cu e _ _ the value of Kc in water 4;)l].l]1?"5158l]6 H T (0.182484193H x 1.591863e— = 602.9? Value of Kc in water 603 505 Temperature atwhich Kt is valid 20.5 C 20.5 C ...
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