Lab Report 6 - -4 M Final [SCN-] = 1.0 x 10-4 M - 8.4 x...

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John Xu 4/6/07 Lab Report Experiment 6: Spectroscopic Determination of an Equilibrium Constant 1. [FeSCN 2+ ] for testtube 1 Moles KSCN = Molarity * Volume= 2.0 x 10 -4 M KSCN * 0.005 = 1.0 x 10 -6 Moles of SCN - = moles of FeSCN 2+ [SCN - ] = 1.0 x 10 -6 / 0.01 L = 1.0 x 10 -4 M [SCN - ] = [FeSCN 2+ ] = 1.0 x 10 -4 M 2. [FeSCN 2+ ] for testtube 2 Using Beer’s Law A 1 /C 1 = A 2 /C 2 0.443 / 1.0 x 10 -4 M = 0.373 / C 2 C 2 = 8.4 x 10 -5 M 3. [Fe 3+ ] for testtube 2 Initial [Fe 3+ ] = 0.040 M Final [Fe 3+ ] = 0.040 M - 8.4 x 10 -5 M = 0.040 M 4. [SCN - ] for testtube 2 Initial [SCN - ] = 1.0 x 10
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Unformatted text preview: -4 M Final [SCN-] = 1.0 x 10-4 M - 8.4 x 10-5 M = 1.6 x 10-5 M 5. K = [FeSCN 2+ ] / [Fe 3+ ][SCN-] = (8.4 x 10-5 M) / (0.040 M * 1.6 x 10-5 M) = 130 6. Equilibrium Concentrations Testtube [FeSCN 2+ ] [Fe 3+ ] [SCN-] K 1 1.0 x 10-4 M 2 8.4 x 10-5 M 0.040 M 1.6 x 10-5 M 130 3 7.8 x 10-5 0.016 M 2.2 x 10-5 M 220 4 6.2 x 10-5 0.0063 M 3.8 x 10-5 M 260 5 4.2 x 10-5 0.0026 M 5.8 x 10-5 M 280 6 2.0 x 10-5 0.00098 M 8.0 x 10-5 M 280 7. Average K = (130 + 220 + 260 + 280 + 280) / 5 = 230...
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This note was uploaded on 04/07/2008 for the course CHEM 211 taught by Professor Crane, b during the Spring '06 term at Cornell University (Engineering School).

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