Post Lab #7 - Jacob Sutker Peter Widger Chem 211 Post Lab#7...

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Jacob Sutker Peter Widger October 31, 2006 Chem 211 Post Lab #7 Part C: Sample calculation - mL NaOH pH ΔpH ΔmL ΔpH/ΔmL 2.02 1.30 1.30 – 1.25 = .05 2.02-1.01= 1.01 .05/1.01 = .050 3.01 1.38 1.38-1.30 = .08 3.01-2.02= 0.99 .08/0.99 = .081 pH vs. ml NaOH 0 2 4 6 8 10 12 14 0 10 20 30 ml NaOH pH Series1 change in pH vs. mL NaOH 0 10 20 30 40 50 60 0 10 20 30 mL NaOH change in pH Series1
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Part D: Sample Calculation -
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Unformatted text preview: mL NaOH pH ΔpH ΔmL ΔpH/ΔmL 1.10 2.86 2.86-2.73= .13 1.10-0 = 1.10 .13/1.10 = .118 2.11 2.92 2.92-2.86= .06 2.11 – 1.10 = 1.01 .06/1.01 = .059 pH vs. mL NaOH 2 4 6 8 10 12 14 5 10 15 20 mL NaOH pH Series1 change in pH vs. mL NaOH 5 10 15 20 25 5 10 15 20 mL NaOH change in pH Series1...
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  • Fall '06
  • CRANE, B
  • Chemistry, pH, Harshad number, mL NaOH, 06 mL, 8 6 4 2 0 0 5 10 mL, 8 6 4 2 0 0 10 ml

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