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Unformatted text preview: 9.76 15.45 9.94 15.6 10.11 16.1 10.3 16.5 10.5 17.21 10.72 18.3 10.92 19.8 11.12 22.2 11.32 25.1 11.52 31.3 Calculations: To find the half equivalence point, I divided the equivalence point volume by 2. I then looked at my graph to find the pH at the half equivalence point. The pK a is equal to the pH, which was 4.18. The Ka is equal to the concentration of [H 3 O + ] which is also equal to 10pH . 104.18 = 6.6x105 To find the number of equivalents of NaOH, I took the volume at the equivalence point and multiplied by the molarity. 7.55mL(1/1000)(.1477/1) = .001115 To find the equivalent mass of the acid, I divided the mass of the acid used by the number of equivalents of acid. .4007g #44 / .001115 = 359.3287g/equivalent Conclusion: My final Ka value was 6.6x105 , my pKa value was 4.18, and my molar mass/equivalent turned out to be 359.3287g. I feel these numbers aren’t perfect, and I could have improved while taking pH reading, taking them at a smaller more precise measure throughout the experiment. I also feel error could have occurred while trying to read the buret readings....
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 Spring '11
 MCC
 pH, Pk, Equivalence point, half equivalence point

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