Monoprotic weak acid - 9.76 15.45 9.94 15.6 10.11 16.1 10.3...

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The K a & Molar Mass of a Monoprotic Weak Acid Elisha Steiner CHM152 26277 November 11 th , 2015. Dr Weide
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Abstract The purpose of the experiment was to determine the pKa, Ka, and molar mass of the unknown acid # 44. The pKa was found to be 4.18, the Ka was found to be 6.6x10 -5 and the molar mass was found to be 359.3287g/mol Introduction The difference between a weak acid, and strong acids can be as much as 10 orders of magnitude. Strong acids dissociate more completely than weak acids, meaning they produce higher concentrations of the conjugate base, and hydronium cation in solution. Another way to distinguish and acid is by its equivalent mass, because the equivalent mass of an acid or base equals the mass of the acid or base titrated divided by the number of equivalents. In this lab, the pK a of an unknown acid is determined through titration with NaOH, and graphing its pH vs. Volume. Data KPH .6037gg KPH.6089g KHP.6006g 20.20mL 20.10mL 19.80M .1463M .1483M .1485M Avg M of NaOH = .1477 #44 unknown = .4007g PH Volume in mL 2.91 0 3.33 1.98 3.6 3.1 3.84 4.75 4.04 6.45 4.24 8.2 4.44 9.9 4.85 11.3 5.05 12.5 5.43 13.4 5.62 14.25
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6.05 14.53 6.37 14.83 6.68 15 7.36 15 7.93 15.1 8.52 15.1 8.88 15.15 9.06 15.2 9.3 15.2 9.55 15.3
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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 10-pH . 10-4.18 = 6.6x10-5 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 6.6x10-5 , 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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