Acetic acid and naoh lies when approximately 26 ml of

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acetic acid and NaOH lies when approximately 26 mL of NaOH has been added to the acetic acid. We also measured pH for one trial of each set of titrations. For our titration of KHP with NaOH, it can be assumed that the equivalence point lies around when about 21 mL of NaOH has been added to the KHP solution. We can see this because the pH value for 20 mL is 6.30 and the pH value for 22 mL is 7.15. A pH of 7.00 would be somewhere in between 20 mL and 22 mL. We can also see from our pH calculations that the equivalence point for the titration of acetic acid with NaOH lies around when 25 mL of NaOH is added. This agrees with our other data because we have already established that, since the average mL of NaOH added is 25.875 mL, the equivalence point must be somewhere around this value. We also converted the molarity of the original sample before dilution into % V/V and the value we obtained was close to the 5% declared on the bottle. We got a value of 3.829%. This difference suggests some errors or dilutions that occurred during the experiment. We can conclude from this experiment that the
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equivalence point of a titration of KHP with NaOH is around 21 mL of NaOH and that the equivalence point of a titration of acetic acid with NaOH is around 25 mL of NaOH. Post-Lab Questions 1. Calculate and report molarity of NaOH solution for each trial of KHP, using color change as an indication of equivalence point. Calculate the average molarity of NaOH titrant (from a least 3 titrations; do not include in this average titrations where the equivalence point was not determined reliably).
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Molarity of NaOH for titration #1: 0.49 g KHP × = 0.0024 mol KHP 0.0024 mol NaOH/0.018 L NaOH = 0.133 M NaOH Molarity of NaOH for titration #2: 0.59 g KHP × = 0.00245 mol KHP 0.00245 mol NaOH/0.0238 L NaOH = 0.103 M NaOH Molarity of NaOH for titration #3: 0.51 g KHP × = 0.0025 mol KHP 0.0025 mol NaOH/0.0238 L NaOH = 0.105 M NaOH Molarity of NaOH for titration #4: 0.50 g KHP × = 0.0024 mol KHP 0.0024 mol NaOH/0.0235 L NaOH = 0.104 M NaOH Averages molarities (disregarding #1 because it is an outlier): (0.103 + 0.105 + 0.104)/3 = 0.104 M 2.
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