10 4 350 x 10 4 000125 00050 See attached graphs Calculations C Claims We found

# 10 4 350 x 10 4 000125 00050 see attached graphs

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10^-4 3.50 x 10^-4 0.00125 0.0050
See attached graphs Calculations C. Claims We found through this experiment that the initial concentration of H3PO4 was 0.04463 M. We found pKa1 to be 1.77 and pKa2 to be 1.85 and from there we were able to determine Ka1 to be 1.70 x 10^ -2 and Ka2 to be 1.41 x 10^-2. After we finally found the concentration of H3PO4 to be 0.005 M. D. Evidence For this experiment, we were given the molarity of NaOH to be 0.103. Following the procedure, we titrated 0.0500 M of H3PO4 with .103 M of NaOh and plotted the curve.
The curve told us where the equivalence points were, and using those points we multiplied liters by molarity to find moles of NaOH. Since the stoichiometry of the balanced reaction showed a 1:1 ratio of NaOH and H3Po4, we were able to assume that the moles would be equal to each other. Then, we used the equation pH=pKa + log (base/acid) to find the Ka value. First, we have to use pKa= - log(Ka). This allowed us to find the strength of the acid. For the cola, we used the same procedures to find the strength of the aid H3PO4 in the cola. E. Reflection a. No, we didn’t see the expected number of equivalence points for the triprotic acid because the third equivalence point is not detectable because it occurs at such a high pH that the value of Ka3 would be very very small. We also did not really find the second equivalence point because there wasn’t enough data collect, so calculations and conclusion could be off due to that error.

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• Fall '11
• HeatherRankin
• Chemistry, pH, Pk, Equivalence point, Phosphoric acid