20 mL of 0.1 M glycylglycine hydrochloride (aq) was placed into a beaker, and 0.

Question

20 mL of 0.1 M glycylglycine hydrochloride (aq) was placed into a beaker, and 0.2 M NaOH (aq) solution was added in 1 mL aliquots until either 25 mL of NaOH was added or the pH of the glycylglycine hydrochloride and NaOH mixture was 12. From the graph below, the pKa value for the carboxylic acid group and the pKa value for the amine group must be derived. I am unsure how I would do this

The chart below is the data used for the graph:

mL of 0.2 M NaOH Added pH Reading 0 2.1 1 2.3 2 2.6 3 2.9 4 3.1 5 3.3 6 3.6 7 4.0 8 4.7 9 7.3 10 7.7 11 8.0 12 8.2 13 8.5 14 8.7 15 9.0 16 9.5 17 11.0 18 11.9 19 12.1 Titration of 0.1M Glycylglycine 14 12 10 o 1 2 3 4 5 67 8 9 10 11 12 13 14 15 16 17 18 19 Volume of NaOH Added (mL)

Explanation / Answer

vol of base used at first equivalence point = 9.0 mL

vol of base at half way to first equivalence point = 4.5 mL

pKa1 =pH = 3.2

vol of base used at second equivalence point = 16.5 mL

vol of base at half way to second equivalence point = 12.75 mL

pKa2 =pH = 8.5

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