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# H 2 po 4 10 ph pk a h 2 po 4 10 720 720 h 2 po 4 and

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] = [H 2 PO 4 ] × 10 ( pH - pK a ) = [ H 2 PO 4 ] × 10 (7.20-7.20 ) = [ H 2 PO 4 ] And, since [HPO 4 2 ] + [H 2 PO 4 ] = 0.05 M [HPO 4 2 ] = [H 2 PO 4 ] = 0.025 M In 100ml we have 100ml × 1L 1000 ml × 0.025 M = 0.0025 mole of each.

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Now, addition of 50 ml of 0.1M HCl accomplishes two things: (1) It dilutes the solution; and, (2) It introduces protons that will convert HPO 4 2- to H 2 PO 4 - . The moles of protons is given by: 50 ml × 1L 1000ml × 0.01M = 0.0005 mole Thus, 0.0005 mole of HPO 4 2 will be converted to H 2 PO 4 or There will be 0.0025 - 0.0005 mole HPO 4 2 and 0.0025 + 0.0005 mole H 2 PO 4 pH = pK a + log 10 [HPO 4 2 ] [H 2 PO 4 ] = 7.2 + log 10 0.0025 0.0005 0.0025 + 0.0005 = 7.02 (1) And [HPO 4 2 ] = (0.0025 0.0005) mole 100 ml + 50 ml × 1000 ml 1 L = 0.0133 M [H 2 PO 4 ] = (0.0025 + 0.0005) mole 100 ml + 50 ml × 1000 ml 1 L = 0.0200 M Note: This amount of acid added to 100 ml of water gives pH = 2.5. b. If 50 ml of 0.01 M NaOH is added to 100 ml of 0.05 M phosphate buffer at pH 7.2, what is the resultant pH? What are the concentrations of H 2 P0 4 - and HPO 4 2- in this final solution? Answer: For NaOH, the same equations apply with one important difference: HPO 4 2- is increased and H 2 PO 4 - is decreased. Thus, the fraction in the first equation (above) is inverted. This gives pH = 7.38. Note: If added to water instead, this amount of NaOH would result in a solution with pH = 11.5.