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−] = [H2PO4−]×10( pH - pKa)= [ H2PO4−]×10(7.20-7.20 )=[ H2PO4−]And, since [HPO42−]+[H2PO4−]=0.05 M[HPO42−] = [H2PO4−]=0.025 MIn 100ml we have 100ml×1L1000 ml×0.025 M=0.0025mole 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 HPO42-to H2PO4-. The moles of protons is given by: 50 ml×1L1000ml×0.01M= 0.0005 moleThus, 0.0005 mole of HPO42−will be converted to H2PO4−orThere will be 0.0025- 0.0005 mole HPO42−and 0.0025+0.0005 mole H2PO4−pH=pKa+log10[HPO42−][H2PO4−]= 7.2+log100.0025−0.00050.0025+0.0005=7.02 (1)And[HPO42−] =(0.0025−0.0005) mole100 ml+ 50 ml×1000 ml1 L=0.0133 M[H2PO4−] =(0.0025+0.0005) mole100 ml+ 50 ml×1000ml1 L=0.0200 MNote: 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 H2P04- and HPO42- in this final solution?Answer: For NaOH, the same equations apply with one important difference: HPO42-is increased and H2PO4-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.