soln 5 bis 102

# soln 5 bis 102 - K a ’ x K o ’ K a x K o = 6.6 x 10-9 x...

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Solutions for Assignment 5 - Segel Book: Page 93 Problem 53 (Note: This problem is similar to Problem 1-52 in the Segel book, Page 84.) The dissociation of H 2 CO 3 is shown below. H + + CO 3 = H + + HCO 3 - H 2 CO 3 CO 2 + H 2 O pKa 2 = 10.25 pKa 1 = 6.1 At pH 7.4: HCO 3 - would be the Bronsted base = .024 M CO 2 or H 2 CO 3 would be the Bronsted acid = .0012 M If add .0032 M H + , the base would be .024 - .0032 = .0208 M, but the concentration of the acid would remain the same because of the release of CO 2 . The pH at the end would be = pKa 1 + log 0208/.0012 pH = 6.1 + log 17.3 = 6.1 +1.24 = 7.34

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Problem 55
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Unformatted text preview: K a ’ x K o ’ K a x K o = 6.6 x 10-9 x 1 = 2.4 x 10-7 x K o ’ K o ’ = .0275 Problem 56 HHgbO 2 HHgb + O 2 Hgb HgbO 2 + H + + H + + O 2 pKa = 8.18 pKa = 6.62 In presence of oxygen: 7.4 = 6.62 + log HgbO 2 /HHgbO 2 .78 = log HgbO 2 /HHgbO 2 6 = HgbO 2 /HHgbO 2 After release of oxygen: 7.4 = 8.18 + log Hgb/HHgb-.78 = log Hgb/HHgb .166 = Hgb/HHgb x = HHgb .166x = Hgb 1.166 x = 1 mole x = .858 mole HHgbO 2- HHgb .858 - .143 = .715 mole x = HHgbO 2 6x = HgbO 2 7x = 1 mole x = .143 mole...
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soln 5 bis 102 - K a ’ x K o ’ K a x K o = 6.6 x 10-9 x...

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