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Chapter 17 Aqueous Ionic Equilibrium 359 Because [H₂CO₃] = 50.11872 = (0.0060 (0.12 + x) x) solve for x. Note that the ratio of moles is the same as the ratio of concentrations because the volume for both terms is the same. 50.11872(0.0060 x) = (0.12 + 0.30071 - 50.11872x = 0.12 + x 51.11872x = 0.18071 x = 0.0035351 mol NaOH then 0.0035351 mol NaOH X 40.00 NaOH = 0.14140 g NaOH = 0.14 g NaOH 1 NaOH Check: The units (g) are correct. The amount of base needed is small because the concentrations of the buffer components are very low. 17.57 Given: HC₂H₃O₂/KC₂H₃O₂, NH₃/NH₄Cl, and HCIO/KCIO potential buffer systems to create buffer at pH = 7.20 Find: best buffer system and ratio of component masses Other: = 1.8 X 10⁻⁵, = 1.8 X 10⁻⁴, = 1.76 10⁻⁵, = 2.9 X 10⁻⁸ Conceptual Plan: Calculate pKₐ of all potential buffer acids for the base Kb and = -log Kb 14 = + pKb choose the pKₐ that is closest to 7.20 then pH, [base]/[acid] mass base/mass acid pH = + [base] M M (base) (acid) Solution: For = -log Kₐ = -log (1.8 X 10⁻⁵) = 4.74 For = -log = -log (1.8 X 10⁻⁴) = 3.74 For NH₃/NH₄Cl: pKb = -log Kb = -log (1.76 X 10⁻⁵) = 4.75 Because 14 = pKₐ + pKb, pKₐ = 14 pKb = 14 4.75 = 9.25. And for HCIO/KCIO: pKₐ = -log -log (2.9 X 10⁻⁸) = 7.54 So the HCIO/KCIO buffer system has the pKₐ that is closest to 7.20. So pH = pKₐ + log [base] [acid] = 7.54 + [HCIO] [KCIO] = 7.20. Solve for log [KCIO] [HCIO] = 7.20 7.54 = -0.34 [HCIO] [KCIO] = = 0.457088 Then convert to mass ratio using 90.55 KCIO M (base) KCIO 0.457088 0.79 M (acid) 52.46 g HCIO HCIO Check: The units (none and g base/g acid) are correct. The buffer system with the closest to 10⁻⁷ is the best choice. The magnitude of the answer makes physical sense because the buffer needs more acid than base (and this fact is not overcome by the heavier molar mass of the base). 17.59 Given: 500.0 mL of 0.100 M HNO₂/0.150 M KNO₂ buffer and (a) 250 mg NaOH, (b) 350 mg (c) 1.25 g HBr, and (d) 1.35 g HI Find: whether buffer capacity is exceeded Conceptual Plan: mL L then [HNO₂], L mol and [KNO₂], L mol 1L M 1000 mL L then calculate moles of acid or base to be added to the buffer mol then M 1000 mg compare the added amount to the buffer amount of the opposite component. Ratio of base/acid must be between 0.1 and 10 to maintain the buffer integrity. Solution: 500.00 mL X 1000 1L mL = 0.5000 L then 0.100 mol HNO₂ X 0.5000 = 0.0500 mol HNO₂ and 0.150 mol KNO₂ 1L X 0.5000 L = 0.0750 mol KNO₂ 1 For 1 mol NaOH (a) NaOH: 250 X X 1000 40.00 = 0.00625 mol NaOH. Because the buffer contains 0.0500 mol acid, the amount of acid is reduced by 0.00625/0.0500 = 0.125 and the ratio of base/acid is still between 0.1 and 10. The buffer capacity is not exceeded. Copyright © 2017 Pearson Education, Inc.