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358 Chapter 17 Aqueous Ionic Equilibrium (b) Given: 5.0 L of blood buffer Find: mass to lower pH to 7.0 Conceptual Plan: Final pH, then [HCO₃⁻],L mol and [base] pH = + L [H₂CO₃], L mol H₂CO₃ then write balanced equation then M + L mol mol mol HCI g HCI 36.46 HCI set up stoichiometry table 1 mol HCI Solution: pH = = 6.1 + = 7.0. Solve for log = 7.0 6.1 = 0.9 = = 7.9433. Then [H₂CO₃] [H₂CO₃] 0.024 mol X = 0.12 mol and 0.0012 mol X 5.0 L = 0.0060 mol H₂CO₃. L Because is a strong acid, [HCI] = [H⁺]. Set up a table to track changes: + H₂CO₃(aq) Before addition ≈ 0.00 mol 0.12 mol 0.0060 mol Addition - x After addition ≈ 0.00 mol (0.12 x) mol (0.0060 + x) mol Because [HCO₃⁻] [H₂CO₃] = 7.9433 = (0.0060 (0.12 + x) solve for Note that the ratio of moles is the same as the ratio of concentrations because the volume for both terms is the same. 7.9433(0.0060 + x) = (0.12 x) 0.0476598 + 7.9433x = 0.12 x 8.9433x = 0.07234 x = 0.0080888 mol then 0.0080888 X 36.46 g = 0.29492 g HCI = 0.3 g 1 Check: The units (g) are correct. The amount of acid needed is small because the concentrations of the buffer components are very low and the buffer starts only 0.4 pH unit above the final pH. (c) Given: 5.0 L of blood buffer Find: mass NaOH to raise pH to 7.8 Conceptual Plan: Final pH, pKₐ then mol and M L mol then write balanced equation then + + mol mol mol NaOH g NaOH 40.00 g NaOH set up stoichiometry table 1 mol NaOH Solution: pH = log = 6.1 + [H₂CO₃] = 7.8 Solve for [HCO₃⁻] = 7.8 6.1 = 1.7 [H₂CO₃] = 10¹.⁷ = 50.11872. Then 0.024 mol X = 0.12 mol and 0.0012 mol 1L X = 0.0060 mol H₂CO₃. Because NaOH is a strong base, [NaOH] = Set up a table to track changes: OH⁻(aq) + H₂CO₃(aq) + Before addition ≈ 0.00 mol 0.0060 mol 0.12 mol Addition x After addition ≈ 0.00 mol (0.0060 x) mol (0.12 + x) mol Copyright © 2017 Pearson Education, Inc.