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Chapter 17 Aqueous Ionic Equilibrium 353 17.45 Given: 150.0 mL buffer of 0.15 M benzoic acid at pH = 4.25 Find: mass sodium benzoate Other: = 6.5 X 10⁻⁵ Conceptual Plan: Identify acid and base components then pH, [HC₇H₅O₂] [NaC₇H₅O₂] acid = base = C₇H₅O₂⁻ mL L then [NaC₇H₅O₂], L mol NaC₇H₅O₂ g NaC₇H₅O₂ 1L 1000 mL 144.10 NaC₇H₅O₂ M 1 mol NaC₇H₅O₂ Solution: pH = + log [base] = -log (6.5 X 10⁻⁵) + log [NaC₇H₅O₂] 0.15 = 4.25. Solve for [NaC₇H₅O₂]. [NaC₇H₅O₂] log [NaC₇H₅O₂] = 4.25 - 0.15 4.1870866 = 0.06291 = = 0.15 M 1.1559 [NaC₇H₅O₂] = 0.17338 M Convert to moles using M 0.17338 mol NaC₇H₅O₂ 1 L X 0.150 = 0.026007 mol NaC₇H₅O₂ 0.026007 X 144.10 1 NaC₇H₅O₂ = 3.7 NaC₇H₅O₂ Check: The units (g) are correct. The magnitude of the answer makes physical sense because the volume of solution is small and the concentration is low; much less than a mole is needed. 17.47 (a) Given: 250.0 mL buffer of 0.250 M HC₂H₃O₂ and 0.250 M NaC₂H₃O₂ Find: initial pH Other: = 1.8 X 10⁻⁵ Conceptual Plan: Identify acid and base components then M NaC₂H₃O₂ M C₂H₃O₂⁻ then acid = base = NaC₂H₃O₂(aq) Na⁺(aq) + pH Solution: Acid = HC₂H₃O₂, so [acid] = [HC₂H₃O₂] = 0.250 M. Base = C₂H₃O₂⁻. Because one C₂H₃O₂⁻ ion is generated for each NaC₂H₃O₂, [C₂H₃O₂⁻] = 0.250 C₂H₃O₂⁻ = [base]. Then Check: The units (none) are correct. The magnitude of the answer makes physical sense because the pH is equal to the pKₐ of the acid because there are equal amounts of acid and base. (b) Given: 250.0 mL buffer of 0.250 M HC₂H₃O₂ and 0.250 M NaC₂H₃O₂, add 0.0050 mol HCI Find: pH Other: = 1.8 X 10⁻⁵ Conceptual Plan: Part I: Stoichiometry: mL L then [NaC₂H₃O₂], L mol NaC₂H₃O₂ and [HC₂H₃O₂], L mol 1L 1000 mL write balanced equation then + NaC₂H₃O₂ HC₂H₃O₂ + NaCl mol NaC₂H₃O₂, mol mol HCI mol NaC₂H₃O₂, mol then set up stoichiometry table Part II: Equilibrium: mol NaC₂H₃O₂, mol HC₂H₃O₂, L, pH Copyright © 2017 Pearson Education, Inc.