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Chapter 17 Aqueous Ionic Equilibrium 357 final pH, [NaF]/[HF] and mL L then [HF], L mol HF and [NaF], L mol NaF 1L 1000 mL M L then write balanced equation then NaOH + HF NaF + mol HF, mol NaF, [NaF]/[HF] mol NaOH g NaOH. 40.00 g NaOH set up stoichiometry table 1 mol NaOH Finally, when the buffer concentrations are raised to 0.350 M, simply multiply the g NaOH by the ratio of concentrations Solution: initial pH = = -log (3.5 X 10⁻⁴) = 3.46 then pH = + log [base] [acid] = -log (3.5 X 10⁻⁴) + log [NaF] = 4.00. Solve for [NaF]/[HF]. log [NaF] [HF] = 4.00 3.46 = 0.54 [NaF] [HF] = = 3.5. 350.0 mL X 1000 1L = then 0.150 mol HF 0.150 mol NaF X 0.3500 = 0.0525 mol HF and 1L 0.3500 = 0.0525 mol NaF 1L Set up a table to track changes: NaOH(aq) HF(aq) NaF(aq) + H₂O(aq) Before addition =0.00 mol 0.0525 mol 0.0525 mol Addition x After addition =0.00 mol (0.0525 x) mol (0.0525 + x) mol - Because [NaF] [HF] = 3.5 = (0.0525 (0.0525 + x) x) mot' 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. 3.5(0.0525 - x) = (0.0525 + x) 0.18375 - 3.5x = 0.0525 + x 0.13125 = 4.5x x = 0.029167 mol NaOH then 0.029167 X 40.00 1 NaOH = 1.1667 g NaOH = 1.2 g NaOH To scale the amount of NaOH up to a 0.350 M HF and NaF solution, multiply the NaOH mass by the ratio of 0.350 M concentrations. 1.1667 g NaOH X 0.150 M = 2.7 g NaOH Check: The units (g) are correct. The magnitude of the answers makes physical sense because there is much less than a mole of each of the buffer components; so there must be much less than a mole of NaOH. The higher the buffer con- centrations, the higher the buffer capacity and the mass of NaOH it can neutralize. 17.53 (a) Yes, this will be a buffer because is a weak base and is its conjugate acid. The ratio of base to acid is 0.10/0.15 = 0.67, so the pH will be within 1 pH unit of the pKₐ. (b) No, this will not be a buffer solution because HCI is a strong acid and NaOH is a strong base. (c) Yes, this will be a buffer because HF is a weak acid and the NaOH will convert (20.0/50.0) X 100% = 40% of the acid to its conjugate base. (d) No, this will not be a buffer solution because both components are bases. (e) No, this will not be a buffer solution because both components are bases. 17.55 (a) Given: blood buffer 0.024 M and 0.0012 M = 6.1 Find: initial pH Conceptual Plan: Identify acid and base components then M HCO₃⁻, pH acid = base = Solution: Acid = H₂CO₃, so [acid] = [H₂CO₃] = 0.0012 M. Base = [base] = [HCO₃⁻] = 0.024 M Then pH = + log [base] = 6.1 + log 0.0012 0.024 M = 7.4. Check: The units (none) are correct. The magnitude of the answer makes physical sense because the pH is greater than the pKₐ of the acid because there is more base than acid. Copyright © 2017 Pearson Education, Inc.