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Chapter 17 Aqueous Ionic Equilibrium 363 17.69 Given: 25.0 mL of 0.115 M RbOH titrated with 0.100 M (a) Find: initial pH Conceptual Plan: Because RbOH is a strong base, it will dissociate completely, [RbOH] = [OH⁻] [H₃O⁺] pH Kw = pH = -log [H₃O⁺] Solution: Because RbOH is a strong base, [RbOH] excess = Kw = so [H₃O⁺] = Kw = 1.0 0.115 10⁻¹⁴ = 8.69565 X 10⁻¹⁴ M and pH = -log[H₃O⁺] = -log (8.69565 X 10⁻¹⁴) = 13.06. Check: The units (none) are correct. The pH is reasonable because the concentration is greater than 0.1 M, and when the base dissociates completely, the pH becomes greater than 13. (b) Find: volume of acid to reach equivalence point Conceptual Plan: Write balanced equation then mL L then [RbOH], L mol RbOH then + RbOH RbCl + 1000 mL M = mol L set mol base (RbOH) = mol acid (HCI) and [HCI], mol HCI L HCI mL HCI balanced equation has 1:1 stoichiometry mol 1000 mL M = L 1L Solution: 25.0 RbOH X 1L = 0.0250 L RbOH then 1000 mL 0.115 mol RbOH L X 0.0250 L = 0.002875 mol RbOH. So mol base = mol RbOH = 0.002875 mol = mol HCI then 0.002875 X 0.100 1L HCI = 0.02875 X 1000 mL = 28.8 mL Check: The units (mL) are correct. The volume of acid is greater than the volume of base because the concen- tration of the base is a little greater than that of the acid. (c) Find: pH after adding 5.0 mL of acid Conceptual Plan: Use calculations from part (b). Then mL L then [HCI], L mol HCI then 1L 1000 mL M = L mol RbOH, mol mol excess RbOH and L RbOH, L HCI total L then set up stoichiometry table L RbOH + L HCI = total L mol excess RbOH, L [RbOH] = [OH⁻] [H₃O⁺] pH. L Kw = pH = -log [H₃O⁺] 1L 0.100 mol Solution: 5.0 X = 0.0050 L then 1000 mL X 0.0050 L = 0.00050 mol Because HCI is a strong acid, = [H₃O⁺]. Set up a table to track changes: HCl(aq) + RbOH(aq) RbCl(aq) + Before addition 0.00 mol 0.002875 mol 0.00 mol - Addition 0.00050 mol After addition ≈ =0.00 mol 0.002375 mol 0.00050 mol Then 0.0250 L RbOH + 0.0050 L = 0.0300 L total volume. So mol excess base = mol RbOH = 0.002375 mol RbOH 0.002375 mol in 0.0300 L, [RbOH] = = 0.0791667 M. Because RbOH is a strong 0.0300 L base, [RbOH] excess = Kw = [H₃O⁺] [H₃O⁺] = Kw = 1.0 0.0791667 10⁻¹⁴ Copyright © 2017 Pearson Education, Inc.