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364 Chapter 17 Aqueous Ionic Equilibrium = 1.26316 X 10⁻¹³ M and pH = -log [H₃O⁺] = -log (1.26316 X 10⁻¹³) = 12.90 Check: The units (none) are correct. The pH is a little lower than the initial pH, which is expected because this is a strong base. (d) Find: pH at equivalence point Solution: Because this is a strong acid-strong base titration, the pH at the equivalence point is neutral, or 7. (e) Find: pH after adding 5.0 mL of acid beyond the equivalence point Conceptual Plan: Use calculations from parts (b) and (c). Then the pH is only dependent on the amount of excess acid and the total solution volumes. Then mL excess L excess then [HCI], L excess mol HCI excess 1L mol M = 1000 mL L then L RbOH, L HCI to equivalence point, L HCI excess total L then L RbOH + L HCI to equivalence point + L HCI excess = total L mol excess total L [HCI] = [H₃O⁺] pH. Solution: 5.0 HCI X 1000 1L = 0.0050 L excess then 0.100 mol X 0.0050 = 0.00050 mol excess. Then 0.0250 L RbOH + 0.0288 L + 0.0050 L = 0.0588 L total volume. = 0.00050 mol HCI excess = 0.0085034 M excess 0.0588 L Because is a strong acid, [HCI] excess = [H₃O⁺]. Finally, pH = [H₃O⁺] = -log (0.0085034) = 2.07. Check: The units (none) are correct. The pH is dropping sharply at the equivalence point, the pH after 5 mL past the equivalence point should be quite acidic. 17.71 Given: 20.0 mL of 0.105 M HC₂H₃O₂ titrated with 0.125 M NaOH Other: = 1.8 X 10⁻⁵ (a) Find: initial pH Conceptual Plan: Because is a weak acid, set up an equilibrium problem using the initial concentration. So M [H₃O⁺] pH ICE table pH = -log Solution: HC₃H₃O₂(aq) + + C₂H₃O₂⁻(aq) [HC₂H₃O₂] [H₃O⁺] Initial 0.105 0.00 Change -x +x +x Equil 0.105 +x +x = [HC₂H₃O₂] = 1.8 = 0.105 Assume that x is small (x 0.105), so 0.105 = X = 0.105 and = 1.3748 X M = [H₃O⁺]. Confirm that the assumption is valid. 1.3748 10⁻³ X 100% = 1.3%