University:
Amherst College
Course:
CHEM 371 | Inorganic Chemistry
Academic year:

2019
Views:

178

Pages:

3
Author:

Michael Haley

Acid/Base Titration Practice Problems 1. What is the pH of the solution when 25.00 mL 0.016 M lactic acid (HC3H5O3, Ka = 1.4x10-4) is titrated with the following volumes of 0.010 M NaOH? A. 0.00 mL 1.4x10-4 = (x)(x)/(0.016-x) x = 0.00143 pH = -log(0.00143) = 2.84 B. 10.00 mL HC3H5O3 + OH— C3H5O3— + H2O B 0.00040 0.00010 0 --+x --C -x -x A 0.00040-x 0.00010-x x --0.00030 0 0.00010 --pH = 3.45 (3.38… Henderson-Hasselbalch answer) C. 20.00 mL HC3H5O3 + OH— C3H5O3— + H2O B 0.00040 0.00020 0 --+x --C -x -x A 0.00040-x 0.00020-x x --0.00020 0 0.00020 --pH = 3.88 (3.85… Henderson-Hasselbalch answer) D. 30.00 mL HC3H5O3 + OH— C3H5O3— + H2O B 0.00040 0.00030 0 --+x --C -x -x A 0.00040-x 0.00030-x x --0.00010 0 0.00030 --pH = 4.35 (4.33… Henderson-Hasselbalch answer) E. 40.00 mL HC3H5O3 + OH— C3H5O3— + H2O B 0.00040 0.00040 0 ---x +x --C -x A 0.00040-x 0.00040-x x --0 0 0.00040 --ICE table with C3H5O3— + H2O  HC3H5O3 + OH— (Kb) 7.14x10-11 = (x)(x)/([0.00040 mol/0.0650 L] – x) x = 6.63x10-7; so pOH = 6.18 and pH = 7.82 2. F. 50.00 mL HC3H5O3 + OH— C3H5O3— + H2O B 0.00040 0.00050 0 --C -x -x +x --A 0.00040-x 0.00050-x x --0 0.00010 0.00040 --Excess strong base, so pOH = -log(0.00010 mol/0.0750 L) = 2.88 and pH = 11.12 G. 60.00 mL HC3H5O3 + OH— C3H5O3— + H2O B 0.00040 0.00060 0 --C -x -x +x --A 0.00040-x 0.00060-x x --0 0.00020 0.00040 --Excess strong base, so pOH = -log(0.00020 mol/0.0850 L) = 2.63 and pH = 11.37 What is the pH of the solution when 25.00 mL 0.024 M morphine (C17H19NO3, Kb = 1.6x10-6) is titrated with the following volumes of 0.030 M HCl? A. 0.00 mL 1.6x10-6 = (x)(x)/(0.024-x) x = 0.000195 pOH = -log(0.000195) = 3.71, so pH = 14 – 3.71 = 10.29 B. 5.00 mL C17H19NO3 + H3O+ HC17H19NO3 + H2O B 0.00060 0.00015 0 --C -x -x +x --A 0.00060-x 0.00015-x x --0.00045 0 0.00015 --pH = 8.68 (8.68… Henderson-Hasselbalch answer) C. 10.00 mL C17H19NO3 + H3O+ HC17H19NO3 + H2O B 0.00060 0.00030 0 --C -x -x +x --A 0.00060-x 0.00030-x x --0.00030 0 0.00030 --pH = 8.20 (8.20… Henderson-Hasselbalch answer) D. 15.00 mL C17H19NO3 + H3O+ HC17H19NO3 + H2O B 0.00060 0.00045 0 --C -x -x +x --A 0.00060-x 0.00045-x x --0.00015 0 0.00045 --pH = 7.73 (7.73… Henderson-Hasselbalch answer) E. 20.00 mL C17H19NO3 + H3O+ HC17H19NO3 + H2O B 0.00060 0.00060 0 ---x -x +x --C A 0.00060-x 0.00060-x x --0 0 0.00060 --ICE table with HC17H19NO3 + H2O  C17H19NO3 + H3O+ (Ka) 6.25x10-9 = (x)(x)/([0.00060 mol/0.0450 L] – x) x = 9.126x10-6; so pH = 5.04 F. 25.00 mL C17H19NO3 + H3O+ HC17H19NO3 + H2O B 0.00060 0.00075 0 --C -x -x +x --A 0.00060-x 0.00075-x x --0 0.00015 0.00060 --Excess strong acid, so pH = -log(0.00015 mol/0.0500 L) = 2.52

AcidBase Titration Practice Problems

Related Documents

Description

Related Documents

Want to Download This Document?

Free up your schedule!

Take 5 seconds to unlock