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Chem 113 Week 12

by: Caroline Hurlbut

Chem 113 Week 12 Chem 113

Caroline Hurlbut
GPA 3.7

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Acid/base titrations, the common ion effect, the Henderson-Hasselbalch equation, and buffers.
General Chemistry II
Ingrid Marie Laughman
Class Notes
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This 2 page Class Notes was uploaded by Caroline Hurlbut on Friday April 15, 2016. The Class Notes belongs to Chem 113 at Colorado State University taught by Ingrid Marie Laughman in Spring 2016. Since its upload, it has received 12 views. For similar materials see General Chemistry II in Chemistry at Colorado State University.


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Date Created: 04/15/16
Acid/Base Titrations • progress of an acid/base titration can be shown on a titration curve • titration - solution of known concentration is used to determine concentration of unknown solution • a reaction with an acid and a base produces water and a salt • Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) —use when system is a buffer that undergoes no change • ex. You have a solution that is 0.20 M HC2H3O2 and 0.50 M C2H3O2- with Ka = 1.8 x 10^-5. Calculate the pH of the solution. pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.745 pH = 4.745 + log(0.50/0.20) = 5.14 • common ion effect - suppresses ionization of a weak acid by adding more of an ion that is a product of that equilibrium • ex. when NaCl and KCl are dissolved in the same solution, the Cl- ions are common to both salts Buffers • buffer - solution of weak acid and conjugate base or weak base and conjugate acid that resists minor changes in pH by neutralizing small amounts of acid/base • 2 types of buffers —weak acid/base + salt (buffer solution is given) —titration forms system of weak acid/base + salt (buffer solution is made) if a solution is a buffer, you can use the Henderson-Hasselbalch equation to calculate • pH • ex. Add 5 mL of 0.1 M HCl to a normal person’s blood. There are about 5.0 L of blood in the human body and [H2CO3] = 0.0024 M. Calculate the pH of the buffer solution. pH buffer of blood: H2O + H2CO3⁶HCO3- + H3O+ pKa H2CO3 = 6.1 ratio of HCO3-:H2CO3 = 20:1 blood pH= 7.4 pH = pKa + log([HCO3-]/[H2CO3]) Step 1: Calculate [H3O+] HCl + H2O→H3O+ + Cl- 5.0 x 10^-3 L x 0.1 mol HCl x 1 mol H3O+ = 0.0005 mol H3O+ L 1 mol HCl Step 2: Calculate [HCO3-] before and after adding HCl 20 x [H2CO3] = 20 x 0.0024 M= 0.048 M before 5.0 L x 0.048 mol HCO3- = 0.24 mol HCO3- L 0.24 mol - 0.0005 mol = 0.0479 M HCO3- after 5.005 L Step 3: Calculate [H2CO3] H3O+ + HCO3-→H2CO3 + H2O 0.0005 mol H3O+ x 1 mol H2CO3 = 0.0005 mol H2CO3 1 mol H3O+ 5.0 L x 0.0024 mol/L HCO3- + 0.0005 mol H2CO3 = 0.0025 M 5.005 L pH = 6.1 + log(0.0479 M/0.00250 M) pH = 6.1 + log(19.16) = 7.38 • Ratio log(K) pH [A-] = [HA] 1 0 =pKa [A-] = 10[HA] 10 1 =pKa + 1


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