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Chem 113 Week 11

by: Caroline Hurlbut

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Chem 113 Week 11 Chem 113

Caroline Hurlbut
CSU
GPA 3.7

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These notes go through some calculations involving Ka and Kb values, such as how to determine whether a solution with a given Ka and Kb value is acidic, basic, or neutral, and finding the pH of a s...
COURSE
General Chemistry II
PROF.
Ingrid Marie Laughman
TYPE
Class Notes
PAGES
1
WORDS
CONCEPTS
Chemistry
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This 1 page Class Notes was uploaded by Caroline Hurlbut on Friday April 8, 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 18 views. For similar materials see General Chemistry II in Chemistry at Colorado State University.

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Date Created: 04/08/16
Calculations with Ka & Kb Values • for a conjugate acid-base pair, Ka x Kb = Kw, where Kw = 1.0 x 10^-14 at 25˚C ex. Is a solution of NH4HC2O4 at 25˚C acidic, basic, or neutral? • Ka C2H2O4 = 5.6 x 10^-2 Kb NH3 = 1.8 x 10^-5 Solve for new Ka for NH4+ + H2O⁶NH3 + H3O+ Ka* = 1.0 x 10^-14 = 5.56 x 10^-10 *Use Ka* = Kw 1.8 x 10^-5 Kb Solve for new Kb for C2H2O4 + H2O⁶HC2H2O4+ + OH- Kb* = 1.0 x 10^-14 = 1.79 x 10^-13 *Use Kb* = Kw 5.6 x 10^-2 Ka Choose bigger K value 5.56 x 10^-10 > 1.79 x 10^-13 Since Ka* > Kb*, NH4HC2O4 is acidic (at 25˚C). • ex. Calculate pH of a 0.10 M NH4NO3 solution. Kb NH3 = 1.8 x 10^-5 NH4+(aq) + H2O(l) ⁶ NH3(aq) + H3O+(aq) I 0.10 M 0 M 0 M C -x +x +x E 0.10 - x x x Ka* = 1 x 10^-14 = 5.56 x 10^-10 1.8 x 10^-5 5.56 x 10^-10 = x^2 0.1 - x Neglect x? 5.56 x 10^-10 (0.1) = x^2 x = 7.46 x 10^-6 7.46 x 10^-6 x 100 = 0.0075% < 5% ✓ 0.10 pH = -log[H3O+] = -log(7.46 x 10^-6) = 5.13

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