Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 16 - Questions and Problems - Page 773: 16.72

Answer

Original base molarity: $0.1547M$

Work Step by Step

1. Find $[OH^-]$ pH + pOH = 14 11.22 + pOH = 14 pOH = 2.78 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 2.78}$ $[OH^-] = 1.66 \times 10^{- 3}$ 2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [N{H_4}^+] = x$ -$[NH_3] = [NH_3]_{initial} - x$ 3. Now, use the Kb and x values and equation to find the initial concentration value. $Kb = \frac{[OH^-][N{H_4}^+]}{ [Initial NH_3] - x}$ $ 1.8\times 10^{- 5}= \frac{[x^2]}{ [Initial NH_3] - x}$ $ 1.8\times 10^{- 5}= \frac{( 1.66\times 10^{- 3})^2}{[Initial NH_3] - 1.66\times 10^{- 3}}$ $[Initial NH_3] - 1.66\times 10^{- 3} = \frac{ 2.754\times 10^{- 6}}{ 1.8\times 10^{- 5}}$ $[Initial NH_3] - 1.66\times 10^{- 3} = 0.153$ $[Initial NH_3] = 0.153 + 1.66\times 10^{- 3}$ $[Initial NH_3] = 0.1547$
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