## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 50

#### Answer

a. $[H_3O^+] = 2.8 \times 10^{-9} \space M$; $[OH^-] = 3.6 \times 10^{-6} \space M$ b. $[H_3O^+] = 5.9 \times 10^{-12} \space M$; $[OH^-] = 1.7 \times 10^{-3} \space M$ c. $[H_3O^+] = 1.3 \times 10^{-3} \space M$; $[OH^-] = 7.7 \times 10^{-12} \space M$

#### Work Step by Step

We find: a. $$[H_3O^+] = 10^{-pH} = 10^{-8.55} = 2.8 \times 10^{-9} \space M$$ $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 2.8 \times 10^{-9} } = 3.6 \times 10^{-6} \space M$$ b.$$[H_3O^+] = 10^{-pH} = 10^{-11.23} = 5.9 \times 10^{-12} \space M$$ $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 5.9 \times 10^{-12} } = 1.7 \times 10^{-3} \space M$$ c.$$[H_3O^+] = 10^{-pH} = 10^{-2.87} = 1.3 \times 10^{-3} \space M$$ $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 1.3 \times 10^{-3} } = 7.7 \times 10^{-12} \space M$$

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