## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 51

#### Answer

\begin{matrix} [H_3O^+] & [OH^-] & pH & Acidic \space or \space Basic \\ 7.1 \times 10^{-4} & 1.4 \times 10^{-11} & 3.15 & Acidic \\ 3.7 \times 10^{-9} & 2.7 \times 10^{-6} & 8.43 & Basic \\ 7.9 \times 10^{-12} & 1.3 \times 10^{-3} & 11.1 & Basic \\ 6.3 \times 10^{-4} & 1.6 \times 10^{-11} & 3.20 & Acidic \end{matrix}

#### Work Step by Step

We find: 1. $$[H_3O^+] = 10^{-pH} = 10^{-3.15} = 7.1 \times 10^{-4} \space M$$ $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 7.1 \times 10^{-4} } = 1.4 \times 10^{-11} \space M$$ $[H_3O^+] \gt [OH^-] : Acidic$ 2.$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 3.7 \times 10^{-9} } = 2.7 \times 10^{-6} \space M$$ $$pH = -log[H_3O^+] = -log( 3.7 \times 10^{-9} ) = 8.43$$ $[H_3O^+] \lt [OH^-] : Basic$ 3.$$[H_3O^+] = 10^{-pH} = 10^{-11.1} = 7.9 \times 10^{-12} \space M$$ $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 7.9 \times 10^{-12} } = 1.3 \times 10^{-3} \space M$$ $[H_3O^+] \lt [OH^-] : Basic$ 4. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.6 \times 10^{-11} } = 6.3 \times 10^{-4} \space M$$ $$pH = -log[H_3O^+] = -log( 6.3 \times 10^{-4} ) = 3.20$$ $[H_3O^+] \gt [OH^-] : Acidic$

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