## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 47

#### Answer

a. $[OH^-] = 8.3 \times 10^{-7}$, the solution is basic. a. $[OH^-] = 1.2 \times 10^{-10}$, the solution is acidic. a. $[OH^-] = 2.9 \times 10^{-13}$, the solution is acidic.

#### Work Step by Step

1. The $K_w$ at $25 ^oC$ is equal to $10^{-14}$. Using that, calculate $[OH^-]$ a. $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 1.2 \times 10^{-8} } = 8.3 \times 10^{-7} \space M$$ b. $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 8.5 \times 10^{-5} } = 1.2 \times 10^{-10} \space M$$ c. $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 3.5 \times 10^{-2} } = 2.9 \times 10^{-13} \space M$$ 2. If $[H_3O^+] \gt [OH^-]$, the solution is acidic; if $[H_3O^+] \lt [OH^-]$, the solution is basic. a. $[H_3O^+] \lt [OH^-]$: Basic b. $[H_3O^+] \gt [OH^-]$: Acidic c. $[H_3O^+] \gt [OH^-]$: Acidic

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