## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 53

#### Answer

$[H_3O^+] = 1.5 \times 10^{-7}$ $pH = 6.81$

#### Work Step by Step

1. Write the $K_w$ expression: $$K_w = [H_3O^+][OH^-]$$ 2. In pure water: $[H_3O^+] = [OH^-]$. $$K_w = [H_3O^+][H_3O^+]$$ $$\sqrt {K_w} = [H_3O^+]$$ $$[H_3O^+] = \sqrt{2.4 \times 10^{-14}} = 1.5 \times 10^{-7}$$ 3. Calculate the pH: $$pH = -log(\sqrt {2.4 \times 10^{-14}}) = 6.81$$

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