## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 16 - Questions and Problems - Page 771: 16.23

#### Answer

$[OH^-] = 7.14 \times 10^{- 12}M$

#### Work Step by Step

1. Since HCl is a strong acid: $[HCl] = [H_3O^+]$ 2. Calculate $[H_3O^+]$ using $[OH^-]$: $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $1.4 \times 10^{- 3} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 1.4 \times 10^{- 3}}$ $[OH^-] = 7.14 \times 10^{- 12}$

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