## Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning

# Chapter 13 - Exercises - Page 575d: 94

#### Answer

$[Sr(OH)_2] = 1.581\times 10^{- 4}M$

#### Work Step by Step

1. Calculate $[OH^-]$: pH + pOH = 14 10.5 + pOH = 14 pOH = 3.5 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 3.5}$ $[OH^-] = 3.162 \times 10^{- 4}$ 2. Since $Sr(OH)_2$ is a strong base with 2 OH in each molecule: $[OH^-] = 2 * [Sr(OH)_2]$ $3.162\times 10^{- 4} = 2 * [Sr(OH)_2]$ $\frac{ 3.162\times 10^{- 4}}{ 2} = [Sr(OH)_2]$ $1.581\times 10^{- 4}M = [Sr(OH)_2]$

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