## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652b: 32

#### Answer

$[OH^-] = 4.571 \times 10^{- 4}M$ 0.008817g of $Ba(OH)_2$.

#### Work Step by Step

1. Calculate [OH^-]: pH + pOH = 14 10.66 + pOH = 14 pOH = 3.34 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 3.34}$ $[OH^-] = 4.571 \times 10^{- 4}$ 2. Since $Ba(OH)_2$ is a strong base with 2 OH in each molecule: $[OH^-] = 2 * [Ba(OH)_2]$ $4.571\times 10^{- 4} = 2 * [Ba(OH)_2]$ $\frac{ 4.571\times 10^{- 4}}{ 2} = [Ba(OH)_2]$ $2.285\times 10^{- 4}M = [Ba(OH)_2]$ 3. Calculate the number of moles: $(Ba(OH)_2)$ $n(moles) = concentration(M) * volume(L)$ - 250ml = 0.250L $n(moles) = 2.285\times 10^{- 4} * 0.25$ $n(moles) = 5.714\times 10^{- 5}$ 4. Find the mass value in grams: $(Ba(OH)_2)$ 137.3* 1 + 2 * ( 16* 1 + 1.01* 1 ) = 154.31g/mol $mass(g) = mm(g/mol) * n(moles)$ $mass(g) = 154.31 * 5.714\times 10^{- 5}$ $mass(g) = 8.817\times 10^{- 3}$

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