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

Published by Cengage Learning

# Chapter 15 - Additional Aqueous Equilibria - Problem Solving Practice 15.15 - Page 691: a

#### Answer

$AgCl$ will be the first to precipitate.

#### Work Step by Step

For $AgCl$ to precipitate: 1. Write the $K_{sp}$ expression: $AgCl(s) \lt -- \gt 1Ag^{+}(aq) + 1Cl^{-}(aq)$ $1.8 \times 10^{-10} = [Ag^{+}]^ 1[Cl^{-}]^ 1$ $1.8 \times 10^{-10} = (0.01)^ 1( 1S)^ 1$ 2. Find the necessary chloride ion concentration: $1.8 \times 10^{-10}= 0.01 \times ( 1S)^ 1$ $\frac{1.8 \times 10^{-10}}{0.01} = ( 1S)^ 1$ $1.8 \times 10^{-8} = S$ ------- 1. Write the $K_{sp}$ expression: $PbCl_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Cl^{-}(aq)$ $1.6 \times 10^{-5} = [Pb^{2+}]^ 1[Cl^{-}]^ 2$ $1.6 \times 10^{-5} = (0.1)^ 1([Cl^-])^ 2$ 2. Find the necessary chloride ion concentration: $1.6 \times 10^{-5}= (0.1)^ 1 \times ([Cl^-])^ 2$ $\frac{1.6 \times 10^{-5}}{0.1} = ([Cl^-])^ 2$ $1.6 \times 10^{-4} = ([Cl^-])^ 2$ $\sqrt [ 2] {1.6 \times 10^{-4}} = [Cl^-]$ $0.013 = [Cl^-]$ Since $AgCl$ needs less chloride ions, it will precipitate first.

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