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$
$1.8 \times 10^{-8} = S$
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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.
