Answer
(a) $9.1 \times 10^{-9}M$
(b) $7.4 \times 10^{-8}M$
Work Step by Step
(a)
1. Write the $K_{sp}$ expression:
$ AgI(s) \lt -- \gt 1Ag^{+}(aq) + 1I^-(aq)$
$8.3 \times 10^{-17} = [Ag^{+}]^ 1[I^-]^ 1$
$8.3 \times 10^{-17} = (9.1 \times 10^{-9})^ 1( 1S)^ 1$
2. Find the molar solubility.
$8.3 \times 10^{-17}= (9.1 \times 10^{-9})^ 1 \times ( 1S)^ 1$
$ \frac{8.3 \times 10^{-17}}{9.1 \times 10^{-9}} = ( 1S)^ 1$
$9;1 \times 10^{-9} = ( 1S)^ 1$
$9.1 \times 10^{-9} = S$
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(b)
1. Write the $K_{sp}$ expression:
$ Al(OH)_3(s) \lt -- \gt 3OH^{-}(aq) + 1Al^{3+}(aq)$
$1.8 \times 10^{-33} = [OH^{-}]^ 3[Al^{3+}]^ 1$
$1.8 \times 10^{-33} = (2.9 \times 10^{-9})^ 3( 1S)^ 1$
2. Find the molar solubility.
$1.8 \times 10^{-33}= (2.9 \times 10^{-9})^ 3 \times ( 1S)^ 1$
$1.8 \times 10^{-33}= 2.4 \times 10^{-36} \times ( 1S)^ 1$
$ \frac{1.8 \times 10^{-33}}{2.4 \times 10^{-36}} = ( 1S)^ 1$
$7.4 \times 10^{-8} = ( 1S)^ 1$
$7.4 \times 10^{-8} = S$
