Answer:
Silver
chromate:
\[\underset{s}{\mathop{A{{g}_{2}}Cr{{O}_{4}}}}\,\to
2\underset{2s}{\mathop{A{{g}^{+}}}}\,+\underset{s}{\mathop{CrO_{4}^{2-}}}\,\]
\[{{K}_{sp}}={{[A{{g}^{+}}]}^{2}}[CrO_{4}^{2-}]\]
\[={{[2s]}^{2}}[s]=4{{s}^{3}}\]
\[s={{\left[ \frac{{{K}_{sp}}}{4} \right]}^{1/3}}={{\left[
\frac{1.1\times {{10}^{-12}}}{4} \right]}^{1/3}}\]
\[=6.5\times {{10}^{-5}}M\]
\[[A{{g}^{+}}]=2\times 6.5\times {{10}^{-5}}M\]
\[=13\times {{10}^{-5}}M\]
\[[CrO_{4}^{2-}]=s=6.5\times {{10}^{-5}}M\]
Similarly solubility of barium chromate can be calculated.
Ferric hydroxide:
\[\underset{s}{\mathop{Fe{{(OH)}_{3}}}}\,\to
\underset{s}{\mathop{F{{e}^{3+}}}}\,+3\underset{3s}{\mathop{O{{H}^{-}}}}\,\]
\[{{K}_{sp}}=[F{{e}^{3+}}]{{[3O{{H}^{-}}]}^{3}}\]
\[=[s]{{[3s]}^{3}}=27{{s}^{4}}\]
\[s={{\left[ \frac{{{K}_{sp}}}{27} \right]}^{1/4}}={{\left[
\frac{1\times {{10}^{-38}}}{27} \right]}^{1/4}}\](Solve using log table)
\[=1.38\times {{10}^{-10}}M\]
\[[F{{e}^{3+}}]=1.38\times {{10}^{-10}}M\]
\[[O{{H}^{-}}]=3\times 1.38\times {{10}^{-10}}\]
\[=4.14\times {{10}^{-10}}M\]
Lead chloride:
\[\underset{s}{\mathop{PbC{{l}_{2}}}}\,\to
\underset{s}{\mathop{P{{b}^{2+}}}}\,+\underset{2s}{\mathop{2C{{l}^{-}}}}\,\]
\[{{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}\]
\[1.6\times {{10}^{-5}}=[s]{{[2s]}^{2}}\]
\[1.6\times {{10}^{-5}}=4{{s}^{3}}\]
\[s={{\left[ \frac{1.6\times
{{10}^{-5}}}{4} \right]}^{1/3}}\]
\[=0.0158\,M\]
\[\therefore \]\[[P{{b}^{2+}}]=0.0158M,\,\,\,[C{{l}^{-}}]=2\times
0.0158\]
\[=0.0316\,M\]
Mercurous iodide:
\[\underset{s}{\mathop{H{{g}_{2}}{{I}_{2}}}}\,\to
\underset{s}{\mathop{Hg_{2}^{2+}}}\,+\underset{2s}{\mathop{2{{I}^{-}}}}\,\]
\[{{K}_{sp}}=[H{{g}^{2+}}]{{[{{I}^{-}}]}^{2}}\]
\[4.5\times {{10}^{-29}}=s{{[2s]}^{2}}=4{{s}^{3}}\]
\[s={{\left[ \frac{4.5\times
{{10}^{-29}}}{4} \right]}^{1/3}}\]
\[=2.24\times {{10}^{-10}}M\]
\[[Hg_{2}^{2+}]=2.24\times {{10}^{-10}}M\]
\[[{{I}^{-}}]=2\times 2.24\times
{{10}^{-10}}M\]
\[=4.48\times {{10}^{-10}}M\]
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