Question

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at $298K$ from their solubility product constants. Determine also the molarities of individual ions.
The solubility product constants are:
Silver chromate, $1.1 \times {10^{ - 12}}$
Barium chromate, $1.2 \times {10^{ - 10}}$
Ferric hydroxide, $1.0 \times {10^{ - 38}}$
Lead dichloride, $1.6 \times {10^{ - 5}}$
Mercurous chloride, $4.5 \times {10^{ - 29}}$

Answer 1

Hint: ${K_{sp}}$ represents the level at which a solute dissolves in a solution. More soluble substances have higher ${K_{sp}}$ value, each concentration is raised to the power of the respective coefficient of ion in a balanced chemical equation to calculate ${K_{sp}}$.

Complete step by step answer:
Silver chromate, $A{g_2}Cr{O_4}$
\[A{g_2}Cr{O_4} \to 2A{g^ + } + Cr{O_4}^{2 - }\]
We can see that as $xM$ silver chromate dissociates the obtained concentration of both the ions will be ${\text{2x and x}}$.
$ \Rightarrow \left[ {A{g^ + }} \right] = 2x$ and \[\left[ {Cr{O_4}^{2 - }} \right] = x\]
And the ${K_{sp}}$ will be
${K_{sp}} = \;{\left[ {A{g^ + }} \right]^2}\left[ {Cr{O_4}^{2 - }} \right]$
${K_{sp}} = \;{\left( {2x} \right)^2}\left( x \right) = 4{x^3}$
And we have already provided the value of ${K_{sp}}$ so we will compare it with our result.
$ \Rightarrow 4{x^3} = 1.1 \times {10^{ - 12}}$
$ \Rightarrow x = 6.5 \times {10^{ - 5}}$
Therefore the molarities of individual ions are,
$\left[ {A{g^ + }} \right] = 2x = 2 \times 6.5 \times {10^{ - 5}} = 1.30 \times {10^{ - 4}}$ and \[\left[ {Cr{O_4}^{2 - }} \right] = 6.5 \times {10^{ - 5}}\]
Barium chromate,$BaCr{O_4}$
\[BaCr{O_4} \to B{a^{ + 2}} + Cr{O_4}^{2 - }\]
We can see that as $xM$ barium chromate dissociates the obtained concentration of both the ions will be ${\text{x}}$.
$ \Rightarrow \left[ {B{a^{ + 2}}} \right] = x$ and \[\left[ {Cr{O_4}^{2 - }} \right] = x\]
And the ${K_{sp}}$ will be
${K_{sp}} = \;\left[ {B{a^{ + 2}}} \right]\left[ {Cr{O_4}^{2 - }} \right]$
${K_{sp}} = \;\left( x \right)\left( x \right) = {x^2}$
And we have already provided the value of ${K_{sp}}$ so we will compare it with our result.
$ \Rightarrow {x^2} = 1.2 \times {10^{ - 10}}$
$ \Rightarrow x = 1.09 \times {10^{ - 5}}$
Therefore the molarities of individual ions are,
$ \Rightarrow \left[ {B{a^{ + 2}}} \right] = \left[ {Cr{O_4}^{2 - }} \right] = 1.09 \times {10^{ - 5}}$
Similarly for ferric hydroxide,$Fe{(OH)_3}$
\[Fe{(OH)_3} \to F{e^{ + 3}} + 3O{H^ - }\]
$\left[ {F{e^{ + 3}}} \right] = x,\left[ {O{H^ - }} \right] = 3x$
${K_{sp}} = x \times {\left( {3x} \right)^3} = 27{x^4} = 1.0 \times {10^{ - 38}}$
Therefore the molarities of individual ions are,
\[[F{e^{ + 3}}] = x = 1.38 \times {10^{ - 10}}M\] and \[[O{H^ - }] = 3x = 4.14 \times {10^{ - 10}}M\]
Lead dichloride,$PbC{l_2}$
\[PbC{l_2} \to P{b^{ + 2}} + 2C{l^ - }\]
$\left[ {P{b^{ + 2}}} \right] = x,\left[ {C{l^ - }} \right] = 2x$
${K_{sp}} = x \times {\left( {2x} \right)^2} = 4{x^3} = 1.6 \times {10^{ - 5}}$
Therefore the molarities of individual ions are,
\[[P{b^{ + 2}}] = x = 0.0159M\] and \[[C{l^ - }] = 2x = 0.0318M\]
Mercurous iodide,$H{g_2}{I_2}$
\[H{g_2}{I_2} \to 2H{g^ + } + 2{I^ - }\]
$\left[ {H{g^ + }} \right] = \left[ {{I^ - }} \right] = 2x$
${K_{sp}} = {\left( {2x} \right)^2} \times {\left( {2x} \right)^2} = 16{s^4} = 4.5 \times {10^{ - 29}}$
$x = 4.09 \times {10^{ - 8}}M$
Therefore the molarities of individual ions are,
$ \Rightarrow \left[ {H{g^ + }} \right] = \left[ {{I^ - }} \right] = 2 \times 4.09 \times {10^{ - 8}} = 8.18 \times {10^{ - 8}}M$.


Note:
The concentration used of each and every ion in the calculation of solubility product constant should be in molarity or mole per liter. Sometimes in numerical a term \[Q\](reaction quotient) is introduced, it is used to determine the precipitate formation, if $Q = {K_{sp}}$ , a precipitate will form.