Question

The solubility product of \[A{{g}_{2}}{{C}_{2}}{{O}_{4}}\] at 25℃ is \[1.29\times {{10}^{-11}}mo{{l}^{3}}{{L}^{-3}}\]. A solution of
\[{{K}_{2}}{{C}_{2}}{{O}_{4}}\] containing 0.1520 mol in 500 mL of water is shaken with excess of
\[A{{g}_{2}}C{{O}_{3}}\] till the following equilibrium is reached.
\[A{{g}_{2}}C{{O}_{3}}+{{K}_{2}}{{C}_{2}}{{O}_{4}}\rightleftharpoons A{{g}_{2}}{{C}_{2}}{{O}_{4}}+{{K}_{2}}C{{O}_{3}}\]
At equilibrium, the solution contains 0.0358 mole of \[{{K}_{2}}C{{O}_{3}}\]. Assuming the degree
of dissociation of \[{{K}_{2}}{{C}_{2}}{{O}_{4}}\] and \[{{K}_{2}}C{{O}_{3}}\] to be equal, calculate the solubility product of \[A{{g}_{2}}C{{O}_{3}}\].

Answer 1

Hint: Solubility product constant, \[{{K}_{sp}}\]​, is an equilibrium constant for a solid substance which is dissolving in an aqueous solution. It tells us the level at which a solute can dissolve in solution. The more soluble a substance is, the higher will be the value \[{{K}_{sp}}\].


Complete step-by-step answer:
We have been given that \[A{{g}_{2}}C{{O}_{3}}\] is present in excess and it reacts with \[{{K}_{2}}{{C}_{2}}{{O}_{4}}\]. Initially, before the reaction the amount of products will be zero. As the reaction proceeds, the reactants will be depleted to form products. The same thing is depicted in the below equation.

\[\begin{align}
  & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;A{{g}_{2}}C{{O}_{3}}\;\;\;+\;\;\;{{K}_{2}}{{C}_{2}}{{O}_{4}}\;\;\;\;\;\;\;\rightleftharpoons \;\;\;\;\;\;\;A{{g}_{2}}{{C}_{2}}{{O}_{4}}+{{K}_{2}}C{{O}_{3}} \\
 & Moles\,at\,start\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;Excess\;\;\;\;\;\;\;\;\;\;\;\;\;0.1520\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,0 \\
 & Moles\,after\,reation\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0.1520-0.0358)\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;\;\;\;0.0358\,\,\,\,\,\,\,\,\,0.0358 \\
 & \,\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=0.1162 \\
\end{align}\]

The molar concentration of \[{{K}_{2}}{{C}_{2}}{{O}_{4}}\] or \[{{C}_{2}}{{O}_{4}}^{2-}\] left unreacted will be;
\[\left[ {{C}_{2}}{{O}_{4}}^{2-} \right]=\dfrac{0.1162}{0.5}=0.2324\,M\]
The molar concentration of \[{{K}_{2}}C{{O}_{3}}\] or \[C{{O}_{3}}^{2-}\] left unreacted will be;

\[\left[ C{{O}_{3}}^{2-} \right]=\dfrac{0.0358}{0.5}=0.0716\,M\]
We have been given that the solubility product (\[{{K}_{sp}}\]) of \[A{{g}_{2}}{{C}_{2}}{{O}_{4}}\]at 25℃ is \[1.29\times {{10}^{-11}}mo{{l}^{3}}{{L}^{-3}}\]
Therefore, we can write from the above statement that,
\[{{\left[ A{{g}^{+}} \right]}^{2}}\left[ {{C}_{2}}{{O}_{4}}^{2-} \right]={{K}_{sp}}\] of \[(A{{g}_{2}}{{C}_{2}}{{O}_{4}})\]
We will now substitute the value of concentration of \[\left[ {{C}_{2}}{{O}_{4}}^{2-} \right]\] found earlier.
\[{{\left[ A{{g}^{+}} \right]}^{2}}\left[ 0.2324 \right]=1.29\times {{10}^{-11}}\]
On rearranging,
\[{{\left[ A{{g}^{+}} \right]}^{2}}=\dfrac{1.29\times {{10}^{-11}}}{0.2324}=5.55\times {{10}^{-11}}\]
Now the solubility product (\[{{K}_{sp}}\]) of \[A{{g}_{2}}C{{O}_{3}}\] can be written as,
\[{{K}_{sp}}={{\left[ A{{g}^{+}} \right]}^{2}}\left[ C{{O}_{3}}^{2-} \right]\]
Finally, substituting the values of concentration of \[{{\left[ A{{g}^{+}} \right]}^{2}}\]and \[\left[ C{{O}_{3}}^{2-} \right]\] respectively, we get
\[\begin{align}
  & {{K}_{sp}}=5.55\times {{10}^{-11}}\times 0.0716 \\
 & \,\,\,\,\,\,\,\,=3.974\times {{10}^{-12}}\,mo{{l}^{3}}{{L}^{-3}} \\
\end{align}\]

Note: We can also find out which compound will precipitate from the data found in the solution. It can be given as:
\[\begin{align}
  & {{\left[ A{{g}^{+}} \right]}^{2}}\left[ {{C}_{2}}{{O}_{4}}^{2-} \right]={{\left( \dfrac{2\times 0.0358}{0.5} \right)}^{2}}\left( \dfrac{0.1520}{0.5} \right) \\
 & \,\,\,\,=\,6.23\times {{10}^{-3}} \\
\end{align}\]
Since, \[{{\left[ A{{g}^{+}} \right]}^{2}}\left[ {{C}_{2}}{{O}_{4}}^{2-} \right]>{{K}_{sp}}\] of\[A{{g}_{2}}{{C}_{2}}{{O}_{4}}\]\[(1.29\times {{10}^{-11}}mo{{l}^{3}}{{L}^{-3}})\]
\[A{{g}_{2}}{{C}_{2}}{{O}_{4}}\] will precipitate out.