Question

100.0 mL of a clear saturated solution of $A{{g}_{2}}S{{O}_{4}}$ is added to 250.0 mL of a clear saturated solution of $PbCr{{O}_{4}}$. Will any precipitate form and if so what? Given, ${{K}_{sp}}$ values of $A{{g}_{2}}S{{O}_{4}}$, $A{{g}_{2}}Cr{{O}_{4}}$, $PbCr{{O}_{4}}$ are $1.4\times {{10}^{-5}}$,$2.4\times {{10}^{-12}}$,$2.8\times {{10}^{-13}}$ and $1.6\times {{10}^{-8}}$ respectively.

Answer 1

Hint: The solubility product expression can be defined as the product of the concentrations of the ions which are involved in equilibrium, each raised to the power of its coefficient in the equilibrium expression. We could compare the different values of solubility products of given compounds and identify whether any precipitate will be formed or not.

Complete step by step answer:
First, let’s write the equilibrium expression for $A{{g}_{2}}S{{O}_{4}}$
\[A{{g}_{2}}S{{O}_{4}}\rightleftharpoons 2A{{g}^{+}}+SO_{4}^{2-}\]
The solubility product ${{K}_{sp}}$ for $A{{g}_{2}}S{{O}_{4}}$ can be written as
\[{{K}_{sp}}=\left[ A{{g}^{+}} \right]\left[ SO_{4}^{2-} \right]=1.4\times {{10}^{-5}}\]
Let S mol dissolves and we can write as $S=\left[ SO_{4}^{2-} \right]$ and $2S=\left[ A{{g}^{+}} \right]$. From the equilibrium expression we could write as
\[4{{S}^{3}}=1.4\times {{10}^{-5}}=\left[ A{{g}^{+}} \right]=0.0304M\]
\[S=0.0152M\]
- For mixed solutions we can write as ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
\[\left[ A{{g}^{+}} \right]=\dfrac{0.0304\times 100}{350}=8.68\times {{10}^{-3}}M\]
\[\left[ SO_{4}^{2-} \right]=4.34\times {{10}^{-3}}M\]
Let’s write the equilibrium expression for $PbCr{{O}_{4}}$
 \[PbCr{{O}_{4}}\rightleftharpoons P{{b}^{2+}}+CrO_{4}^{2-}\]
- The solubility product ${{K}_{sp}}$ for $PbCr{{O}_{4}}$ can be written as:
\[{{K}_{sp}}=\left[ P{{b}^{2+}} \right]\left[ CrO_{4}^{2-} \right]=2.8\times {{10}^{-13}}\]
\[{{S}^{2}}=2.8\times {{10}^{-13}}\]
\[S=5.29\times {{10}^{-7}}M\]
- Upon dilution we can write the equation as follows:
\[\dfrac{5.29\times {{10}^{-7}}\times 250}{350}=3.78\times {{10}^{-7}}M=\left[ P{{b}^{2+}} \right]=\left[ CrO_{4}^{2-} \right]\]
 We are asked to find whether the precipitation will occur or not.
The solubility product ${{K}_{sp}}$ for $A{{g}_{2}}Cr{{O}_{4}}$ can be written as :
\[{{K}_{sp}}={{\left[ A{{g}^{+}} \right]}^{2}}\left[ CrO_{4}^{2-} \right]=1.1\times {{10}^{-12}}\]
\[{{\left[ 8.86\times {{10}^{-3}} \right]}^{2}}\left[ 3.78\times {{10}^{-7}} \right]=2.85\times {{10}^{-11}}\]
As we know, $2.85\times 1{{0}^{-11}}$ > $1.1\times {{10}^{-12}}$.
As we can see the solubility product is greater than ${{K}_{sp}}$ of $A{{g}_{2}}Cr{{O}_{4}}$ and as a result $A{{g}_{2}}Cr{{O}_{4}}$ will precipitate.

Note: Keep in mind that the solubility differs from the solubility product constant in which solubility is the amount of a substance which dissolves to form a saturated solution and it is often expressed as grams of solute per liter of solution or as number of moles of solute per liter solution (molar solubility). If the concentrations of other ions or pH changes, the solubility can also change.