Question

The solubility product constant of $ A{{g}_{2}}Cr{{O}_{4}} $ and $ AgBr $ is $ 1.1 \times {{10}^{-12}} $ and $ 5.0 \times {{10}^{-13}} $ respectively. Calculate the ratio of the molarities of the solutions.

Answer 1

Hint: First, we will write the equation of $ A{{g}_{2}}Cr{{O}_{4}} $ and $ AgBr $
The equation of silver chromate
 $ A{{g}_{2}}Cr{{O}_{4}}\overset{{}}{\leftrightarrows}2A{{g}^{+}}+CrO_{4}^{2-} $
The equation of silver bromide
 $ AgBr\overset{{}}{\leftrightarrows}A{{g}^{+}}+B{{r}^{-}} $

Complete answer:
Let $ {{S}_{1}} $ be the solubility of the silver chromate
The solubility of $ A{{g}_{2}}Cr{{O}_{4}}\overset{{}}{\leftrightarrows}2A{{g}^{+}}+CrO_{4}^{2-} $
First, we calculate for $ 2A{{g}^{+}} $
Solubility of $ 2A{{g}^{+}} $ is $ {{(2{{S}_{1}})}^{2}} $
Solubility of $ CrO_{4}^{2-} $ is $ {{S}_{1}} $
Totally for silver chromate, $ {{k}_{sp}}={{(2{{S}_{1}})}^{2}} \times {{S}_{1}}=4S_{1}^{3} $
The given solubility product constant is $ 1.1 \times {{10}^{-12}} $
Calculation: to find the value of $ {{S}_{1}} $
Follow the below steps to find the value
First, we are substituting the given product constant value for silver chromate
Then we are dividing them by $ 4 $ (moving the coefficient $ 4 $ from the reactant side to the product side )
Further, we are taking the cubic root on both sides
At last, we are moving the decimal point to get the final solution
 $ 4S_{1}^{3}=1.1 \times {{10}^{-12}} $
 $ S_{1}^{3}=1.1 \times {{10}^{-12}}/4 $
 $ S_{1}^{3}=0.275 \times {{10}^{-12}} $
 $ {{S}_{1}}=0.65029 \times {{10}^{-4}} $
 $ {{S}_{1}}=6.5 \times {{10}^{-5}} $
Therefore, we have calculated the value of $ {{S}_{1}} $
 $ {{S}_{1}}=6.5 \times {{10}^{-5}} $
Let $ {{S}_{2}} $ be the solubility of silver bromide
The solubility of $ AgBr\overset{{}}{\leftrightarrows}A{{g}^{+}}+B{{r}^{-}} $
Solubility of $ A{{g}^{+}} $ is $ {{S}_{2}} $
Solubility of $ B{{r}^{-}} $ is $ {{S}_{2}} $
Totally for silver bromide, $ {{K}_{sp}}={{S}_{2}} \times {{S}_{2}}={{S}_{2}}^{2} $
Calculation: to find the value of $ {{S}_{2}} $
Follow the below steps to find the value
First, we are substituting the given product constant value for silver bromide
Further, we are taking the square root on both sides
 $ \begin{align}
  & S_{2}^{2}=5. \times {{10}^{-13}} \\
 & {{S}_{2}}=7.07 \times {{10}^{-7}} \\
\end{align} $
Therefore, we have calculated the value of $ {{S}_{2}} $
Now by dividing the $ {{S}_{1}} $ by $ {{S}_{2}} $ we can identify the ratio of molarities of the given saturated solution
 $ \dfrac{{{S}_{1}}}{{{S}_{2}}}=\dfrac{6.50 \times {{10}^{-5}}}{7.07 \times {{10}^{-7}}}=91.9 $
Therefore, we have found out the final solution.

Note:
 $ A{{g}_{2}}Cr{{O}_{4}}\overset{{}}{\leftrightarrows}2A{{g}^{+}}+CrO_{4}^{2-} $
silver chromate, $ {{k}_{sp}}={{(2{{S}_{1}})}^{2}} \times {{S}_{1}}=4S_{1}^{3} $
 $ {{S}_{1}}=6.5 \times {{10}^{-5}} $
 $ AgBr\overset{{}}{\leftrightarrows}A{{g}^{+}}+B{{r}^{-}} $
 $ S_{2}^{2}=5. \times {{10}^{-13}} $
 $ {{S}_{2}}=7.07 \times {{10}^{-7}} $