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Question: The solubility product constant of \( A{{g}_{2}}Cr{{O}_{4}} \) and \( AgBr \) is \( 1.1 \times {{10}...

The solubility product constant of Ag2CrO4A{{g}_{2}}Cr{{O}_{4}} and AgBrAgBr is 1.1×10−121.1 \times {{10}^{-12}} and 5.0×10−135.0 \times {{10}^{-13}} respectively. Calculate the ratio of the molarities of the solutions.

Explanation

Solution

First, we will write the equation of Ag2CrO4A{{g}_{2}}Cr{{O}_{4}} and AgBrAgBr
The equation of silver chromate
Ag2CrO4⇆2Ag++CrO42−A{{g}_{2}}Cr{{O}_{4}}\overset{{}}{\leftrightarrows}2A{{g}^{+}}+CrO_{4}^{2-}
The equation of silver bromide
AgBr⇆Ag++Br−AgBr\overset{{}}{\leftrightarrows}A{{g}^{+}}+B{{r}^{-}}

Complete answer:
Let S1{{S}_{1}} be the solubility of the silver chromate
The solubility of Ag2CrO4⇆2Ag++CrO42−A{{g}_{2}}Cr{{O}_{4}}\overset{{}}{\leftrightarrows}2A{{g}^{+}}+CrO_{4}^{2-}
First, we calculate for 2Ag+2A{{g}^{+}}
Solubility of 2Ag+2A{{g}^{+}} is (2S1)2{{(2{{S}_{1}})}^{2}}
Solubility of CrO42−CrO_{4}^{2-} is S1{{S}_{1}}
Totally for silver chromate, ksp=(2S1)2×S1=4S13{{k}_{sp}}={{(2{{S}_{1}})}^{2}} \times {{S}_{1}}=4S_{1}^{3}
The given solubility product constant is 1.1×10−121.1 \times {{10}^{-12}}
Calculation: to find the value of S1{{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 44 (moving the coefficient 44 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
4S13=1.1×10−124S_{1}^{3}=1.1 \times {{10}^{-12}}
S13=1.1×10−12/4S_{1}^{3}=1.1 \times {{10}^{-12}}/4
S13=0.275×10−12S_{1}^{3}=0.275 \times {{10}^{-12}}
S1=0.65029×10−4{{S}_{1}}=0.65029 \times {{10}^{-4}}
S1=6.5×10−5{{S}_{1}}=6.5 \times {{10}^{-5}}
Therefore, we have calculated the value of S1{{S}_{1}}
S1=6.5×10−5{{S}_{1}}=6.5 \times {{10}^{-5}}
Let S2{{S}_{2}} be the solubility of silver bromide
The solubility of AgBr⇆Ag++Br−AgBr\overset{{}}{\leftrightarrows}A{{g}^{+}}+B{{r}^{-}}
Solubility of Ag+A{{g}^{+}} is S2{{S}_{2}}
Solubility of Br−B{{r}^{-}} is S2{{S}_{2}}
Totally for silver bromide, Ksp=S2×S2=S22{{K}_{sp}}={{S}_{2}} \times {{S}_{2}}={{S}_{2}}^{2}
Calculation: to find the value of S2{{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
S22=5.×10−13 S2=7.07×10−7 \begin{aligned} & S_{2}^{2}=5. \times {{10}^{-13}} \\\ & {{S}_{2}}=7.07 \times {{10}^{-7}} \\\ \end{aligned}
Therefore, we have calculated the value of S2{{S}_{2}}
Now by dividing the S1{{S}_{1}} by S2{{S}_{2}} we can identify the ratio of molarities of the given saturated solution
S1S2=6.50×10−57.07×10−7=91.9\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:
Ag2CrO4⇆2Ag++CrO42−A{{g}_{2}}Cr{{O}_{4}}\overset{{}}{\leftrightarrows}2A{{g}^{+}}+CrO_{4}^{2-}
silver chromate, ksp=(2S1)2×S1=4S13{{k}_{sp}}={{(2{{S}_{1}})}^{2}} \times {{S}_{1}}=4S_{1}^{3}
S1=6.5×10−5{{S}_{1}}=6.5 \times {{10}^{-5}}
AgBr⇆Ag++Br−AgBr\overset{{}}{\leftrightarrows}A{{g}^{+}}+B{{r}^{-}}
S22=5.×10−13S_{2}^{2}=5. \times {{10}^{-13}}
S2=7.07×10−7{{S}_{2}}=7.07 \times {{10}^{-7}}