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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 s...

100.0 mL of a clear saturated solution of Ag2SO4A{{g}_{2}}S{{O}_{4}} is added to 250.0 mL of a clear saturated solution of PbCrO4PbCr{{O}_{4}}. Will any precipitate form and if so what? Given, Ksp{{K}_{sp}} values of Ag2SO4A{{g}_{2}}S{{O}_{4}}, Ag2CrO4A{{g}_{2}}Cr{{O}_{4}}, PbCrO4PbCr{{O}_{4}} are 1.4×10−51.4\times {{10}^{-5}},2.4×10−122.4\times {{10}^{-12}},2.8×10−132.8\times {{10}^{-13}} and 1.6×10−81.6\times {{10}^{-8}} respectively.

Explanation

Solution

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 Ag2SO4A{{g}_{2}}S{{O}_{4}}
Ag2SO4⇌2Ag++SO42−A{{g}_{2}}S{{O}_{4}}\rightleftharpoons 2A{{g}^{+}}+SO_{4}^{2-}
The solubility product Ksp{{K}_{sp}} for Ag2SO4A{{g}_{2}}S{{O}_{4}} can be written as
Ksp=[Ag+][SO42−]=1.4×10−5{{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=[SO42−]S=\left[ SO_{4}^{2-} \right] and 2S=[Ag+]2S=\left[ A{{g}^{+}} \right]. From the equilibrium expression we could write as
4S3=1.4×10−5=[Ag+]=0.0304M4{{S}^{3}}=1.4\times {{10}^{-5}}=\left[ A{{g}^{+}} \right]=0.0304M
S=0.0152MS=0.0152M
- For mixed solutions we can write as M1V1=M2V2{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}
[Ag+]=0.0304×100350=8.68×10−3M\left[ A{{g}^{+}} \right]=\dfrac{0.0304\times 100}{350}=8.68\times {{10}^{-3}}M
[SO42−]=4.34×10−3M\left[ SO_{4}^{2-} \right]=4.34\times {{10}^{-3}}M
Let’s write the equilibrium expression for PbCrO4PbCr{{O}_{4}}
PbCrO4⇌Pb2++CrO42−PbCr{{O}_{4}}\rightleftharpoons P{{b}^{2+}}+CrO_{4}^{2-}
- The solubility product Ksp{{K}_{sp}} for PbCrO4PbCr{{O}_{4}} can be written as:
Ksp=[Pb2+][CrO42−]=2.8×10−13{{K}_{sp}}=\left[ P{{b}^{2+}} \right]\left[ CrO_{4}^{2-} \right]=2.8\times {{10}^{-13}}
S2=2.8×10−13{{S}^{2}}=2.8\times {{10}^{-13}}
S=5.29×10−7MS=5.29\times {{10}^{-7}}M
- Upon dilution we can write the equation as follows:
5.29×10−7×250350=3.78×10−7M=[Pb2+]=[CrO42−]\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 Ksp{{K}_{sp}} for Ag2CrO4A{{g}_{2}}Cr{{O}_{4}} can be written as :
Ksp=[Ag+]2[CrO42−]=1.1×10−12{{K}_{sp}}={{\left[ A{{g}^{+}} \right]}^{2}}\left[ CrO_{4}^{2-} \right]=1.1\times {{10}^{-12}}
[8.86×10−3]2[3.78×10−7]=2.85×10−11{{\left[ 8.86\times {{10}^{-3}} \right]}^{2}}\left[ 3.78\times {{10}^{-7}} \right]=2.85\times {{10}^{-11}}
As we know, 2.85×10−112.85\times 1{{0}^{-11}} > 1.1×10−121.1\times {{10}^{-12}}.
As we can see the solubility product is greater than Ksp{{K}_{sp}} of Ag2CrO4A{{g}_{2}}Cr{{O}_{4}} and as a result Ag2CrO4A{{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.