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Question
100 mL of a saturated solution of Ag2SO4 is added to 250 mL of a saturated solution of PbCrO4. Will any precipitate form and if so what?
Given that : Ksp for Ag2SO4, Ag2CrO4,PbCrO4 and PbSO4 are 1.4×10−5, 2.4×10−12, 2.8×10−13 and 1.6×10−8 respectively.
Given that : Ksp for Ag2SO4, Ag2CrO4,PbCrO4 and PbSO4 are 1.4×10−5, 2.4×10−12, 2.8×10−13 and 1.6×10−8 respectively.
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Solution
The correct option is A Ag2CrO4
For Ag2SO4⇌2Ag⊕+SO2−4
2S S
Ksp=4S3 or S=3√Ksp4=3√1.4×10−54=1.52×10−2M
For PbCrO4⇌Pb2++CrO2−4
Ksp=S21 or S1=√Ksp=√2.8×10−13=5.29×10−7M
In solution, conentration. of each ion can be given as:
Thus, [Ag⊕]=2S×100350=2×1.52×10−2×100350 =0.869×10−2M
[SO2−4]=S×100350=1.52×10−2×100350=0.43×10−2
[Pb2+]=S1×250350=5.29×10−7×250350=3.78×10−7
[CrO2−4]=S1×250350=5.29×10−7×250350=3.78×10−7
It is the evident that,
[Ag⊕][CrO2−4]=(0.869×10−2)2×(3.78×10−7)
=2.85×10−11>Ksp(Ag2CrO4)
The product is greater than Ksp of Ag2CrO4 and thus, Ag2CrO4 will precipitate out.
For Ag2SO4⇌2Ag⊕+SO2−4
2S S
Ksp=4S3 or S=3√Ksp4=3√1.4×10−54=1.52×10−2M
For PbCrO4⇌Pb2++CrO2−4
Ksp=S21 or S1=√Ksp=√2.8×10−13=5.29×10−7M
In solution, conentration. of each ion can be given as:
Thus, [Ag⊕]=2S×100350=2×1.52×10−2×100350 =0.869×10−2M
[SO2−4]=S×100350=1.52×10−2×100350=0.43×10−2
[Pb2+]=S1×250350=5.29×10−7×250350=3.78×10−7
[CrO2−4]=S1×250350=5.29×10−7×250350=3.78×10−7
It is the evident that,
[Ag⊕][CrO2−4]=(0.869×10−2)2×(3.78×10−7)
=2.85×10−11>Ksp(Ag2CrO4)
[Ag⊕][CrO2−4]=(0.869×10−2)2×(3.78×10−7)
=2.85×10−11>Ksp(Ag2CrO4)
The product is greater than Ksp of Ag2CrO4 and thus, Ag2CrO4 will precipitate out.
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