Question

# The solubility product of a saturated solution of Ag2CrO4 in water at 298 K will be................., if the EMF of the cell:Ag|Ag⊕(satAg2CrO4sol)||Ag(0.1M)|Ag is 0.164V at 298 K.

A
2.287×1012M3
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B
2.659×1012M3
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C
1.965×1012M3
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D
Noneofthese
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Solution

## The correct option is C 2.287×10−12M3Cell circuit is given as below:Ag|Ag⊕ (saturated) Ag2CrO4||Ag⊕(0.1M)|AgEcell=0.164 at 298 K.For the given cell (concentration cell)Ecell=0.0591nlogc2c1c1=[Ag⊕ in Ag2CrO4 (left side)c2=(Ag⊕) (right side)N = 1 (number of exchange electron)So, 0.164=0.05911log0.1c1Or log0.1c1=0.1640.0591 = 2.774On solvingc1=[Ag⊕]inAg2CrO4=1.66×10−4MWe get Ag2CrO4⇌2Ag⊕+CrO2−4[CrO2−4]=[Ag⊕]2So, [CrO2−4=1.66×10−42M=0.83×10−4M∴KspofAg2CrO4=[Ag⊕]2[CrO2−4]=(1.66×10−4)(0.83×10−4)=2.287×10−12M3

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