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Solubility of `AgBr` in water is `S_(1),` in `"0.01 M "CaBr_(2)` is `S_(2)` in `0.01 M NaBr` is `S_(3)` and in `0.05 M AgNO_(3)` is `S_(4)`. The correct order of these solubilities is `:`
A. `S_(1)gt S_(2)gtS_(3)gtS_(4)`
B. `S_(1)gt S_(2)=S_(3)gtS_(4)`
C. `S_(4)gt S_(3)gtS_(2)gtS_(1)`
D. `S_(1)gt S_(3)gtS_(2)gtS_(4)`

1 Answer

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Correct Answer - 4
In water, no common ion effect will act hence `S_(1)` will be maximum .
`{:([CaBr_(2)]=0.01,,,[Br^(-)]=0.02,,),([NaBr]=0.01,,,[Br^(-)]0.01,,),([AgNO_(3)]=0.05,,,[Ag^(+)]=0.05,,):}`
More the concentration of common ion, less will be the solubility. `S_(1)gtS_(3)gtS_(2)gtS_(4)`

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