Let the solubilities of `AgBr` in water and in `0.01M CaBr_(2), 0.01M KBr`, and `0.05M AgNO_(3)`be `S_(1), S_(2),S_(3)` and `S_(4)`, respectively. Give the decreasing order of the solubilities.

Options

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)`

Answer

4

Explanation

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)`