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.

Explanation

i. `AgBr + H_(2)O rarr Ag^(o+) (aq) + Br^(Θ) (aq) …S_(1)`
ii. `AgBr` in `0.01 M CaBr_(2) …S_(2)`
`[Br^(Θ)]` added `= 0.01 xx 2 =0.02M`
iii. `AgBr` in `0.01 M KBr …S_(3)`
`[Br^(Θ)` added `= 0.01M`
iv. `AgBr` in `0.05 M AgNO_(3) ..S_(4)`
`[Ag^(o+)]` added `= 0.05M`
Since both `Br^(Θ)` ions and `Ag^(o+)` ions act as common ions, so larger the concentration of `Br^(Θ)` or `Ag^(o+)` ion added, more is the superssion of ionisation of `AgBr` and hence less will the solubility of `AgBr`.
Therefore, the decreasing solubility order.
`S_(1) gt S_(3) gt S_(2) gt S_(4)`