Suppose the solubilities of AgCl in water, in 0.01 M `CaCl_(2)` solution, in 0.01 M NaCl solution and in 0.05 M `AgNO_(3)` solution are `S_(1), S_(2), S_(3) and S_(4)` respectively. The correct order of these solubilities is

Options

A. `S _(1) gt S _(2) gt S _(3) gt S _(4)`
B. `S _(1) gt S _(2) = S _(3) gt S _(4)`
C. `S _(1) gt S _(3) gt S _(2) gt S _(4)`
D. `S _(4) gt S _(2) gt S _(3) gt S _(1)`

Answer

D

Explanation

AgCl in water `[Ag ^(+)] [CL ^(-)] = 10 ^(-10)`
`S xx S = S ^(2) = 10 ^(-10)`
`S = 10 ^(-5) mol dm ^(-3) = S _(1)`
`AgCl i n 0.01 M Ca Cl _(2) ,[Ag ^(+)] [0.02] = 19 ^(-10)`
More the concentration of common ion less the solubility of AgCl.
`[Ag ^(+)] = (10 ^(-10))/( 2 xx 1 ^(-2)) = 0. 5 xx 10 ^(-8) = 5 xx 10 ^(-9) mol dm ^(-3) = S _(2).`
Agcl in `0.01 M NaCl ,[Ag ^(+)] [10 ^(-2)] = 10 ^(-10)` ,br> `[Ag ^(+)] = (10 ^(-10))/( 10 ^(-2)) = 10 ^(-8) mol dm ^(-3) = S _(3), AgCl ` in `0.05 MAgCO _(3) [0.05][Cl] = 10 ^(-10)`
`[Cl^(-)] = (10 ^(-10))/( 5 xx 10 ^(-2)) = 0.2 xx 10 ^(-8) = 2 xx 10 ^(-9) mol dm ^(-3) = S _(4)` Hence, `S _(1) gt S _(3) gt S _(2) gt S _(4)`