Correct Answer - A
If `K_(sp)` is `3.2 xx10^(-17)`
In pure water `K_(sp)=4s^(3)=32xx10^(-18)`
`s_(4)=2xx10^(-6)`
in `NH_(3)` solubility increases due to complex formation.
ln 0.1M `AgNO_(3)`
`Ag_(2)CrO_(4)iff2Ag^(+) + CrO_(4)^(-2)`
`2S_(2)+0.1 " "S_(2)`
`S_(2)=32 xx 10^(-18)`
`ln 0.2 M" "K_(2)CrO_(4)`
`Ag_(2)CrO_(4)iff2Ag^(+)+CrO_(4)^(-2)`
`2S_(1)" "S_(1)+0.2`
`3.2 xx10^(-17)=4S_(1)^(2)`
`16xx10^(-17)=4S_(1)^(2)`
`S_(1)=6.5 xx10^(-9)`
