Given that , the solubility product `K_(sp)` , of AgCl is `1.8 xx 10^(-10)` , the concentration of `Cl^(-)` ions that must just be exceeded before AgCl will precipitate from a solution containing `4 xx 10^(-3)` M `Ag^(+)` ions isA. `4.5 xx 10^(-8) M`B. `4 xx 10^(-8) M`C. `1.8 xx 10^(-8) M`D. `1 xx 10^(-8) M`

Given that , the solubility product `K_(sp)` , of AgCl is `1.8 xx 10^(

1.	Given that , the solubility product `K_(sp)` , of AgCl is `1.8 xx 10^(-10)` , the concentration of `Cl^(-)` ions that must just be exceeded before AgCl will precipitate from a solution containing `4 xx 10^(-3)` M `Ag^(+)` ions isA. `4.5 xx 10^(-8) M`B. `4 xx 10^(-8) M`C. `1.8 xx 10^(-8) M`D. `1 xx 10^(-8) M`
Answer» Correct Answer - A `[Ag^(+)][Cl^(-)]=1.8 xx 10^(-10)` `[Cl^(-)]=(1. 8 xx 10^(-10))/(4 xx 10^(-3))=4.5 xx 10^(-8)` Thus, if conc. Of `Cl^(-)` ions exceeds `4.5 xx 10^(-8)` M precipitation will occur because ionic product will exceed `K_(sp)`.

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