Question

Given `K_"sp"*AgCl=1.8xx10^(-10)`, if `[Ag^+]=10^(-10)` and `[Cl^-]=10^(-4)`, will silver chloride precipitate?

Answer 1

`Ag^(+) + Cl^(-) rightleftharpoonsAgCl(s)darr`

`K_"sp"=1.8xx10^-10`

Explanation:

We have the `K_"sp"` value, and thus we know that if the ion product `[Ag^+][Cl^-]>K_"sp"`, silver chloride will precipitate until `[Ag^+][Cl^-]-=K_"sp"`.

For `A.` `[Ag^+]=10^-10*mol*L^-1`; `[Cl^-]=10^-4*mol*L^-1`.

The ion product `[Ag^+][Cl^-]=10^-10xx10^-4=10^-14`,

Since this ion-product is LESS than `K_"sp"` silver chloride will not precipitate. I think you are due to do the next problems.

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