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

1 Answer
anor277
Dec 11, 2016

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.

See [here for similar treatments.](https://socratic.org/questions/what-is-ksp-in-chemistry)

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