Given Ksp⋅AgCl=1.8×10−10, if [Ag+]=10−10 and [Cl−]=10−4, will silver chloride precipitate?

1 Answer
anor277
Dec 11, 2016

Ag++Cl−⇌AgCl(s)⏐↓

Ksp=1.8×10−10

Explanation:

We have the Ksp value, and thus we know that if the ion product [Ag+][Cl−]>Ksp, silver chloride will precipitate until [Ag+][Cl−]≡Ksp.

For A. [Ag+]=10−10⋅mol⋅L−1; [Cl−]=10−4⋅mol⋅L−1.

The ion product [Ag+][Cl−]=10−10×10−4=10−14,

Since this ion-product is LESS than Ksp 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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