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

Dec 11, 2016

$A {g}^{+} + C {l}^{-} r i g h t \le f t h a r p \infty n s A g C l \left(s\right) \downarrow$

${K}_{\text{sp}} = 1.8 \times {10}^{-} 10$

#### Explanation:

We have the ${K}_{\text{sp}}$ value, and thus we know that if the ion product $\left[A {g}^{+}\right] \left[C {l}^{-}\right] > {K}_{\text{sp}}$, silver chloride will precipitate until $\left[A {g}^{+}\right] \left[C {l}^{-}\right] \equiv {K}_{\text{sp}}$.

For $A .$ $\left[A {g}^{+}\right] = {10}^{-} 10 \cdot m o l \cdot {L}^{-} 1$; $\left[C {l}^{-}\right] = {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$.

The ion product $\left[A {g}^{+}\right] \left[C {l}^{-}\right] = {10}^{-} 10 \times {10}^{-} 4 = {10}^{-} 14$,

Since this ion-product is LESS than ${K}_{\text{sp}}$ silver chloride will not precipitate. I think you are due to do the next problems.