What is the solubility in mol/L of silver iodide, "AgI"AgI ?

Its K_(sp)Ksp value is 8.3*10^-178.3⋅10−17.

1 Answer
Stefan V.
Feb 17, 2017

9.11 * 10^(-9)"M"9.11⋅10−9M

Explanation:

Silver iodide, "AgI"AgI, is an insoluble ionic compound, which basically means that when this compound is added to water, an equilibrium is established between the undissolved solid and the dissolved ions.

"AgI"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "I"_ ((aq))^(-)AgI(s)⇌Ag+(aq)+I−(aq)

Notice that every mole of silver iodide that dissolves produces 11 mole of silver cations and 11 mole of iodie anions.

The expression for the solubility product constant, K_(sp)Ksp, of silver iodide looks like this

K_(sp) = ["Ag"^(+)] * ["I"^(-)]Ksp=[Ag+]⋅[I−]

If you take ss to be the molar solubility of silver iodide, you can say that you will have

K_(sp) = s * s = s^2Ksp=s⋅s=s2

This means that the molar solubility of this salt will be equal to

s = sqrt(K_(sp)) = sqrt(8.3 * 10^(-17)) = color(darkgreen)(ul(color(black)(9.11 * 10^(-9)"M")))

In other words, you can only hope to dissolve 9.11 * 10^(-9) moles of silver iodide for every liter of water at room temperature.

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