Question

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

Its `K_(sp)` value is `8.3*10^-17`.

Answer 1

`9.11 * 10^(-9)"M"`

Explanation:

Silver iodide, `"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))^(-)`

Notice that every mole of silver iodide that dissolves produces `1` mole of silver cations and `1` mole of iodie anions.

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

`K_(sp) = ["Ag"^(+)] * ["I"^(-)]`

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

`K_(sp) = s * s = s^2`

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.