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

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

1 Answer
Stefan V.
Feb 17, 2017

#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.

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