Ksp+and+ICE

Solubility Equilibrium and ICE
Example: what would the equilibrium concentrations of the products be if 0.10 gram of solid silver phosphate was added to 500. mL of water?

Silver phosphate (Ag 3 PO 4 ) is only slightly soluble in water (Ksp = 2.8 x 10 -18 ), and would dissociate as follows:

Ag 3 PO 4 (s) <--> 3 Ag + (aq) + PO 4 3- (aq)

Its equilibrium constant expression is as follows:

Ksp = [Ag + ] 3 [PO 4 3- ]

Set up an ICE chart right under the chemical equation.

Since Ag 3 PO 4 is a solid, it is ignored; initially, there are no products:


 * || Ag 3 PO 4 (s) || <--> || 3 Ag + (aq) || + || PO 4 3- (aq) ||
 * Initial (I): || -- ||  || 0 ||   || 0 ||
 * Change (C): || -- ||  ||   ||   ||   ||
 * Equilibrium (E): || -- ||  ||   ||   ||   ||

As the reaction proceeds, a quantity of silver phoshpate is used up, and is converted according to mole ratio into the products:


 * || Ag 3 PO 4 (s) || <--> || 3 Ag + (aq) || + || PO 4 3- (aq) ||
 * Initial (I): || -- ||  || 0 ||   || 0 ||
 * Change (C): || -- ||  || + 3x ||   || + x ||
 * Equilibrium (E): || -- ||  ||   ||   ||   ||

The equilibrium position is reached:


 * || Ag 3 PO 4 (s) || <--> || 3 Ag + (aq) || + || PO 4 3- (aq) ||
 * Initial (I): || -- ||  || 0 ||   || 0 ||
 * Change (C): || -- ||  || + 3x ||   || + x ||
 * Equilibrium (E): || -- ||  || 3x ||   || x ||

Now use the E row by inserting its values into the equilibrium constant expression.


 * || Ag 3 PO 4 (s) || <--> || 3 Ag + (aq) || + || PO 4 3- (aq) ||
 * Equilibrium (E): || -- ||  || 3x ||   || x ||

Ksp = 2.8 x 10 -18 = [Ag + ] 3 [PO 4 3- ] = (3x) 3 (x) 2.8 x 10 -18 = 27x 4

and solving for x yields the **molar solubility** for silver phosphate:

x = 1.8 x 10 -5 M

Now refer to the ICE chart; x is the value for the molar concentrations of Ag + and PO 4 3-, so therefore:

[Ag + ] = 3x = 3(1.8 x 10 -5 M) = 5.4 x 10 -5 M

[PO 4 3- ] = x = 1.8 x 10 -5 M

From here, you could do some work to figure out how many moles and grams of silver phosphate actually dissolved.