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The solubility product constant can be used to compute the equilibrium concentration of ions in solution in the same way that you use other equilibrium constant expressions to compute equilibrium concentrations.

Use the Initial, Change, Equilibrium method to compute the concentrations: write the equilibrium expression, the initial amounts of each thing in the expression, how much they change by, then the equilibrium amounts. From that, you can compute the concentrations of all species.

Example: BaF2 has a Ksp of 1.8*10-7. What is the concentration of barium in a saturated solution of BaF2?

Solution:. First, write the balanced equation and the solubility constant expression:

BaF2(s) < = > Ba+2(aq) + 2F-(aq)
Ksp = [Ba+2][F-]2
We start with no barium or fluoride ions in solution. If we set the amount of barium ions formed to x, then the number of flouride ions formed is just 2x

[Ba+2] (M) [F-] (M)
Initial 0 0
Change +x +2x
Equilibrium x 2x


Now plug in the equilibrium values shown above and solve for x
Ksp = [Ba+2][F-]2
Ksp = (x)(2x)2
x = 3.6*10-3
Since we defined the concentration of Ba+2 to be x, we've solved the problem
[Ba+2] = 3.6*10-3


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