The Common-Ion Effect
The solubility of a substance is affected not only by temperature but also by the presence of other solutes. We now consider the case in which the other solute has an ion in common with the substance involved in the equilibrium, in other words we are going to look at the commopn-ion effect in solubility. Just as in the case of the common-ion effect in acid-base equilibria, the common-ion effect in solubility is just a special case of LeChattelier's principle. Consider the equilibrium of CaF2(s) with its saturated solution, if we add a solute containing Ca2+ or a solute containing F- to the saturated solution, the equilibrium will shift to the left making CaF2 less soluble.
CaF2(s) <==> Ca2+(aq) + 2 F-(aq)
<----Addition of Ca2+ or F- shifts equilibrium left decreasing solubility--------<<
Example 1
The Ksp for CaF2 is 4.0 x 10-11. What is the molar solubility of CaF2 in a 0.10 M solution of Ca(NO3)2?
First set up an equilibrium table and let x = change concentration of F-.
Concentrations(M) CaF2(s) <==> Ca2+(aq) + 2 F-(aq)
Starting---------------------------0.10--------- 0
Change --------------------------0.10+x -----+2x
Equilibrium-----------------------0.10+x-------+2x
Ksp = [Ca2+][F-]^2 = (0.10 +x)(2x)^2 = (0.10)[(2x)^2] = 4.0 x 10-11 (assume that x is negligible compared to 0.10 M)
x^2 = (4.0 x 10-11 / (4)(0.10) = 1.0 x 10-10
x= 1.0 x 10-5
PLEASE CHECK MY WORK, I DID THIS QUICKLY SO I MIGHT HAVE MADE A MISTAKE
