1. 16.5-16.6 Solubility Equilibria and the Solubility Product Constant - Precipitation

2. Solubility
Remember, solubility means that the solute dissolves in water, and also that there are solubility rules that according to the AP Board, you do not need to memorize.
However, the college board does say:
Just as some acid and bases don’t fully dissociate, that same goes for ionic compounds. In order to determine the degree of dissociation, we can use, or solve for, the solubility product constant: Ksp
example: The expression for the solubility constant for CaF2(s)
is: Ksp = [Ca2+][F-]2
Ksp values are on pg. 784 of your text.
Just a reminder, solids are omitted from equilibrium expressions because their concentrations are constant.

3. Molar Solubility
Molar solubility, the solubility in units of mol/L or M
Ksp is not molar solubility. Ksp has only one value at a given temperature, but the solubility can be a different value in different kinds of solutions.
For example, AgCl in water is going to have a different solubility than AgCl in a saline solution.
In a saline solution, there is a common ion… the Cl- ion, which is going to make AgCl less soluble, and according to Le Chatelier’s Principle, it will shift the equilibrium in which direction?
Left, producing more precipitate
In order to figure out molar solubility from Ksp, we need to set up an ICE table, and instead of x, we will use S to represent the concentration that dissolves (which is the molar solubility).

4. Let’s Try a Practice Problem!
Calculate the molar solubility of Fe(OH)2 in pure water. (the Ksp of Fe(OH)2 at 25oC is 4.87x10-17) Fe(OH)2(s) Fe2+(aq) + 2OH-(aq) Ksp = [Fe2+][OH-]2 4.87x10-17 = S(2S)2 4S3 = 4.87x x10-17 S3 = = 1.22x10-17 S = 𝑥10−17 = 2.30x10-6 mol / L or 2.30x10-6 M 4
[Fe2+] M
[OH-]2
Initial
0.00
Change +S
+2S
Equilibrium S 2S

5. The Effect of a Common Ion on Solubility
In general, the solubility of an ionic compound is lower in a solution containing a common ion than in pure water. This is because according to Le Chatelier’s Principle, the presence of a common ion causes the equilibrium to shift left, which means less solid dissolves (decreasing solubility).
Let’s Try a Practice Problem!
Calculate the molar solubility of CaF2 in a solution containing M Ca(NO3)2. (Here you can tell that the initial [Ca2+] = M (The Ksp of CaF2 = 1.46x10-10)
CaF2(s) Ca2+(aq) + 2F-(aq)
Ksp = [Ca2+][F-] Continued
[Ca2+] M
[F-]2
Initial
0.250
0.00
Change +S
+2S
Equilibrium S 2S

6. Ksp = [Ca2+][F-]2
1.46x10-10 = ( S)(2S)2
S2 = 1.46x10-10
S = 1.21x10-5 M
Let’s Try a Quick Question That’s Test Your Understanding of the Common Ion Effect.
In which solution is BaSO4 most soluble?
In a solution that is 0.10 M in BaNO3
In a solution that is 0.10 M in Na2SO4
In a solution that is 0.10 M in NaNO3
(c) In a solution that is 0.10 M in NaNO3 , because there are no
common ions.

7. The Effect of pH on Solubility
In general, the solubility of an ionic compound with a strongly basic or weakly basic anion increases with increasing acidity (decreasing pH)
Common basic anions include: OH-, S2-, and CO32-.
Which compound, FeCO3 or PbBr2, is more soluble in acid than base?
FeCO3 will dissolve faster in acidic solution because the CO32- ion is a basic anion whereas, Br- is the conjugate base of a strong acid and is therefore pH neutral.

8. Precipitation
The Relationship between Q (the value of the product under any condition) and Ksp (the value of the product at equilibrium only)
If Q < Ksp, the solution is unsaturated and more of the solid ionic compound can dissolve in the solution.
If Q = Ksp, the solution is saturated. The solution is holding the equilibrium amount of the dissolved ions and additional solid does not dissolve in solution.
If Q > Ksp, the solution is supersaturated. Under most circumstances, the excess solid would precipitate out of a supersaturated solution.  the only condition where a precipitate will form.

9. Let’s Try a Practice Problem!
A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is M in Pb(NO3)2 and M in NaBr, will a precipitate form in this newly mixed solution.
First, we have to look at the cross products of the reaction:
Pb(NO3)2(aq) + 2NaBr(aq) PbBr2(s) + 2NaNO3(aq)
(Remember, according to our solubility guidelines sodium nitrate is soluble, but lead(II) bromide could precipitate out of solution…this all depends on if Q > Ksp)
Let’s find out. The Ksp of PbBr2 is 4.67x So we need to find Q which is based on the given concentrations above:
PbBr2(s) Pb2+(aq) + 2Br-(aq)
Q = [Pb2+][Br-]2 = (0.0600)(0.0158)2 = 1.50x10-5
Q > Ksp, therefore precipitate forms.

10. Selective Precipitation
Selective precipitation: a process involving the addition of reagent that forms a precipitate with one of the dissolved cations but not the other.
This goes along with what we did when we performed the gravimetric analysis lab.
The difference in Ksp values required for selective precipitation is a factor of at least The precipitating reagent will cause the ionic compound with the smaller Ksp to precipitate out.

11. Let’s Try a Practice Problem!
The magnesium and calcium ions present in seawater ([Mg2+] = M and [Ca2+] = M) can be separated by selective precipitation with KOH. What minimum [OH-] triggers the precipitation of the Mg2+ ion? Mg(OH)2 has a Ksp of 2.06x10-13 and Ca(OH)2 has a Ksp of 4.68x10-6. (also remember, the precipitation will commence when the value of Q for the precipitating compound just equals the value of Ksp) KspMg(OH)2 = QMg(OH)2 Mg(OH)2(S) Mg2+(aq) + 2OH-(aq) 2.06x10-13 = [Mg2+][OH-]2 = (0.059)(OH-)2 [OH-]2 = (2.06x10-13)/(0.059) = 3.49x10-12 [OH-] = 1.9X10-6 M

12. Study for quiz on chapter 16!