Solubility Equilibria
I. Brief Review: Stoichiometry of Ionic Solutes, and K and Q expressions revisited
A. Reminder: When ionic solids dissolve in water, they generally are assumed to dissociate
completely. (i.e., 100% of the dissolved formula units separate into ions.)
Ex. Write the net ionic equation that describes the dissolution and dissociation of FeCl3(s):
ANS. FeCl3(s) → Fe3+(aq) + 3 Cl-(aq)
Ex. What are the concentrations of Fe3+ and Cl- in a solution of 0.50 M FeCl3(aq)?
ANS.
[Fe3+] = 0.50 M, because there is one mole of Fe3+ ions for every one mole of FeCl3.
[Cl-] = 1.50 M, because there are three moles of Cl- for every mole of FeCl3:
⎛ 0.50 mol FeCl3 ⎞ ⎛ 3 mol Cl ⎞ 1.50 mol Cl
=
⇒ 1.50 M Cl⎜
⎟
⎜
⎟
L of sol'n ⎠ ⎝ 1 mol FeCl3 ⎠
L of sol'n
⎝
B. Link to equilibrium ideas:
FeCl3 is a soluble ionic compound; this means that quite a bit of FeCl3(s) will dissolve (per liter of
solution) before the solution becomes saturated. But when the solution is saturated, there is a
dynamic equilibrium between undissolved solid and dissolved ions. At this point, the rate of
forward "reaction" (dissolution) and reverse "reaction" (crystallization) are equal, so that
concentrations remain constant even though both processes are always occurring. We can thus treat
this process exactly as we do any other equilibrium process—we can write out the balanced
chemical equation and an associated equilibrium constant expression:
FeCl3(s) p Fe3+(aq) + 3 Cl-(aq); Ksp = [Fe3+]eq[Cl-]eq3
As for all equilibrium equations, there is also an associated Q expression: Qsp = [Fe3+][Cl-]3
The "sp" subscript is just a way to indicate what kind of process is involved—like the subscript "a"
indicates an acid ionization equation.
"sp" stands for "solubility product", because 1) its value is related to the substance's solubility (amount
of solid that dissolves in a given volume of water or solution to reach saturation), and 2) the
equilibrium constant expression in these kinds of processes is always a product of concentration factors
(the denominator is always "one" because the reactant is a solid).
C. Solubility of a substance is related to its Ksp!
Ex. What is the Ksp for FeCl3 at 20 °C, given that its solubility is 92 g per 100 mL at 20 °C?
Don't let this problem trick you! Recall the first example up above! If we know the molar
concentration of FeCl3, we can easily determine the [Fe3+] and [Cl-]. That's all we need! 92 g
per 100 mL equals 920 g per 1000 mL (= 1 L), and the molar mass of FeCl3 is 162 g/mol, so the
concentration of FeCl3 in a saturated solution (i.e., at equilibrium) will be:
⎛ 920 g FeCl3 ⎞ ⎛ 1 mol ⎞ 5.67 mol FeCl3
= 5.67 M FeCl3
⎜
⎟ ⎜ 162 g ⎟ =
1L
L
⎝
⎠⎝
⎠
So, [Fe3+]eq = 5.67 M, and [Cl-]eq = 3(5.67 M) = 17.0 M (!!), and
Kap = (5.67)(17.0)3 = 2.8 x 104 (MUCH GREATER THAN 1)
The K value here is much greater than one, which means this process favors products at
equilibrium. For solubility equilibria, this corresponds to a substance being "soluble"!
**This example was a bit silly; generally we do not calculate or tabulate Ksp's for soluble ionic compounds—
we do so for insoluble or slightly soluble ones. But I hope this helped make the connection.....
II. Reviewing Equilibrium Problem Types and Relating them to a Ksp context (Ksp problems).
NOTE: All Ksp equations will have the form:
Mn A m (s)
⎯⎯
→
←⎯
⎯
n Mx + (aq) + m A y- (aq)
and so all Ksp expressions will have the form:
K sp =
n
⎡⎣Mx + (aq) ⎤⎦ ⎡⎣ A y- (aq) ⎤⎦
m
where Mx+ represents a metal cation, and Ay- represents an anion of any kind.
A. Question: "In which direction will net reaction occur to reach equilibrium?"
To answer, need to calculate Q for the state given, and compare its value to K:
If Q (current state) < K (state at equilibrium), then forward reaction occurs (net)
If Q (current state) > K (state at equilibrium), then reverse reaction occurs (net)
In a Ksp
context
Will a precipitate form? (when solutions providing Mx+ and Ay- are mixed); Ksp is given.
YES, if Qsp > Ksp, because reverse reaction in this context means "solid is formed"
NOTE: if Qsp < Ksp, then the solution is simply unsaturated—forward reaction needs to occur
to reach equilibrium, but there is no solid, so equilibrium cannot be established!
• You need to calculate the concentrations of Mx+ and Ay- in the solution after mixing
(**This often means you have to consider dilution!**)
Example: [Like Exp. 26 in lab (if Ksp were given)!] If you mix 5.00 mL of 0.0120 M
Pb(NO3)2(aq) and 5.00 mL of 0.0300 M KI(aq) [at room T], will a precipitate
form? Assume that Ksp ~ 1 x 10-7
B. Recall the three "Types" of Equilibrium Problems Discussed earlier
1. Type A (or I): "Find K if [ ]eq's are given"
In a Ksp
context
(or find one [ ]eq if other [ ]eq's and K are given)
The solubility of MnAm is [given]. What is Ksp?
(Find Ksp if given solubility)
Example: The solubility of PbBr2(s) at 25 °C is 7.9 g/L. What is Ksp for PbBr2 at 25 °C
In a Ksp
context
Variation:
Find one [ ]eq if the other [ ]eq and Ksp are given
Example: What is the [Br-] in a solution in which [Pb2+] = 0.050 M? Ksp = 4.0 x 10-5 (at a given T)
2. Type B (or II): "Find all [ ]eq's if given [ ]o's and K"
This is when you use a "standard" ICE type of approach: let x represent the amount of a reactant
or product that changes as the system approaches equilibrium, express the equilibrium
concentrations in terms of that x, and then use the equilibrium constant expression equation,
along with algebra, to solve for the value of x, and then the [ ]eq's.
In a Ksp
context
Ksp for MnAm is [given]. What is the solubility of MnAm in ________?
(at a given T)
• could be (mass) solubility (units of g/L or g/100 mL) or molar solubility (units of mol/L)
• solubility in pure water just means that [ ]o's are all zero!
• solubility in a solution of "y molar aqueous potassium [anion name]" OR "z molar aqueous
[metal ion name] nitrate" just means that [ ]o's are NOT all zero! This is a
"common ion" situation!
Example: What is the solubility of FeF3(s) (a) in pure water at 25 °C? (b) in 0.10 M NaF(aq) at
25 °C? (c) in 0.10 M Fe(NO3)3(aq) at 25 °C? Ksp = 1.1 x 10-7 for FeF3(s) at 25 °C.
In general, I ALWAYS write out the dissolution/dissociation equilibrium chemical equation and
associated Ksp expression before doing anything else:
FeF3(s) p Fe3+ + 3 F-;
Ksp = [Fe3+][F-]3
(I've left off the (aq) designations
and "eq's" for simplicity)
(a) If solid is added to pure water, then there is no Fe3+ or F- in the solution to begin with.
Let x = the number of moles of FeF3(s) that dissolve per liter of solution to reach
saturation (equilibrium). Then x also equals the [Fe3+] that forms as dissolution occurs.
[Fe3+] (M)
[F-] (M)
I (initial)
0
0
C (change)
+x
+ 3x
E (at equilibrium)
x
3x
Ksp = [Fe3+][F-]3 => 1.1 x 10-7 = (x)(3x)3 = 27x4 => x4 = 4.07 x 10-9 => x = 0.0080
(+ root)
So 0.0080 mol FeF3 dissolves per liter => the molar solubility = 0.0080 M
And the solubility in g/L = 0.0080 mol/L x 112.85 g/mol = 0.90 g/L (= 0.090 g/100 mL)
(b) If solid is added to a solution of 0.10 M NaF(aq) then there is already 0.10 M F- in
solution (from a source OTHER THAN the FeF3) before the solid is added.
→ This doesn't affect how x is defined! It just affects the initial concentrations:
Let x = the number of moles of FeF3(s) that dissolve per liter of solution to reach
saturation (equilibrium). Then x also equals the [Fe3+] that forms as dissolution occurs.
[Fe3+] (M)
[F-] (M)
I (initial)
0
0.10
C (change)
+x
+ 3x
E (at equilibrium)
x
3x + 0.10
Ksp = [Fe3+][F-]3 => 1.1 x 10-7 = (x)(3x + 0.10)3
Because of the common-ion effect (and the fact that Ksp is small), x [and therefore 3x] will be very
small compared to 0.10, allowing for the (very good) approximation that 3x + 0.10 ≈ 0.10:
1.1 x 10-7 = (x)(0.10)3 = (10-3)x => x = 1.1 x 10-4 (approximation was valid)
So 1.1 x 10-4 mol FeF3 dissolves per liter => the molar solubility = 1.1 x 10-4 M (in 0.10 M
NaF(aq))
And the solubility in g/L = 1.1 x 10-4 M x 112.85 g/mol = 0.012 g/L (= 0.0012 g/100 mL)
→ NOTE: The solubility is over 70 times smaller in the 0.10 M F-(aq) solution due to
the common ion effect (a product ion is already present to start with)
(b) If solid is added to a solution of 0.10 M Fe(NO3)3(aq) then there is already 0.10 M Fe3+
in solution (from a source OTHER THAN the FeF3) before the solid is added.
Let x = the number of moles of FeF3(s) that dissolve per liter of solution to reach
saturation (equilibrium). Then x also equals the [Fe3+] that forms as dissolution occurs.
[Fe3+] (M)
[F-] (M)
I (initial)
0.10
0
C (change)
+x
+ 3x
E (at equilibrium)
x + 0.10
3x
Ksp = [Fe3+][F-]3 => 1.1 x 10-7 = (x + 0.10)(3x)3 ; but x << 0.10 (as above) so:
1.1 x 10-7 = (0.10)(3x)3 = (0.1)(27x3) => x3 = 4.1 x 10-8 => x = 0.0034
molar solubility = 0.0034 M
solubility = 0.39 g/L = 0.039 g/100 mL
(approximation was valid,
although it's getting a bit close)
(Solubility about 2.5 times smaller
than in pure water—common
ion effect)
NOTE: 0.10 M F- does not have the same exact effect as 0.10 M Fe3+--this is due to the
different stoichiometry (and thus exponent). Be VERY careful with your ICE
tables when dealing with common-ion effect type problems of this type!
3. Type C (III): "Find K from [ ]o's and ONE [ ]eq
In a Ksp
context
Find Ksp from [ ]o's and ONE [ ]eq (This is like Exp. 26 in lab!)
Example: You mix 5.00 mL of 0.0120 M Pb(NO3)2 and 5.00 mL of KI, and after equilibrium is
reached, the concentration of I- is 0.0075 M. What is Ksp?
III. LeChatelier's Principle Applied to Solubility Equilibria (and the effect of acidity/basicity)
To decrease the solubility of an ionic solid:
• Add (or start off with) a common ion.
Example: Adding CaCl2 or Na2CO3 decreases the solubility of CaCO3.
To increase the solubility of an ionic solid:
• Add something that will react with either the dissolved cation or anion
-- Addition of an acid (or adjusting/maintaining the pH to a lower value using a buffer, etc.)
(If the anion in the ionic compound is NOT a negligible base)
Examples: CaCO3, PbS, Fe(OH)3, etc. will be more soluble at lower pH's, because the
anions will be converted into their conjugate acids: HCO3- (or even H2CO3),
HS- (or even H2S), H2O, etc.
***But AgCl or PbI2 will be unaffected because Cl- and I- are negligible bases***
→ Effects of acid rain!!!
◊ Marble statues/structures (CaCO3(s)) dissolve more readily
◊ Heavy metals are released in soil (e.g., galena (PbS) is a lead containing mineral whose
solubility will increase as [H3O+] increases)
-- Addition of a species (later, called a "ligand") that will form a complex ion with the metal ion
Example: AgCl can be made more soluble by adding NH3 because the following reaction
occurs, shifting the dissolution equilibrium to the "right":
AgCl(s) p
Ag+
+
2 NH3
Ag(NH3)2+
+ Cl-