If you have a salt in a solution of another salt having the same cation or anion, the salt will be less soluble in this solution than it would be in pure water. For example, when adding calcium oxalate, CaC2O4, to a solution of calcium chloride, CaCl2, both salts will contribute a Ca2+ ion. This change in solubility is explained by Le Chatelier's principle:
CaC2O4 (s) ⇌ Ca2+ (aq) + C2O42- (aq)
Calcium chloride will also contribute Ca2+ ion, which will act as a stress on the right hand side of the equation and the equilibrium will shift to the left, and calcium oxalate will precipitate from the solution.
Example:
MgC2O4is added to a 0.437 M solution of Na2C2O4. What is the solubility of MgC2O4 in this solution?
MgC2O4 ⇌ Mg2+ (aq) + C2O42- (aq)
Notice that there is an initial value for C2O42-, contributed by the Na2C2O4.
Ksp = (x)(0.437 + x) = 8.5 x 10-5
(always check, but the approximation should hold for Ksp problems)
x = 1.95 x 10-4 M
Remember the question asked for solubility, not molar solubility. So we have to convert mols to grams using the periodic table:
( 1.95 x 10-4 mol / 1 L) x (112 g MgC2O4 / 1 mol MgC2O4) = 0.022 g/L
Example:
Find the molar solubility for Ca3(PO4)2 in 0.115 M solution of Ca(NO3)2.
(Ksp for Ca3(PO4)2 = 1 x 10-26)
Ca3(PO4)2 (s) ⇌ 3Ca2+ (aq) + 2PO43- (aq)
Ksp = (0.115 + 3x)3(2x)2 → approximation holds
Ksp = (0.115)3(4x2) = 1 x 10-26
x = 1.28 x 10-12
Precipitation Reactions:
To try to find out whether or not precipitation will occur in a reaction, we need to compare the Qc to the Kc, or in this case, the Ksp. (We went over Qc and comparing it to Kc in Chapter 14). The Qc, or ion product as it is called when considering solubility, is found the same way as the Ksp, except the Ksp is found at equilibrium and the Qc is not.
Example:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
[Ag+][Cl-] = Ksp at equilibrium
[Ag+][Cl-] = Qc when not at equilibrium
When Qc is less than Ksp, there is not enough product yet to be at equilibrium, so the reaction will shift to the right, and no ppt (precipitate) will form.
When Qc is greater than Ksp, there is too much product, so the reaction will shift to the left, and a ppt will form.
When Qc is equal to Ksp, the reaction is at equilibrium.
Example:
Will there be a ppt if 50.0mL of 0.450 M AgNO3 is mixed with 50.0mL of 0.350 M NaCl?
The first step is to think about what the possible ppt might be from these compounds. Write out what happens when these two compounds break up:
AgNO3 (s) ⇌ Ag+ (aq) + NO3-(aq)
NaCl (s) ⇌ Na+ (aq) + Cl- (aq)
When these four ions are floating around in the solution, the only possible ppt is AgCl.
Because we are mixing, the first thing we need to do is convert to mols then divide by the new volume to find the new concentrations of AgNO3 and NaCl:
(0.500 L)(0.450 mol / 1 L) = (0.0225 mol) / (0.1 L) = 0.225 M AgNO3
(0.500 L)(0.350 mol / 1 L) = (0.0175 mol) / (0.1 L) = 0.175 M NaCl
Qc = [Ag+][Cl-] = (0.225)(0.175) = 0.0394
Ksp = 1.8 x 10-10
Qc is greater than Ksp, the reaction will shift to the left and a ppt will form.
Fractional Precipitation:
A technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth.
Example:
Na2C2O4 (s) ⇌ 2Na+ (aq) + C2O42- (aq)
Three compounds with low solubility will form: CdC2O4, CaC2O4, MgC2O4.
Ksp for CdC2O4 = 1.5 x 10-8
Ksp for CaC2O4= 2.3 x 10-9
Ksp for MgC2O4 = 8.5 x 10-5
A precipitate will form when Qc is greater than Ksp. In other words, just past the equilibrium. So an easy way to see when a ppt will form is to set Qc = Ksp. To see which ppt will form first, look at which Ksp will be reached first, in other words, compare to see which Ksp is smaller.
In this case, CaC2O4 has the smallest Ksp, so it will be reached first, and CaC2O4 will form ppt first. The second largest is CdC2O4, so this will ppt out second, and MgC2O4 last.
So this question is extremely easy! When asked from a list of compounds which will precipitate out first, or the order of precipitation, look at the Ksp values and go from smallest to largest.
A harder question would be one that asks when each of the ions fall out of solution. In other words, at what [C2O42-] does each metal fall out?
First look at each compound and build the Ksp equation. Solve for the unknown, [C2O42-].
Example:
For CaC2O4:
Ksp = [Ca2+][C2O42-]
2.3 x 10-9 = (0.200)[C2O42-]
[C2O42-] = 1.15 x 10-8 M
Note that the concentration for Ca2+ was given at the beginning of the problem.
For CdC2O4:
Ksp = [Cd2+][C2O42-]
1.5 x 10-8 = (0.200)[C2O42-]
[C2O42-] = 7.5 x 10-8
For MgC2O4:
Ksp = [Mg2+][C2O42-]
8.5 x 10-5 = (0.200)[C2O42-]
[C2O42-] = 4.25 x 10-4
Example:
In this same situation, what is the percent Ca2+ ion left when Cd2+ begins to ppt?
Remember that we figured out earlier than Cd2+ begins to ppt when [C2O42-] = 7.5 x 10-8. To see how much Ca2+ has fallen out at this point, substitute this value for [C2O42-] into the following equation and solve for Ca2+:
Ksp = [Ca2+][C2O42-]
2.3 x 10-9 = [Ca2+](7.5 x 10-8)
[Ca2+] = 0.0307 M
To find the percent Ca2+ that has precipitated out at this point, divide this concentration by the original concentration of Ca2+ and multiply by 100:
(0.0307 M) / (0.200 M) x 100 = 15.3 %
Effect of pH on Solubility:
What happens to the solubility of a compound when a strong acid is added?
Consider a solution of the solids CaSO4 and CaCO3 (according to solubility rules, these are insoluble, so imagine them as solids resting at the bottom of a solution while a tiny percent of their ions escape into solution).
CaSO4 (s) ⇌ Ca2+ (aq) + SO42-(aq)
CaCO3 (s) ⇌ Ca2+ (aq) + CO32- (aq)
If a strong acid is added, H3O+ ion is added to the solution. The anions will react with it:
SO42- (aq) + H3O+ (aq) ⇌ HSO4- (aq) + H2O (l)
The small amount of free SO42- ions will react with H3O+. According to Le Chatelier's, this will drive more SO42- ion out of the CaSO4, but only by a very small margin. The other compound will be more effected by the change in [H3O+]:
CO32- (aq) + H3O+ (aq) ⇌ HCO3- (aq) + H2O (l)
HCO3- reacts with more of the H3O+:
HCO3- (aq) + H3O+ (aq) ⇌ H2CO3 (aq) + H2O (l)
H2CO3 is unstable and reacts further:
H2CO3 (aq) → H2O (l) + CO2 (g)
If the container is open, the CO2 gas will escape, and according to Le Chatelier's, the HCO3- will continue to replenish out of the CaCO3. This compound will be more soluble in the presence of a strong acid.
Complex Ion Formation:
In Chapter 15 we learned the definition of a Lewis Acid/Base:
Lewis Acid: a species that can form a covalent bond by accepting an electron pair from another species.
Lewis Base: a species that can form a covalent bond by donating an electron pair to another species.
And in previous classes we learned that a coordinate covalent bond is a form of bonding between two atoms in which both electrons shared in the bond come from the same atom.
Complex Ion: an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond. A complex is a compound containing complex ions.
Example:
Ligand: a Lewis base that bonds to a metal ion to form a complex ion.
Example:
The overall equation for when aqueous silver ions form a complex ion with ammonia:
Ag+ (aq) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq)
Ag+ is the Lewis acid. NH3 is the Lewis base, and ligand.
The formation constant, or stability constant: Kf, is yet another form of Kc. The equilibrium constant for the formation of the complex ion from aqueous metal ion and the ligands.
For Ag(NH3)2+:
Kf = [Ag(NH3)2+] / [Ag+][NH3]2
Kf values are usually very large. For example,
Kf for Ag(NH3)2+ = 1.7 x 107
When doing calculations with Kf values, it would be harder to work with these values, because the approximation method will never hold. An easy way around this is to remember the mathematical rules that apply to Kc.
When a reaction is reversed so that the reactants are products and the products are reactants, the new Kc is the inverse (1 / Kc) of the original Kc.
Dissociation Constant: Kd, the inverse of Kf.
Example:
Reverse the reaction and take the new Kf, which will be Kd:
Ag(NH3)2+ (aq) ⇌ Ag+ (aq) + 2NH3 (aq)
Kd = [Ag+][NH3]2 / [Ag(NH3)2+] =( 1 / Kf)
Example:
For the formation of the complex ion:
Cu2+ (aq) + 4NH3 (aq) ⇌ Cu(NH3)42+ (aq), Kf = 4.8 x 1012
What is the [Cu2+] at equilibrium when you start with 0.100 M Cu(NO3)2 and 1.00 M NH3?
Note that the approximation method will not hold for these numbers, and we'll be left to work through some unnecessarily tedious math. Or, we could use some shortcuts and make it much easier:
First, assume that the reaction goes to completion:
Cu2+ (aq) + 4NH3 → Cu(NH3)42+ (aq)
Note that only 0.100 M is subtracted in the second line of the table because there is an excess of NH3, and less Cu2+.
Now reverse the reaction to let some of the complex ion fall apart and convert Kf to Kd:
Cu(NH3)42+ (aq) ⇌ Cu2+ (aq) + 4NH3 (aq)
Kd = 1 / Kf = 2.08 x 10-13
2.08 x 10-13 = (x)(0.600 + 4x)4 / (0.100 – x)
Approximation method holds,
2.08 x 10-13 = (x)(0.600)4 / (0.100)
x = 1.60 x 10-13
Complex Ions and Solubility:
Example:
Calculate the molar solubility of AgCl in 1.0 M NH3.
This problem involves the solubility equilibrium for AgCl and the complex ion equilibrium:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq), Ksp = 1.8 x 10-10
Ag+ (aq) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq), Kf = 1.7 x 107
The overall equation for the process can be obtained by applying Hess's Law and adding them together:
AgCl (s) + Ag+ (aq) + 2NH3 (aq) ⇌ Ag+ (aq) + Cl- (aq) + Ag(NH3)2+ (aq)
The Ag+ will cancel on each side and the equation will be:
AgCl (s) + 2NH3 (aq) ⇌ Cl- (aq) + Ag(NH3)2+ (aq)
And remember from Chapter 14 that when you add two reactions, you obtain the new equilibrium constant by multiplying their equilibrium constants.
Kc = Ksp x Kf =
(1.8 x 10-10)(1.7 x 107) = 0.00306
0.00306 = (x2) / (1.00 – 2x)2 → approximation holds
0.00306 = (x2) / (1.00)2
x = 0.0553 M = [Ag(NH3)2+] = [AgCl]
