17.2-17.6: Solubility and the Common Ion Effect-Complex Ions and Solubility

Solubility and the Common Ion Effect:

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) + C₂O₄^2- (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) + C₂O₄^2- (aq)

Notice that there is an initial value for C₂O₄^2-, contributed by the Na₂C₂O₄.

K_sp = (x)(0.437 + x) = 8.5 x 10^-5
(always check, but the approximation should hold for K_sp 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.
(K_sp for Ca₃(PO₄)₂ = 1 x 10^-26)

Ca₃(PO₄)₂ (s) ⇌ 3Ca²⁺ (aq) + 2PO₄^3- (aq)



K_sp = (0.115 + 3x)³(2x)² → approximation holds
K_sp = (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 Q_c to the K_c, or in this case, the K_sp. (We went over Q_c and comparing it to K_c in Chapter 14). The Q_c, or ion product as it is called when considering solubility, is found the same way as the K_sp, except the K_sp is found at equilibrium and the Q_c is not.

Example:

AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq)
[Ag⁺][Cl^-] = K_sp at equilibrium
[Ag⁺][Cl^-] = Q_c when not at equilibrium

When Q_c is less than K_sp, 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 Q_c is greater than K_sp, there is too much product, so the reaction will shift to the left, and a ppt will form.

When Q_c is equal to K_sp, 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

Q_c = [Ag⁺][Cl^-] = (0.225)(0.175) = 0.0394
K_sp = 1.8 x 10^-10

Q_c is greater than K_sp, 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:

Start with a solution of 0.200M each of Cd(NO₃)₂, Ca(NO₃)₂, Mg(NO₃)₂. Each will dissociate to give 0.200M each of Cd²⁺, Ca²⁺, and Mg²⁺. If Na₂C₂O₄ is added slowly, what is the order of ppt? Recognize that each of the three ions will react with the oxalate anion in Na₂C₂O₄, which is C₂O₄^2-.

Na₂C₂O₄ (s) ⇌ 2Na⁺ (aq) + C₂O₄^2- (aq)

Three compounds with low solubility will form: CdC₂O₄, CaC₂O₄, MgC₂O₄.

K_sp for CdC₂O₄ = 1.5 x 10^-8
K_sp for CaC₂O₄= 2.3 x 10^-9
K_sp for MgC₂O₄ = 8.5 x 10^-5

A precipitate will form when Q_c is greater than K_sp. In other words, just past the equilibrium. So an easy way to see when a ppt will form is to set Q_c = K_sp. To see which ppt will form first, look at which K_sp will be reached first, in other words, compare to see which K_sp is smaller.

In this case, CaC₂O₄ has the smallest K_sp, 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 K_sp 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 [C₂O₄^2-] does each metal fall out?

First look at each compound and build the K_sp equation. Solve for the unknown, [C₂O₄^2-].

Example:

For CaC₂O₄:

K_sp = [Ca²⁺][C₂O₄^2-]
2.3 x 10^-9 = (0.200)[C₂O₄^2-]
[C₂O₄^2-] = 1.15 x 10^-8 M
Note that the concentration for Ca²⁺ was given at the beginning of the problem.

For CdC₂O₄:

K_sp = [Cd²⁺][C₂O₄^2-]
1.5 x 10^-8 = (0.200)[C₂O₄^2-]
[C₂O₄^2-] = 7.5 x 10^-8

For MgC₂O₄:

K_sp = [Mg²⁺][C₂O₄^2-]
8.5 x 10^-5 = (0.200)[C₂O₄^2-]
[C₂O₄^2-] = 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 [C₂O₄^2-] = 7.5 x 10^-8. To see how much Ca²⁺ has fallen out at this point, substitute this value for [C₂O₄^2-] into the following equation and solve for Ca²⁺:

K_sp = [Ca²⁺][C₂O₄^2-]
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) + SO₄^2-(aq)
CaCO₃ (s) ⇌ Ca²⁺ (aq) + CO₃^2- (aq)

If a strong acid is added, H₃O⁺ ion is added to the solution. The anions will react with it:

SO₄^2- (aq) + H₃O⁺ (aq) ⇌ HSO₄^- (aq) + H₂O (l)

The small amount of free SO₄^2- ions will react with H₃O⁺. According to Le Chatelier's, this will drive more SO₄^2- ion out of the CaSO₄, but only by a very small margin. The other compound will be more effected by the change in [H₃O⁺]:

CO₃^2- (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: K_f, is yet another form of K_c. The equilibrium constant for the formation of the complex ion from aqueous metal ion and the ligands.

For Ag(NH₃)₂⁺:

K_f = [Ag(NH₃)₂⁺] / [Ag⁺][NH₃]²

K_f values are usually very large. For example,

K_f for Ag(NH₃)₂⁺ = 1.7 x 10⁷

When doing calculations with K_f 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 K_c.

When a reaction is reversed so that the reactants are products and the products are reactants, the new K_c is the inverse (1 / K_c) of the original K_c.

Dissociation Constant: K_d, the inverse of K_f.

Example:

Reverse the reaction and take the new K_f, which will be K_d:

Ag(NH₃)₂⁺ (aq) ⇌ Ag⁺ (aq) + 2NH₃ (aq)

K_d = [Ag⁺][NH₃]² / [Ag(NH₃)₂⁺] =( 1 / K_f)

Example:

For the formation of the complex ion:

Cu²⁺ (aq) + 4NH₃ (aq) ⇌ Cu(NH₃)₄²⁺ (aq), K_f = 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 K_f to K_d:

Cu(NH₃)₄²⁺ (aq) ⇌ Cu²⁺ (aq) + 4NH₃ (aq)



K_d = 1 / K_f = 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), K_sp = 1.8 x 10^-10
Ag⁺ (aq) + 2NH₃ (aq) ⇌ Ag(NH₃)₂⁺ (aq), K_f = 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.

K_c = K_sp x K_f =
(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]