Aqueous Equilibria. I. Solubility, Complex Ion Formation.

When a solid is placed in water an equilibrium is set up:

Solid Solute.

Although all dissolving reactions follow this equilibrium expression, experimentally the most useful reactions studied are

1. ionic solids dissolving in water:

AgCl(s) Ag+(aq) + Cl-(aq)

Sr3(PO4)2(s) 3 Sr++(aq) + 2 PO4---(aq)

2. the formation of complex ions:

Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq)

Cu++(aq) + 4 OH-(aq) Cu(OH)42-(aq)

From thermodynamics we have seen that the equilibrium constant, K, involves activities and that it can be experimentally obtained only when the activity for a solution can be expressed in the form: activity(A) = {[A] M / 1 M}

This is only possible for ideal solutions, which for our purposes will be identified as dilute solutions.

Solubility, Ksp.

Ksp is defined as the equilibrium constant for the dissolution of an ionic compound in water to form a dilute solution.

That is, Ksp is valid only for insoluble salts.

Although a Ksp can be written for the dissolution of sodium chloride in water

NaCl(s) Na+(aq) + Cl-(aq)

it cannot be calculated since the resulting solution is not ideal.

Typical Ksp expressions and values (at 25ºC):

AgCl(s) Ag+(aq) + Cl-(aq)

Ksp = a(Ag+)*a(Cl-) / a(AgCl)

==[Ag+][Cl-] = 1.8 x 10-10

(as long as solid AgCl remains)

Sr3(PO4)2(s) 3 Sr++(aq) + 2 PO43-(aq)

Ksp = a(Sr++)3 * a(PO43-)2 / a(Sr3(PO4)2)

=[Sr++]3 [PO43-]2 = 4 x 10-28

(as long as solid Sr3(PO4)2 remains)

The sp subscript stands for "solubility product".

Solubility and the Solubility Product.

The solubility of a compound can be expressed in different ways, the most usual being either g/100mL or mol/L.

Using stoichiometry and following the routine analysis of equilibria, a relationship between the molar solubility and Ksp can be obtained.

Note that relationship obtained depends on the equilibrium expression:

Let the molar solubility be s mol/L.
 AgCl(s) Ag+(aq) + Cl-(aq) I some solid 0 0 C some solid s s E some solid left s s
Ksp = a(Ag+)*a(Cl-) = 1.8 x 10-10

= s * s = s2

(From which s may be calculated as 1.34 x 10-5 mol/L)

 Sr3(PO4)2(s) 3 Sr++(aq) + 2 PO4---(aq) I some solid 0 0 C some solid 3s 2s E some solid left 3s 2s
Ksp = a(Sr++)3 * a(PO43-)2 = 4 x 10-28

= (3s)3 * (2s)2 = 108s5

(From which s may be calculated as 1.3 x 10-6 mol/L)

Example:

In an experiment to determine the solubility and the Ksp of silver sulphate, it was found that the sulphate ion concentration in equilibrium with solid at 25ºC was 0.0155 M. Calculate the solubility and Ksp for silver sulphate.

Step A: Analysis

 Ag2SO4(s) 2 Ag+(aq) + SO42-(aq) I some solid 0 0 C some solid 2s s E some solid left 2s s
Ksp = a(Ag+)2 * a(SO4) / a(Ag2SO4)

= (2s)2 * s = 4s3.

Given: [SO42-] = 0.0155 M = s

Unknowns: solubility (= s) and Ksp.

Step B: Brainstorm (trivial here)

Step C: Calculate

Solubility of silver sulphate = 0.0155 mol/L.

Ksp(Ag2SO4) = 1.49 x 10-5.

Solubility in water containing other ions.

Assume the solution remains ideal.

The solubility of an ionic substance may depend on the presence of other ions if reactions occur to disturb the equilibrium.

1. Addition of a common ion.

For the silver chloride equilibrium

AgCl(s) Ag+(aq) + Cl-(aq)

If either Ag+ or Cl- is added, the equilibrium position will shift right (to minimize the disturbance) and the solubility of the AgCl will drop.

This phenomenon is often called the common ion effect.

Calculations of the solubility in the presence of a common ion follows the same course as all equilibrium calculations.

Example: Common Ion Effect.

Calculate the solubility of silver chloride in a 0.100 M solution of sodium chloride. Assume the solution remains ideal.

Step A: Analysis

NaCl in solution is Na+ and Cl-

 AgCl(s) Ag+(aq) + Cl-(aq) I some solid 0 0.100 C some solid s s E some solid left s 0.100 + s
Ksp = a(Ag+)*a(Cl-) = 1.8 x 10-10

= s * (0.100 + s)

Unknown: solubility (= s)

Step B: Brainstorm (trivial)

Step C : Calculate

This is a quadratic equation, and, though it may be solved exactly, it is quicker to make an approximation:

approximation: s is small compared with 0.100 such that

0.100 + s is approx = 0.100

Using this: Ksp = s * (0.100)

From which s may be calculated as 1.8 x 10-9 mol/L

1.8 x 10-9 is small compared to 0.100: acceptable.

Example 2: Common Ion Effect.

Calculate the solubility of silver carbonate in a 0.0300 M solution of iron(III) carbonate. Assume the solution remains ideal.

Step A: Analysis

Fe2(CO3)3 in solution is 2 Fe3+ and 3 CO32-.

 Ag2CO3(s) 2 Ag+(aq) + CO32-(aq) I some solid 0 0.0900 C some solid 2s s E some solid left 2s 0.0900 + s
Ksp = a(Ag+)2 * a(CO32-) / a(Ag2CO3)

= 8.1 x 10-12 (from tables)

Unknown is solubility (= s)

Step B: Brainstorm (trivial)

Step C: Calculate

8.1 x 10-12 = (2s)2 * (0.0900 + s)

Assume s is very much smaller than 0.0900

8.1 x 10-12 = (2s)2 * 0.0900

From which s = 4.7 x 10-6

Check approximation: (4.7 x 10-6)*100% / 0.0900 = 0.005%

The approximation is valid.

2. Formation of a Complex Ion.

The silver chloride solubility equilibrium is:

AgCl(s) Ag+(aq) + Cl-(aq)

And silver ions will form a complex ion with ammonia:

Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq)

So if silver chloride is being dissolved in an aqueous ammonia solution, then more AgCl will dissolve than in water as the complexing removes Ag+ ions.

Complex Ion Formation Example:

Calculate the solubility of silver chloride in 0.100 M NH3(aq).

Equation:

AgCl(s) Ag+(aq) + Cl-(aq)

Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq)

AgCl(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl-(aq)

Multiply K values to obtain K for this reaction:

K = 1.8 x 10-10 * 1.6 x 107 = 2.9 x 10-3

Step 1: Analysis
 AgCl(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl-(aq) I some 0.100 0 0 C some -2s s s E some left 0.100 - 2s s s
K = a(Ag(NH3)2+) * a(Cl-) / a(AgCl) * a(NH3)2

= s * s / 1 * (0.100-2s)2

2.9 x 10-3 = s2/ (0.100-2s)2

Unknown is solubility, s

Step B: Brainstorm (trivial)

Step C: Calculate

s = 0.051

Solubility of AgCl in 0.100 M ammonia is 0.051 mol/L

3. Addition of acid or base.

Often one of the ions from the insoluble salt is acidic or basic. Consequently there will be a reaction of this ion which will remove it, making the salt more soluble.

Example:

Many metal sulphides are insoluble:

CuS(s) Cu2+(aq) + S2-(aq) Ksp = 6 x 10-36.

Sulphide anion is a weak base:

S2- + H2O HS- + OH-

Consequently, if acid is added, the acid will react with the OH- produced in this second equilibrium which causes this equilibrium to move right, removing S2-.

Removing S2- from the first equilibrium causes this equilibrium to move right also. And this means that more CuS is dissolving.

Sulphides are more soluble in acid than in neutral or basic solution.

Acetates, oxalates, fluorides, phosphates, sulphides, and hydroxides will follow this same principle.

Carbonates are also more soluble in acid but in this case an irreversible reaction occurs:

CaCO3(s) Ca2+(aq) + CO32-(aq)

CO32-(aq) + H3O+ HCO3-(aq) + H2O

HCO3-(aq) + H3O+ CO2(g) + 2 H2O

The loss of CO2(g) is irreversible, since the gas leaves the solution, so in acid, the salt continues to dissolve up to the limit for CaA2, where A- is the anion associated with the added acid.

Checking for Precipitation.

Evaluation of Q and comparison with K.

When two ions are added together they may combine to form a precipitate of an insoluble salt.

For example, when Ba2+ and F- come together in the same solution, they may precipitate out solid BaF2.

BaF2(s) Ba2+(aq) + 2 F-(aq)

Ksp = 1.0 x 10-6

Whether the reaction will move left to produce the solid precipitate will depend on the value of Q.

If Q > K reaction will move left to reduce Q.

If Q < K reaction will not move, since there is no solid to dissolve.

Example: Will a precipitate form when equal quantities of 0.02 M barium nitrate and 0.01 M sodium fluoride are mixed?

Barium nitrate is Ba(NO3)2; sodium fluoride is NaF.

In answering this type of question, don't forget the dilution factor when the two solutions are added to each other.

Q = a(Ba2+) * a(F-)2 / a(BaF2)

= 0.01 * (0.005)2

= 2.5 x 10-7

Because Q < K no precipitate will form.

Another type of precipitation problem!

A solution contains both 0.10 M silver ion, Ag+ and 0.10 M copper ion, Cu2+. 1.0 M potassium oxalate, K2C2O4, is added dropwise to 1.00 L of this mixture such that the added drops effectively do not change the total volume of 1 L.

1. Which ion will precipitate out first?

2. What will be the concentration of this first ion when the second ion just begins to precipitate?

3.Can this process be used to separate silver ions from lead ions in solution?

Step A: Analysis.

Equations:

Ag2C2O4(s) 2 Ag+(aq) + C2O42-(aq) Ksp = 3.5 x 10-11

CuC2O4(s) Cu2+(aq) + C2O42-(aq) Ksp = 3 x 10-8

Unknowns:

which Q is exceeded first?

a(second) when it just begins to ppt.

Step B: Brainstorm.

As C2O42- is added its concentration will rise. If a(C2O42-) at the point of pptn for each is found, the lower will indicate the first to ppt. The higher will indicate the (C2O42-) when the second ion is about to ppt.

Step C: Calculate.

Ksp = 3.5 x 10-11 = a(Ag+)2 * a(C2O42-) = (0.1)2 * a(C2O42-)

thus for Ag2C2O4, the [C2O42-] at equilibrium = 3.5 x 10-9 M

Ksp = 3 x 10-8 = a(Cu2+) * a(C2O42-) = (0.1) * a(C2O42-)

thus for CuC2O4, the [C2O42-] at equilibrium = 3 x 10-7 M

As the C2O42- is added, its concentration will increase from zero, so it will reach 3.5 x 10-9 M first and the silver ion will start to precipitate out.

For the copper to start to precipitate, [C2O42-] will have to reach 3 x 10-7 M. At which time the [Ag+] can be obtained:

Ksp = 3.5 x 10-11 = a(Ag+)2 * a(C2O42-) = a(Ag+)2 * 3 x 10-7

from which [Ag+]2 = 1.2 x 10-4 M

and [Ag+] = 1.1 x 10-2 M