Introduction:

Solubility is the capability of a substance to dissolve in the water. Solubility is measured in terms of concentration of an ion which is present in a smaller ratio in solution. On the other hand, solubility equilibrium refers to the equilibrium between the dissolved salt (that is, ions) and undissolved salt that generally exists in the saturated solution or a solution of a sparingly soluble salt. The term sparingly soluble salt refers to a salt which is partly (that is, not completely) soluble in water, as consequences of which, the equilibrium between dissolved ions and undissolved salt is possible.

Solubility Product:

Define: The maximum product of ionic concentrations or activities of an electrolyte that at one temperature can carry on in equilibrium by the undissolved phase.

Solubility Product Constant, K_sp:

The solubility product constant, K_sp?, is the equilibrium constant for the solid substance dissolving in the aqueous solution. This represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the K_sp value it has.

Take the general dissolution reaction as:

aA (s) ? cC (aq) + dD (aq)

To solve for the K_sp it is essential to take the molarities or concentrations of the products (cC and dD) and multiply them. When there are coefficients in front of any of the products, it is essential to move up the product to that coefficient power (and as well multiply the concentration by that coefficient). This is illustrated below:

K_sp = [C]^c [D]^d

It will be noted that the reactant, aA, is not comprised in the K_sp equation. Solids are not incorporated when computing equilibrium constant expressions, as their concentrations don't change the expression; any change in their concentrations is insignificant, and thus omitted.

Therefore, K_sp symbolizes the maximum amount of solid which can be dissolved in the aqueous solution. The answer will encompass the units of molarity, mol L^-1 and a measure of concentration.

Illustration:

AgCl ↔ Ag⁺ + Cl^-                     K_sp = [Ag⁺][Cl^-]

Cu(OH)₂ ↔ Cu²⁺ + 2OH^-         K_sp = [Cu²⁺][OH^-]²

Common ion effect:

The solubility of one salt is decreased by the presence of the other salt encompassing a common ion. 

For illustration take the reaction: AgCl(s) ↔ Ag⁺ (aq) + Cl^- (aq)

Ions present in the solution are Ag⁺ and Cl^- from the salt silver chloride.

Assume that a soluble salt like ammonium chloride is added to the solution, with reference to Le Chatelier's Principle, what would take place??

The addition of extra chloride ions would increase the [Cl^-] in the equilibrium illustrated above and the system would then try to reduce the [Cl^-] by shifting the equilibrium to the LEFT.

By shifting to the left, this causes an increase in the quantity of solid silver chloride formed, that is, the solubility of the silver chloride reduces in the presence of ammonium chloride. What would be noticed is an increased amount of precipitation if the salt ammonium chloride (or any salt having a common ion) is added to the solution having silver chloride.

How to compute the solubility product of a sparingly soluble salt:

If the molar solubility of CaI₂ at 25 °C is 7.9 x 10^-7 mol dm^-3, then determine its K_sp value?

Step 1: At first, write the equilibrium expression CaI₂ ↔ Ca²⁺ + 2I^-

Step 2: Write down the K_sp expression [Ca²⁺][I^-]²

Step 3: Find out the number of moles of each ion present

Ratio of CaI₂: Ca²⁺: I^- = 1: 1: 2

Thus, number of moles of Ca²⁺ = 7.9 x 10^-7 

And number of moles of I^- = 7.9 x 10^-7 x 2 = 1.58 x 10^-5

Step 4: Utilize the number of moles of each ion to find out the K_sp value

K_sp = 7.9 x 10^-7 x (1.58 x 10^-5)² = 1.97 x 10^-16

To compute the molar solubility from the K_sp value:

K_sp of PbI₂ is 1.39 x 10^-8. Find out its molar solubility?

Step 1: Write down the equilibrium expression: PbI₂ ↔ Pb²⁺ + 2I^-

Step 2: Make use of the letter 'a' to symbolize the number of mol of PbI₂, then a moles of PbI₂ provides a moles of Pb²⁺ and 2a moles of I^-

Step 3: Write down the K_sp expression by employing the algebraic terms

K_sp = [Pb²⁺][I^-]² =  a * (2a)² = 4a³ → 1.39 x 10^-8 = 4a³

Step 4: Solve for a:

³√ (1.39 x 10^-8)/4 = 0.0015

Step 5: Obtain molar solubility,

molar solubility of PbI₂ is 0.0015 mol dm^-3 

How to find out the decrease in solubility of a sparingly soluble salt through common ion effect:

If the molar solubility of CaI₂ is 7.9 x 10^-7 mol dm^-3 in water and then 0.1M NaI is added to the solution, then find out the new solubility of the CaI₂?

Step 1: Write down the K_sp value & expression of CaI₂ (taken from the prior illustration)

K_sp = [Ca²⁺][I^-]² = 1.97 x 10-16

Step 2: Replace the concentration of the common ion having the value given  

NaI is the soluble salt that dissociates FULLY thus the concentration of I^- = 0.1 (avoid the I^- ions coming from the sparingly soluble salt.)

Step 3: Find out the concentration of Ca²⁺

Ca²⁺ = Ksp/[I^-]²  =  (1.97 x 10^-16)/ 0.12  = 1.97 x 10^-14

Step 4: Find out the molar solubility

Ratio of Ca²⁺: CaI₂ = 1: 1

Thus number of moles of CaI₂ = molar solubility

                                              = 1.97 x 10^-14 mol dm^-3

Thus the solubility of CaI₂ reduces from 7.9 x 10^-7 to around 2 x 10^-14

Real life examples of solubility product:

The Kidney stones are crystals of calcium oxalate. Whenever the concentrations of calcium ions and oxalate ions are too high, the K_sp of calcium oxalate is surpassed and this causes the precipitation of the calcium oxalate crystals. 

Note that the kidney stones are usually of the order of 3 millimetres or more to cause the obstacle of the urethra and ultimately pain.  By minimizing the uptake of oxalate ions and to lesser extent calcium ions, this decreases the risk of kidney stone formation.

Qualitative analysis as well exploits the principle of solubility product of the sparingly soluble salts. For a precipitate to be made, the ionic product should surpass the solubility product. Then precipitation takes place until the ionic product no longer surpasses the K_sp value.

Reagents employed for qualitative analysis should be of a certain minimum concentration, if not, the ionic product wouldn't be adequate to equivalent the K_sp value and no precipitate would be seen and a wrong conclusion of the ion present would take place.

Uses of K_sp:

Smaller is the solubility product of a substance, the lower is its solubility.

The solubility product can be employed to predict whether a precipitate will form whenever the two solutions are mixed.

Consider the equilibrium:

PbCl₂ ↔ Pb²⁺ + 2Cl^-

Whenever the concentration of Cl^- was increased by adding HCl, the solubility product would be surpassed and PbCl₂ would be precipitated from the solution till the equilibrium was restored.

Predicting the formation of Precipitate:

Precipitate is basically the solid formed as an outcome of chemical reaction among two or more solutions. Let state that you mix two solutions, how do you know whether the precipitate forms or not? Before you can respond to this question, you got to know what type of precipitate is going to form by employing the solubility rules. From the information provided in the problem, first compute the molar concentrations of the ions responsible for the precipitate. Subsequently, compute the reaction quotient Q (this is precisely similarly as K_sp apart from it is based on the initial (given) concentrations) and compare it by K_sp.

If Q > K_sp precipitate will form or else no.

Illustration:

Precisely 100 ml of 0.05 M AgNO³ are added to precisely 500 ml of 0.05 M HCl. Will a precipitate made?

Answer:

Ions present in the solution are Ag⁺, NO^3-, H⁺, and Cl^-. According to the solubility rules, just AgCl precipitate is possible. From the given information, first we compute the molar concentrations of Ag⁺ and Cl^- ions:

[Ag⁺] = (100 ml/600 ml) x 0.05 M = 8.3 x 10^-3 M

[Cl-] = (500 ml/600 ml) x 0.05 M = 4.2 x 10^-2 M

Note that the 600 ml is the net volume of the solution. Then we compute Q:

Q = [Ag⁺][Cl^-] = (8.3 x 10^-3)(4.2 x 10^-2)

                       = 3.5 x 10^-4

Thus, Q (= 3.5 x 10^-4) > K_sp (= 1.6 x 10^-10)

The precipitate of AgCl will form.

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