SOLUBILITY EQUILIBRIA

Each of the molecular solids dissolves to give an individual aqueous molecule such as

H₂O
C₁₂H₂₂O_11(S) —————> C₁₂H₂₂O₁₁ _(aq)

The ionic solids dissociate to give their respective positive and negative ions :

H₂O
NaCl _(s) ————————-> Na + (aq) + Cl ^– (aq )

The ions formed from the dissociation of the ionic solids can also carry an electrical current which makes the salt solutions good conductors of electricity. However, molecular solids do not dissociate in water to give ions so as no electrical current can be carried.

Solubility

The ratio of the maximum amount of solute to the volume of the solvent in which the solute can dissolve. This is generally expressed in two ways i.e –

Grams of solute per 100 g of water
Moles of solute per litre of solution

A salt refers to be soluble if it dissolves in water to give a solution along with the concentration of at least 0.1 m at the room temperature. A salt is also considered to be insoluble if the concentration of an aqueous solution is less than 0.0001 m at the room temperature. Salts are considered to be slightly soluble; those between 0.0001 m and 0.1 m.

Solubility Equilibria

According to the principles of solubility equilibria, the salts have low solubilities in water. The reaction that can be considered for the dissociation of the salt AgCl is –

AgCl_(s) ————— Ag⁺_(aq) + Cl^– _(aq)

However, the reverse reaction for the dissolving of the salt would be the precipitation of the ions to form a solid –

Ag⁺_(aq) + Cl^– _(aq)—————– AgCl(s)

When the rate at which AgCl dissolves is equal to the rate at which AgCl precipitates, the system has reached the solubility equilibria.

Solubility Product Equilibrium Constant (K_sp)

The product of the equilibrium concentrations of the ions in a saturated solution of a salt. Each concentration in this is raised to the power of the respective coefficient of ion in the balanced equation. For example, the solubility product equilibrium constant for the dissociation of AgCl is:

Another example of a solubility product equilibrium constant where we can consider the reaction for the dissociation of CaF₂ in water is :

The solubility product equilibrium constant for this reaction would be the product of the concentration of Ca²⁺ ion and the concentration of the F^– ion raised to the second power (squared):

Solved Examples for You

Question: In the atmosphere, $S O_{2}$ and NO are oxidised to $S O_{3}$ and $N O_{2}$ , respectively, which react with water to give $H_{2} S O_{4}$ and $H N O_{3}$ . The resultant solution is called acid rain. $S O_{2}$ dissolves in water to form diprotic acid.

SO₂_(g)+H₂O_(l)⇔HSO^⊝₃+H^⊕; K_a1=10⁻²

H $H S O_{3}^{⊝} ⇌ S O_{3}^{2 -} + H^{\oplus}; K_{a_{2}} = 10^{- 7}$

$H S O_{3}^{⊝} ⇌ S O_{3}^{2 -} + H^{\oplus}; K_{a_{2}} = 10^{- 7}$ nd for equilibrium,

$S O_{2} (a q) + H_{2} O (l) ⇌ S O_{3}^{2 -} (a q) + 2 H^{\oplus} (a q)$

$S O_{2} (a q) + H_{2} O (l) ⇌ S O_{3}^{2 -} (a q) + 2 H^{\oplus} (a q)$ $K_{a} = K_{a_{1}} \times K_{a_{2}} = 10^{- 9}$ 300K.

H₂SO₃ is less acidic than H₂SO₄
HNO₃is less acidic than HNO2
SO_2(g) is reduced in the atmosphere during a thunderstorm
CO₂ gas develop more acidity in water than SO₂

Solution: Option A.Sulphurous acid H₂SO₃ is less acidic than sulphuric acid H₂SO₄. When a proton is lost by sulphurous acid, the negative charge is delocalized on 3 O atoms and when a proton is lost by sulphuric acid, the negative charge is delocalized by 4 O atoms. Since the extent of delocalization in HSO^-1 ion is less than the extent of delocalization in HSO^-4, the stability of anion and the acidity of the acid in sulphurous acid is lower than in sulphuric acid.

Solubility equilibrium is base on the assumption that solids dissolve in water to give the basic particles from which they are formed.

Molecular solids dissolve to give individual aqueous molecules.

Ionic solids dissociate to give their respective positive and negative ions:

The ions in formed from the dissociation of ionic solids can carry an electrical current. Salt solutions, therefore, are good conductors of electricity. Molecular solids, however, do not dissociate in water to give ions, so no electrical current can be carried.

Solubility

The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve.

Generally expressed in two ways:

grams of solute per 100 g of water

moles of solute per Liter of solution

- - - - A salt is considered soluble if it dissolves in water to give a solution with a concentration of at least 0.1 M at room temperature.
      - A salt is considered insoluble if the concentration of an aqueous solution is less than 0.0001 M at room temperature.
      - Salts with solubilities between 0.0001 M and 0.1 M are considered to be slightly soluble.
  Salts that have extremely low solubilities dissociate in water according to the principles of equilibrium. For example, the reaction for the dissociation of the salt AgCl is:

The reverse reaction for the dissolving of the salt would be the precipitation of the ions to form a solid:

The system has reached equilibrium when the rate at which AgCl dissolves is equal to the rate at which AgCl precipitates.

- - - - Saturated solution – Contains the maximum concentration of ions that can exist in equilibrium with the solid salt at a given temperature.
  The equilibrium reaction for the dissociation of AgCl is:

Solubility product equilibrium constant (K_sp) – The product of the equilibrium concentrations of the ions in a saturated solution of a salt. Each concentration is raised to the power of the respective coefficient of ion in the balanced equation.

- - - - NOTE: There is no denominator in the solubility product equilibrium constant. The key word to remember is PRODUCT which can remind you that you should have a multiplication (or product) of the concentrations of the ions. The reason that the solid reactant is not written is because its concentration effectively remains constant.
  For example, the solubility product equilibrium constant for the dissociation of AgCl is:

Let’s try another example of a solubility product equilibrium constant. Consider the reaction for the dissociation of CaF₂ in water:

The solubility product equilibrium constant for this reaction would be the product of the concentration of Ca²⁺ ion and the concentration of the F^– ion raised to the second power (squared):

NOTE: Unlike K_a and K_b for acids and bases, the relative values of K_sp cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions.

The solubility product is literally the product of the solubilities of the ions in units of molarity (mol/L)

Sample Calculations

- - - - K_sp can be calculated from the solubility of a salt. Conversely, the solubility of a salt can be calculated from K_sp.
  Let’s try an example calculation problem to demonstrate the relationship between the solubility and the solubility product of a salt.

1) Calculate the solubility of CaF₂ in g/L (K_sp = 4.0 x 10^-8)

First, write the BALANCED REACTION:

Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:

In the above equation, however, we have two unknowns, [Ca²⁺] and [F^–]². So, we have to write one in terms of the other using mole ratios. According to the balanced equation, for every one mole of Ca²⁺ formed, 2 moles of F^– are formed. To simplify things a little, let’s assign the the variable X for the solubility of the Ca²⁺:

If we SUBSTITUTE these values into the equilibrium expression, we now only have one variable to worry about, X:

We can now SOLVE for X:

We assigned X as the solubility of the Ca²⁺ which is equal to the solubility of the salt, CaF₂. However, our units right now are in molarity (mol/L), so we have to convert to grams:

Now, let’s try to do the opposite, i.e., calculate the K_sp from the solubility of a salt.

2) The solubility of AgCl in pure water is 1.3 x 10^-5 M. Calculate the value of K_sp.

First, write the BALANCED REACTION:

Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:

It is given in the problem that the solubility of AgCl is 1.3 x 10^-5. Since the mole ratio of AgCl to both Ag⁺ and Cl^– is 1:1, the solubility of each of the ions is equal to the solubility of AgCl. SUBSTITUTE the solubility given into the equilibrium expression to get K_sp:

Ion Product (Q_sp)

- - - - The product of the concentrations of the ions at any moment in time (not necessarily at equilibrium).
  Imagine we have a saturated solution of AgCl. The equilibrium reaction for the dissociation of this salt is:

[Ag⁺] is equal to [Cl^–] at equilibrium because the mole ratio of Ag⁺ to Cl^– is 1:1. What would happen to the solution if a tiny bit of AgNO₃ (a soluble salt) were added? Since AgNO₃ is soluble, it dissociates completely to give Ag⁺ and NO₃^– ions. There would now be two sources of the Ag⁺ ion, from the AgCl and from the AgNO₃:

Adding AgNO₃ increases the Ag⁺ concentration and the solution is no longer at equilibrium. The ion product (Q_sp) at that moment is bigger than the solubility product (K_sp). The reaction will eventually return to equilibrium but when it does, the [Ag⁺] is no longer equal to the [Cl^–]. Instead, the [Ag⁺] will be larger than the [Cl^–].

Let’s go back to the saturated AgCl solution. What would happen this time if a tiny bit of NaCl (a soluble salt) were added? Since NaCl is soluble, it dissociates completely to give Na⁺ and Cl^–ions. There would now be two sources of the Cl^– ion, from AgCl and from NaCl:

Adding NaCl increases the Cl^– concentration and the solution is no longer at equilibrium. The ion product (Q_sp) at that moment is bigger than the solubility product (K_sp). The reaction will eventually return to equilibrium but when it does, the [Ag⁺] is no longer equal to the [Cl^–]. Instead, the [Cl^–] will be larger than the [Ag⁺].

The ion product (Q_sp) can be used to determine in which direction a system must shift in order to reach equilibrium. There are three possible situations:

- - - Q_sp < K_sp – This means there are not enough ions in the solution. In order to return to equilibrium, more of the solid salt must dissociate into its ions.
    - Q_sp = K_sp – This means that the system is at equilibrium.
    - Q_sp > K_sp – This means that there are too many ions in the solution. In order to return to equilibrium, the excess ions will precipitate to form more solid.

Common-ion effect

^–

Selective precipitation – A technique in which one ion is selectively removed from a mixture of ions by precipitation.

SOLUBILITY EQUILIBRIA