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Solubility Equilibria
Each of the molecular solids dissolves to give an individual aqueous molecule such as
H2O
C12H22O11(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 –
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.
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 (Ksp)
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, SO2 and NO are oxidised to SO3 and NO2, respectively, which react with water to give H2SO4 and HNO3. The resultant solution is called acid rain. SO2 dissolves in water to form diprotic acid.
SO2(g)+H2O(l)⇔HSO⊝3+H⊕; Ka1=10−2
HSO⊝3⇔SO2−3+H⊕; Ka2=10−7
and for equilibrium,
SO2(aq)+H2O(l)⇔SO2−3(aq)+2H⊕(aq)
Ka=Ka1×Ka2=10-9 at 300K.
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.
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 equilibrium reaction for the dissociation of AgCl is:
For example, the solubility product equilibrium constant for the dissociation of AgCl is:
The solubility product is literally the product of the solubilities of the ions in units of molarity (mol/L)
Sample Calculations
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 (Ksp = 4.0 x 10-8)
First, write the BALANCED REACTION:
2) The solubility of AgCl in pure water is 1.3 x 10-5 M. Calculate the value of Ksp.
First, write the BALANCED REACTION:
Ion Product (Qsp)
Imagine we have a saturated solution of AgCl. The equilibrium reaction for the dissociation of this salt is:
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:
The ion product (Qsp) can be used to determine in which direction a system must shift in order to reach equilibrium. There are three possible situations:
Common-ion effect
– The decrease in the solubility of a salt that occurs when the salt is dissolved in a solution that already contains another source of one of its ions. For example, if AgCl is added to a NaCl solution (which contains the common ion, Cl
–
) the solubility of the AgCl decreases.
Selective precipitation – A technique in which one ion is selectively removed from a mixture of ions by precipitation.