Answer:
A. Solubility of calcium sulfite increases
B. Solubility of calcium fluoride increases
C. Solubility of Silver bromide decreases
Explanation:
The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.
Case 1. Calcium sulfite
The dissociation reaction of CaSO3 is given below:
CaSO3 ----> Ca²+ + SO3²-
SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;
H+ + SO3²- ---> HSO3-
This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.
Case 2. Calcium fluoride
The dissociation reaction of calcium fluoride (CaF2) is shown below.
CaF2 ----> Ca²+ + 2F-
Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:
H+ + F- ---->. HF
Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.
Case 3: Silver bromide
The dissociation reaction of AgBr is as follows:
AgBr ----> Ag+ + Br-
The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.
