
Arrange in correct order of increasing solubility of \[AgCl\] in:
A) Water
B) \[0.1{\text{ }}M{\text{ }}NaCl\]
C) \[0.1{\text{ }}M{\text{ }}BaC{l_2}\]
D) \[0.1{\text{ }}M{\text{ }}N{H_3}\]
a. \[A < B < C < D\]
b. \[B < C < D < A\]
c. \[C < B < D < A\]
d. \[C < B < A < D\]
Answer
588.3k+ views
Hint: Solubility depends upon many factors, some of these factors are its dissociation in solvent, the ability to form a complex with the solvent and the presence of a common ion. The presence of common ions decreases the solubility and formation of complexes increases solubility.
Complete step by step answer:
Let’s start with discussing the solubility case wise.
Case 1:
In case 1 we are having the water, Water being a common solvent and \[AgCl\] being an ionic salt it will easily dissolve into the water. So, let’s take water as the base and measure the solubility of other cases with the reference of water.
Case 2:
In case 2 we are having \[0.1{\text{ }}M{\text{ }}NaCl\], in this case due to the presence of a common ion that is \[C{l^ - }\] the solubility of the \[AgCl\] will be less than that of water.
Case 3:
In case 3 we are having \[0.1{\text{ }}M{\text{ }}BaC{l_2}\], in this case due to the presence of a common ion i.e. \[C{l^ - }\] the solubility of \[AgCl\] will be less than that of water. Since the amount of \[C{l^ - }\] ion will be twice as much as \[0.1{\text{ }}M{\text{ }}NaCl\], hence the solubility of \[AgCl\] will be less as compared to solubility of \[AgCl\] in \[0.1{\text{ }}M{\text{ }}NaCl\].
Case 4:
In case 4 we are having \[0.1{\text{ }}M{\text{ }}N{H_3}\], in this case a complex between the \[AgCl\] and \[N{H_3}\] will be formed. The complex will be \[\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]Cl\] and due to this complex formation the solubility of \[AgCl\] will be increased and the solubility of \[AgCl\] in \[0.1{\text{ }}M{\text{ }}N{H_3}\] will be more than that of water.
Hence, the answer to this question is option d. \[C < B < A < D\] .
Note:
The concept of solubility is majorly used to check how much a particular salt or compound is soluble in the solvent or not. Solubility is of utmost importance when you are trying to achieve a solution as if you want a solution of 1M but 1 mole of substance is not soluble in 1L of water then we need to think of some alternative way. Hence the concept of solubility is must to understand.
Complete step by step answer:
Let’s start with discussing the solubility case wise.
Case 1:
In case 1 we are having the water, Water being a common solvent and \[AgCl\] being an ionic salt it will easily dissolve into the water. So, let’s take water as the base and measure the solubility of other cases with the reference of water.
Case 2:
In case 2 we are having \[0.1{\text{ }}M{\text{ }}NaCl\], in this case due to the presence of a common ion that is \[C{l^ - }\] the solubility of the \[AgCl\] will be less than that of water.
Case 3:
In case 3 we are having \[0.1{\text{ }}M{\text{ }}BaC{l_2}\], in this case due to the presence of a common ion i.e. \[C{l^ - }\] the solubility of \[AgCl\] will be less than that of water. Since the amount of \[C{l^ - }\] ion will be twice as much as \[0.1{\text{ }}M{\text{ }}NaCl\], hence the solubility of \[AgCl\] will be less as compared to solubility of \[AgCl\] in \[0.1{\text{ }}M{\text{ }}NaCl\].
Case 4:
In case 4 we are having \[0.1{\text{ }}M{\text{ }}N{H_3}\], in this case a complex between the \[AgCl\] and \[N{H_3}\] will be formed. The complex will be \[\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]Cl\] and due to this complex formation the solubility of \[AgCl\] will be increased and the solubility of \[AgCl\] in \[0.1{\text{ }}M{\text{ }}N{H_3}\] will be more than that of water.
Hence, the answer to this question is option d. \[C < B < A < D\] .
Note:
The concept of solubility is majorly used to check how much a particular salt or compound is soluble in the solvent or not. Solubility is of utmost importance when you are trying to achieve a solution as if you want a solution of 1M but 1 mole of substance is not soluble in 1L of water then we need to think of some alternative way. Hence the concept of solubility is must to understand.
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