
A litre of solution is saturated with AgCl. To this solution, if 1.0×10-4 moles of solid NaCl is added, what will be the [{\rm Ag}^+] assuming no volume change?
A. More
B. Less
C. Equal
D. Zero
Answer
160.8k+ views
Hint: In the given question we are going to apply the concept of the common ion effect and Le Chalelier’s principle. According to the common ion effect in the solution mixture of two electrolytes containing one same ion, the strong electrolyte will suppress the dissociation of the weak electrolyte. To distinguish between the strong electrolyte and weak electrolyte; the easiest way is by determining the difference in the electronegativity of the ions of the given electrolyte, if the difference between the electronegativity of ions of the electrolyte is more then the electrolyte will be strong, and if the difference is less then the electrolyte will be weak.
Complete step by step solution:
The given solution is a one-litre saturated solution of AgCl which contains $Ag^+$ and $Cl^-$
ions in the solution.
$AgCl\ \rightleftharpoons \ {\rm Ag}^++\ {\rm Cl}^-$
Now, adding $1.0\ \times\ {10}^{-4}$ mole of solid NaCl to the saturated solution of AgCL.
$NaCl\rightarrow{\rm Na}^++{\rm Cl}^-$
Since NaCl is dissociating into Na+ and Cl−, it will increase the concentration of Cl− in the solution, and according to the common ion effect, it will suppress the dissociation of AgCl. NaCl is a stronger electrolyte than AgCl because electron affinity of Na is more than Ag which means Ag is more electronegative than Na so the difference between the electronegativity of Na & Cl will be larger than Ag & Cl.
Now, since the concentration of chloride ions is more which will disturb the equilibrium of the solution so, according to the Le Chatelier’s to maintain the equilibrium the Ag+ and Cl− ions will combine and form AgCl.
So the reaction will go backward
${\rm Ag}^++{\rm Cl}^-\rightleftharpoons \ AgCl$; which will result in a decrease in the concentration of Ag+ ions.
Thus, Option (B) is correct
Note: Most of the reaction tries to attain equilibrium to maintain chemical balance which can be determined by the free energy of the reaction. This equilibrium can be affected by temperature, pressure and concentration.
Complete step by step solution:
The given solution is a one-litre saturated solution of AgCl which contains $Ag^+$ and $Cl^-$
ions in the solution.
$AgCl\ \rightleftharpoons \ {\rm Ag}^++\ {\rm Cl}^-$
Now, adding $1.0\ \times\ {10}^{-4}$ mole of solid NaCl to the saturated solution of AgCL.
$NaCl\rightarrow{\rm Na}^++{\rm Cl}^-$
Since NaCl is dissociating into Na+ and Cl−, it will increase the concentration of Cl− in the solution, and according to the common ion effect, it will suppress the dissociation of AgCl. NaCl is a stronger electrolyte than AgCl because electron affinity of Na is more than Ag which means Ag is more electronegative than Na so the difference between the electronegativity of Na & Cl will be larger than Ag & Cl.
Now, since the concentration of chloride ions is more which will disturb the equilibrium of the solution so, according to the Le Chatelier’s to maintain the equilibrium the Ag+ and Cl− ions will combine and form AgCl.
So the reaction will go backward
${\rm Ag}^++{\rm Cl}^-\rightleftharpoons \ AgCl$; which will result in a decrease in the concentration of Ag+ ions.
Thus, Option (B) is correct
Note: Most of the reaction tries to attain equilibrium to maintain chemical balance which can be determined by the free energy of the reaction. This equilibrium can be affected by temperature, pressure and concentration.
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