The Question is from neet-chemistry chemical-equilibrium , and it was created by the author Mr. S M HUSSAIN

Question

The solubility product ($K_{sp}$) of $AgCl$ in water at 25°C is $1.8 	imes 10^{-10}$.  A solution of $AgCl$ in water is in equilibrium with solid $AgCl$. If a small amount of $NaCl$ is added to this saturated solution, what will happen to the solubility ($s$) and $K_{sp}$ of $AgCl$?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Solubility equilibrium and the common ion effect.
 **Related Topics**: Solubility product constant, Le Chatelier's principle.
 **Prerequisites**: Understanding of equilibrium constants and the effect of adding a common ion to a saturated solution.
## SOLUTION PATHWAY
**Given Information**:
	$K_{sp}(AgCl) = 1.8 	imes 10^{-10}$
	Saturated solution of $AgCl$ in equilibrium with solid $AgCl$.
	Addition of $NaCl$ (common ion effect).
**Method & Steps**:
1. **Equilibrium Expression**: The solubility product for $AgCl$ is given by:
	$K_{sp} = [Ag^+][Cl^-] = 1.8 	imes 10^{-10}$
2. **Common Ion Effect**: Adding $NaCl$ increases the concentration of $Cl^-$ ions in the solution.  This is a common ion, as $Cl^-$ is already present in the equilibrium from the dissolution of $AgCl$.
3. **Le Chatelier's Principle**: According to Le Chatelier's principle, the system will shift to relieve the stress caused by the increased $Cl^-$ concentration.  The equilibrium will shift to the left, decreasing the concentration of $Ag^+$ ions in solution.
4. **Solubility Change**: Since the solubility ($s$) of $AgCl$ is directly related to the concentration of $Ag^+$ ions ($s = [Ag^+])$, the solubility of $AgCl$ will decrease.
5. **$K_{sp}$ Remains Constant**: The $K_{sp}$ is a constant at a given temperature. It only changes with temperature. Adding a common ion does not change the $K_{sp}$ value; it only affects the concentrations of the ions in the saturated solution.
 **Result**: The solubility ($s$) of $AgCl$ decreases, and the $K_{sp}$ remains constant.
 **Verification**: The addition of a common ion always reduces the solubility of a sparingly soluble salt.
## OPTION ANALYSIS
A :x: Incorrect.  Adding a common ion decreases solubility, and $K_{sp}$ remains constant.
B ✓: Correct. Adding $NaCl$ introduces a common ion ($Cl^-$), reducing the solubility of $AgCl$ while leaving $K_{sp}$ unchanged.
C :x: Incorrect. $K_{sp}$ is a constant at a given temperature; it does not change upon addition of a common ion.
D :x: Incorrect. Adding a common ion decreases solubility, not increases it.
## LEARNING AIDS
 **Common Mistakes**:
  1.Confusing $K_{sp}$ with solubility: $K_{sp}$ is a constant, while solubility is affected by the common ion effect.
  2.Thinking $K_{sp}$ changes: $K_{sp}$ only changes with temperature.
 **Quick Tips**:
  1.Remember Le Chatelier's principle to predict the shift in equilibrium.
  2.Understand that the common ion effect always reduces the solubility of a sparingly soluble salt.

Mr. S M HUSSAIN · Answer

$s$ increases, $K_{sp}$ increases

Mr. S M HUSSAIN · Answer

The correct Option is "$s$ decreases, $K_{sp}$ remains constant"

Mr. S M HUSSAIN · Answer

$s$ remains constant, $K_{sp}$ decreases

Mr. S M HUSSAIN · Answer

$s$ increases, $K_{sp}$ remains constant

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