A litre of solution is saturated with AgCl. To this solution if `1.0 × 10^(-4)` Mole of solid NaCI is added , what will be the `[Ag^+]` , assuming no volume change

Options

A. increases
B. decreases
C. does not change
D. zero

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Saturated Solution
A saturated solution of AgCl means that the solution is in equilibrium with solid AgCl. The dissociation of AgCl can be represented as:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

At saturation, the concentration of Ag\(^+\) and Cl\(^-\) ions in the solution is determined by the solubility product (Ksp) of AgCl.

### Step 2: Calculate Initial Concentrations
In a saturated solution of AgCl, the concentrations of Ag\(^+\) and Cl\(^-\) ions are equal. Let’s denote this concentration as \( s \):
\[ [\text{Ag}^+] = [\text{Cl}^-] = s \]

### Step 3: Adding NaCl
When we add \( 1.0 \times 10^{-4} \) moles of NaCl to the solution, NaCl dissociates completely:
\[ \text{NaCl (s)} \rightarrow \text{Na}^+ (aq) + \text{Cl}^- (aq) \]

This addition increases the concentration of Cl\(^-\) ions in the solution. The concentration of Cl\(^-\) ions after adding NaCl becomes:
\[ [\text{Cl}^-] = s + 1.0 \times 10^{-4} \]

### Step 4: Common Ion Effect
The increase in Cl\(^-\) concentration due to the addition of NaCl will cause the equilibrium of the AgCl dissociation to shift to the left, according to Le Chatelier's principle. This is known as the common ion effect, which suppresses the dissociation of AgCl. As a result, the concentration of Ag\(^+\) ions will decrease.

### Step 5: Conclusion
Since the addition of NaCl increases the concentration of Cl\(^-\), the concentration of Ag\(^+\) will decrease due to the common ion effect. Therefore, we conclude that:
\[ [\text{Ag}^+] \text{ decreases} \]

### Final Answer
The concentration of Ag\(^+\) ions decreases upon the addition of NaCl.

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