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

Options

A. More
B. Less
C. Equal
D. Zero

Explanation

To solve the problem, we need to determine the concentration of silver ions \([Ag^+]\) in a saturated solution of AgCl after adding a small amount of NaCl. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Saturated Solution
A saturated solution of AgCl means that it is in equilibrium with its solid form. The dissociation of AgCl can be represented as:
\[ AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \]
At equilibrium, the solubility product constant \(K_{sp}\) for AgCl can be expressed as:
\[ K_{sp} = [Ag^+][Cl^-] \]
Let’s denote the solubility of AgCl as \(s\). Therefore, at saturation:
\[ K_{sp} = s^2 \]

### Step 2: Adding NaCl
When we add \(1.0 \times 10^{-4}\) moles of NaCl to the solution, it dissociates completely into:
\[ NaCl \rightarrow Na^+ + Cl^- \]
This means we are adding \(1.0 \times 10^{-4}\) moles of \(Cl^-\) ions to the solution.

### Step 3: Calculate New Concentration of \(Cl^-\)
Before adding NaCl, the concentration of \(Cl^-\) in the saturated solution is equal to \(s\). After adding NaCl, the total concentration of \(Cl^-\) becomes:
\[ [Cl^-] = s + 1.0 \times 10^{-4} \]

### Step 4: Set Up the New Equilibrium Expression
The \(K_{sp}\) expression still holds:
\[ K_{sp} = [Ag^+][Cl^-] \]
Substituting the new concentration of \(Cl^-\):
\[ K_{sp} = [Ag^+](s + 1.0 \times 10^{-4}) \]

### Step 5: Solve for \([Ag^+]\)
Since \(K_{sp}\) is a constant at a given temperature, we can rearrange the equation to find \([Ag^+]\):
\[ [Ag^+] = \frac{K_{sp}}{s + 1.0 \times 10^{-4}} \]
Here, \(s\) is the original concentration of \(Ag^+\) in the saturated solution.

### Step 6: Conclusion
The concentration of silver ions \([Ag^+]\) will decrease due to the common ion effect from the added \(Cl^-\) ions.

### Final Answer
Thus, the final concentration of \([Ag^+]\) after adding \(1.0 \times 10^{-4}\) moles of NaCl is:
\[ [Ag^+] = \frac{K_{sp}}{s + 1.0 \times 10^{-4}} \]