Solubility of AgCl in 0.2 M NaCl is x and that in 0.1 M `AgNO_(3)` is y. Then which of the following is correct?

Options

A. `x=y`
B. `xgty`
C. xlty`
D. We cannot predict

Answer

To solve the problem of comparing the solubility of AgCl in two different solutions (0.2 M NaCl and 0.1 M AgNO3), we can follow these steps:

### Step 1: Write the Dissolution Reaction and Ksp Expression
The dissolution of AgCl can be represented as:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]
The solubility product constant (Ksp) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

### Step 2: Analyze the Effect of NaCl
In the presence of 0.2 M NaCl, the concentration of Cl⁻ ions will be 0.2 M. Thus, if we let the solubility of AgCl in this solution be \( x \), we can express Ksp as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = x \cdot 0.2 \]
This gives us:
\[ K_{sp} = 0.2x \]

### Step 3: Analyze the Effect of AgNO3
In the presence of 0.1 M AgNO3, the concentration of Ag⁺ ions will be 0.1 M. Let the solubility of AgCl in this solution be \( y \). The Ksp expression becomes:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = 0.1 \cdot y \]
This gives us:
\[ K_{sp} = 0.1y \]

### Step 4: Set the Two Ksp Expressions Equal
Since Ksp is a constant for a given temperature, we can set the two expressions equal to each other:
\[ 0.2x = 0.1y \]

### Step 5: Solve for the Relationship Between x and y
Rearranging the equation gives:
\[ \frac{y}{x} = \frac{0.2}{0.1} = 2 \]
This implies:
\[ y = 2x \]

### Conclusion
From our calculations, we find that \( y \) (the solubility of AgCl in 0.1 M AgNO3) is greater than \( x \) (the solubility of AgCl in 0.2 M NaCl). Thus, the correct relationship is:
\[ y > x \]