The solubility product of AgCI is `K_(sp)`. Then the solubility of AgCI in XM KCI is

Options

A. `K_(sp)xx X^2`
B. `X/(K_(sp))`
C. `(K_(sp))/(X^2)`
D. `(K_(sp))/(X)`

Answer

To solve the problem of finding the solubility of AgCl in X molar KCl, we will follow these steps:

### Step 1: Understand the Dissociation of AgCl
AgCl is a sparingly soluble salt that dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Write the Expression for Ksp
The solubility product constant (Ksp) for AgCl can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Since AgCl is a solid, its concentration is not included in the Ksp expression.

### Step 3: Define Solubility (S)
Let the solubility of AgCl in pure water be \( S \) moles per liter. Therefore, at equilibrium:
- The concentration of \(\text{Ag}^+\) will be \( S \).
- The concentration of \(\text{Cl}^-\) will also be \( S \).

Thus, we can write:
\[ K_{sp} = S \cdot S = S^2 \]
From this, we can find:
\[ S = \sqrt{K_{sp}} \]

### Step 4: Consider the Effect of Adding KCl
When we add KCl to the solution, it dissociates completely:
\[ \text{KCl} \rightleftharpoons \text{K}^+ + \text{Cl}^- \]
If we add \( X \) moles of KCl, the concentration of \(\text{Cl}^-\) ions will increase to \( S + X \).

### Step 5: Write the New Ksp Expression
Now, the Ksp expression in the presence of KCl becomes:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S \cdot (S + X) \]

### Step 6: Assume S is Much Smaller than X
Since the solubility \( S \) of AgCl is very small compared to the concentration of Cl^- from KCl (i.e., \( S \ll X \)), we can approximate:
\[ S + X \approx X \]

Thus, the Ksp expression simplifies to:
\[ K_{sp} \approx S \cdot X \]

### Step 7: Solve for S
Rearranging the equation gives:
\[ S = \frac{K_{sp}}{X} \]

### Conclusion
The solubility of AgCl in X molar KCl is:
\[ S = \frac{K_{sp}}{X} \]

### Final Answer
The solubility of AgCl in X molar KCl is \( \frac{K_{sp}}{X} \).

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