What would be the solubility of silver chloride in `0.10M NaCI` solution?
`K_(sp) for AgCI = 1.20 xx 10^(-10)`

Options

A. `0.1M`
B. `1.2 xx 10^(-6)M`
C. `1.2 xx 10^(-9)M`
D. `1.2 xx 10^(-10)M`

Explanation

To find the solubility of silver chloride (AgCl) in a 0.10 M NaCl solution, we can follow these steps:

### Step 1: Understand the Dissociation of NaCl
NaCl is a strong electrolyte, which means it completely dissociates in solution:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
In a 0.10 M NaCl solution, the concentration of Cl⁻ ions is 0.10 M.

### Step 2: Write the Dissociation Reaction of AgCl
Silver chloride (AgCl) also dissociates in water:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]
Let the solubility of AgCl in this solution be \( x \) M. This means that when AgCl dissolves, it produces \( x \) M of Ag⁺ and \( x \) M of Cl⁻.

### Step 3: Set Up the Expression for Ksp
The solubility product constant (Ksp) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
From the previous steps, we know:
- The concentration of Ag⁺ is \( x \).
- The concentration of Cl⁻ is \( 0.10 + x \) (but since \( x \) will be very small compared to 0.10 M, we can approximate it as 0.10 M).

Thus, we can write:
\[ K_{sp} = x \cdot (0.10) \]

### Step 4: Substitute the Given Ksp Value
We know that \( K_{sp} \) for AgCl is \( 1.20 \times 10^{-10} \):
\[ 1.20 \times 10^{-10} = x \cdot 0.10 \]

### Step 5: Solve for x
Now, we can solve for \( x \):
\[ x = \frac{1.20 \times 10^{-10}}{0.10} \]
\[ x = 1.20 \times 10^{-9} \, \text{M} \]

### Conclusion
The solubility of silver chloride (AgCl) in a 0.10 M NaCl solution is:
\[ \text{Solubility} = 1.20 \times 10^{-9} \, \text{M} \]

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