The solubility of AgCI(s) with solubility product `1.6 xx 10^(-10)` in `0.1 M` `NaCl` solution would be :

Options

A. `1.26xx10^(-5)M`
B. `1.6xx10^(-9)M`
C. `1.6xx10^(-11)M`
D. `zero`

Explanation

To find the solubility of AgCl in a 0.1 M NaCl solution, we will follow these steps:

### Step 1: Write the dissociation equation for AgCl
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

### Step 2: Define the solubility (S) of AgCl
Let the solubility of AgCl in the solution be S. Therefore, at equilibrium:
- [Ag⁺] = S
- [Cl⁻] = S (from AgCl) + 0.1 M (from NaCl)

### Step 3: Express the concentration of Cl⁻
The total concentration of Cl⁻ ions in the solution will be:
[Cl⁻] = S + 0.1

### Step 4: Write the expression for the solubility product (Ksp)
The solubility product (Ksp) expression for AgCl is given by:
Ksp = [Ag⁺][Cl⁻]

Substituting the concentrations:
Ksp = S × (S + 0.1)

### Step 5: Substitute the given Ksp value
We know that Ksp for AgCl is 1.6 × 10^(-10):
1.6 × 10^(-10) = S × (S + 0.1)

### Step 6: Assume S is very small compared to 0.1 M
Since Ksp is very small, we can assume that S is much smaller than 0.1 M. Therefore, we can neglect S in comparison to 0.1:
1.6 × 10^(-10) ≈ S × 0.1

### Step 7: Solve for S
Now, rearranging the equation gives:
S = (1.6 × 10^(-10)) / 0.1
S = 1.6 × 10^(-9) M

### Conclusion
The solubility of AgCl in a 0.1 M NaCl solution is 1.6 × 10^(-9) M.

### Final Answer
The correct answer is option B: 1.6 × 10^(-9) M.

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