The molar solubility of AgCl in `1.8 M AgNO_(3)` solution is (`K_(sp)` of `AgCl = 1.8 xx 10^(-10)`)

Options

A. `10^(-5)`
B. `10^(-10)`
C. `1.8 xx 10^(-5)`
D. `1.8 xx 10^(-10)`

Explanation

To find the molar solubility of AgCl in a 1.8 M AgNO₃ solution, we can follow these steps:

### Step 1: Write the dissociation equation for AgCl
AgCl dissociates in water to give:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Write the expression for the solubility product (Ksp)
The solubility product (Ksp) expression for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

### Step 3: Determine the concentration of Ag⁺ in the solution
In a 1.8 M AgNO₃ solution, the concentration of Ag⁺ ions is already 1.8 M. When AgCl dissolves, it will contribute additional Ag⁺ ions, but since the molar solubility (S) of AgCl is expected to be very small compared to 1.8 M, we can approximate:
\[ [\text{Ag}^+] \approx 1.8 + S \approx 1.8 \, \text{M} \]

### Step 4: Set up the Ksp expression with the concentrations
Since the concentration of Cl⁻ ions will be equal to the molar solubility (S) of AgCl, we can substitute into the Ksp expression:
\[ K_{sp} = (1.8)(S) \]

### Step 5: Substitute the known Ksp value
Given that \( K_{sp} \) for AgCl is \( 1.8 \times 10^{-10} \):
\[ 1.8 \times 10^{-10} = (1.8)(S) \]

### Step 6: Solve for S (molar solubility of AgCl)
To find S, rearrange the equation:
\[ S = \frac{1.8 \times 10^{-10}}{1.8} \]
\[ S = 1.0 \times 10^{-10} \, \text{M} \]

### Conclusion
The molar solubility of AgCl in a 1.8 M AgNO₃ solution is:
\[ S = 1.0 \times 10^{-10} \, \text{M} \]

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