The solubility of AgCl in 0.1M NaCI is `(K_(sp) " of AgCl" = 1.2 xx 10^(-10)) `

Options

A. 0.1 M
B. 1.2 `xx 10^(-5)`
C. `1.095 xx 10^(-5)`
D. `1.199 xx 10^(-9)`

Answer

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

### Step 1: Understand the Dissociation of AgCl and NaCl
AgCl dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

NaCl dissociates completely in solution:
\[ \text{NaCl (s)} \rightleftharpoons \text{Na}^+ (aq) + \text{Cl}^- (aq) \]

In a 0.1 M NaCl solution, the concentration of Cl⁻ ions is 0.1 M.

### Step 2: Write the Expression for Ksp
The solubility product constant (Ksp) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given that \( K_{sp} = 1.2 \times 10^{-10} \).

### Step 3: Set Up the Equation
Let the solubility of AgCl in the 0.1 M NaCl solution be \( S \). In this case:
- The concentration of Ag⁺ ions will be \( S \).
- The concentration of Cl⁻ ions will be \( 0.1 + S \) (since we already have 0.1 M from NaCl).

Thus, we can write:
\[ K_{sp} = S(0.1 + S) \]

### Step 4: Substitute the Ksp Value
Substituting the Ksp value into the equation:
\[ 1.2 \times 10^{-10} = S(0.1 + S) \]

### Step 5: Simplify the Equation
This expands to:
\[ 1.2 \times 10^{-10} = 0.1S + S^2 \]
Rearranging gives:
\[ S^2 + 0.1S - 1.2 \times 10^{-10} = 0 \]

### Step 6: Solve the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:
\[ S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = 0.1 \), and \( c = -1.2 \times 10^{-10} \).

Calculating the discriminant:
\[ b^2 - 4ac = (0.1)^2 - 4(1)(-1.2 \times 10^{-10}) \]
\[ = 0.01 + 4.8 \times 10^{-10} \]
\[ = 0.01 + 0.00000048 = 0.01000048 \]

Now substituting into the quadratic formula:
\[ S = \frac{-0.1 \pm \sqrt{0.01000048}}{2} \]
\[ S = \frac{-0.1 \pm 0.1000024}{2} \]

Calculating the two possible values for S:
1. \( S = \frac{-0.1 + 0.1000024}{2} \approx 0.0000012 \)
2. \( S = \frac{-0.1 - 0.1000024}{2} \) (This will give a negative value, which is not feasible)

So, we take:
\[ S \approx 1.2 \times 10^{-6} \]

### Final Answer
The solubility of AgCl in 0.1 M NaCl is approximately \( 1.2 \times 10^{-6} \) M.

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