The solubility product of silver chloride is ` 1.44 xx 10 ^(-4)` at 373 K. The solubility of silver chloride in boiling water will be

Options

A. `0.72xx 10 ^(-4) M~
B. ` 1.20 xx 10 ^(-2) ` M
C. ` 0.72 xx 10^(-2) `M
D. ` 1.20 xx 10 ^(-4) ` M

Answer

To find the solubility of silver chloride (AgCl) in boiling water given its solubility product (Ksp) at 373 K, we can follow these steps:

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

### Step 2: Define the solubility (s)
Let the solubility of AgCl in boiling water be \( s \) mol/L. According to the dissociation, when \( s \) moles of AgCl dissolve, it produces \( s \) moles of \( \text{Ag}^+ \) and \( s \) moles of \( \text{Cl}^- \).

### Step 3: Write the expression for Ksp
The solubility product constant \( K_{sp} \) for AgCl can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = s \cdot s = s^2 \]

### Step 4: Substitute the given Ksp value
We know that \( K_{sp} \) for AgCl at 373 K is given as:
\[ K_{sp} = 1.44 \times 10^{-4} \]
Thus, we can set up the equation:
\[ s^2 = 1.44 \times 10^{-4} \]

### Step 5: Solve for s
To find \( s \), take the square root of both sides:
\[ s = \sqrt{1.44 \times 10^{-4}} \]

### Step 6: Calculate the value of s
Calculating the square root:
\[ s = \sqrt{1.44} \times \sqrt{10^{-4}} = 1.2 \times 10^{-2} \]
Thus, the solubility of AgCl in boiling water is:
\[ s = 0.012 \, \text{mol/L} \]

### Final Answer
The solubility of silver chloride in boiling water is \( 0.012 \, \text{mol/L} \) or \( 1.20 \times 10^{-2} \, \text{mol/L} \).

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