the ratio of solubility of `CH_(3)COOAg` in a buffer solution with `pH = 4` and in `H_(2)O` is
`(Sol Buffer = 2 xx 10^(-6))`

Options

A. `1//2`
B. `2`
C. `1//3`
D. `3`

Explanation

To find the ratio of solubility of CH₃COOAg in a buffer solution with pH = 4 and in H₂O, we can follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of silver acetate (CH₃COOAg) can be represented as:
\[ \text{CH}_3\text{COOAg} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{Ag}^+ \]

### Step 2: Define solubility in terms of S
Let the solubility of CH₃COOAg in water be \( S \). At equilibrium, the concentrations of the ions will be:
- \([CH_3COO^-] = S\)
- \([Ag^+] = S\)

### Step 3: Write the expression for Ksp
The solubility product constant (Ksp) for the dissociation can be expressed as:
\[ K_{sp} = [CH_3COO^-][Ag^+] = S \cdot S = S^2 \]

### Step 4: Use the given Ksp value
Assuming the value of Ksp is given as \( 1 \times 10^{-12} \):
\[ K_{sp} = S^2 = 1 \times 10^{-12} \]

### Step 5: Solve for S in pure water
To find the solubility \( S \) in pure water, we take the square root of Ksp:
\[ S = \sqrt{1 \times 10^{-12}} = 1 \times 10^{-6} \, \text{mol/L} \]

### Step 6: Find solubility in the buffer solution
The solubility of CH₃COOAg in the buffer solution with pH = 4 is given as:
\[ \text{Solubility in buffer} = 2 \times 10^{-6} \, \text{mol/L} \]

### Step 7: Calculate the ratio of solubility
Now, we can find the ratio of solubility in the buffer solution to that in pure water:
\[ \text{Ratio} = \frac{\text{Solubility in buffer}}{\text{Solubility in water}} = \frac{2 \times 10^{-6}}{1 \times 10^{-6}} = 2 \]

### Final Answer
The ratio of solubility of CH₃COOAg in a buffer solution with pH = 4 to that in H₂O is **2**.

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