If `K_(SP)` of `Ag_(2)S` is `10^(-17)`, the solubility of `Ag_(2) S` in 0.1 M solution of `Na_(2)S` will be :

Options

A. `10^(-8)`
B. `5xx10^(-9)`
C. `10^(-15)`
D. `10^(16)`

Explanation

To find the solubility of `Ag2S` in a `0.1 M` solution of `Na2S`, we will follow these steps:

### Step 1: Write the dissociation equation for `Na2S`
`Na2S` dissociates in water to give:
\[ \text{Na}_2\text{S} \rightarrow 2\text{Na}^+ + \text{S}^{2-} \]

From this, we see that the concentration of `S^{2-}` ions from `Na2S` is `0.1 M`.

### Step 2: Write the dissociation equation for `Ag2S`
`Ag2S` dissociates as follows:
\[ \text{Ag}_2\text{S} \rightleftharpoons 2\text{Ag}^+ + \text{S}^{2-} \]

Let the solubility of `Ag2S` in the `0.1 M` `Na2S` solution be `x`. Therefore, at equilibrium:
- The concentration of `Ag^+` will be `2x`.
- The concentration of `S^{2-}` will be `0.1 + x` (but since `x` will be very small compared to `0.1`, we can approximate it as `0.1`).

### Step 3: Write the expression for `Ksp`
The solubility product constant `Ksp` for `Ag2S` is given by:
\[ K_{sp} = [\text{Ag}^+]^2[\text{S}^{2-}] \]

Substituting the equilibrium concentrations into the `Ksp` expression:
\[ K_{sp} = (2x)^2(0.1) \]

### Step 4: Substitute the given value of `Ksp`
We know that \( K_{sp} = 10^{-17} \):
\[ 10^{-17} = (2x)^2(0.1) \]

### Step 5: Simplify the equation
\[ 10^{-17} = 4x^2(0.1) \]
\[ 10^{-17} = 0.4x^2 \]

### Step 6: Solve for `x`
Rearranging gives:
\[ x^2 = \frac{10^{-17}}{0.4} \]
\[ x^2 = 2.5 \times 10^{-17} \]

Taking the square root:
\[ x = \sqrt{2.5 \times 10^{-17}} \]
\[ x = \sqrt{2.5} \times 10^{-8.5} \]
\[ x \approx 1.58 \times 10^{-8.5} \]

### Step 7: Final result
Thus, the solubility of `Ag2S` in `0.1 M` `Na2S` is approximately:
\[ x \approx 1.58 \times 10^{-8} \, \text{M} \]

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