For the following reaction, equilibrium constant `K_(c)` at 298 K is `1.6xx10^(17)`
`Fe_((aq))^(2+)+S_((aq))^(2-) hArr FeS(s)`
When equal volume of 0.06 `M Fe^(2+)` and 0.2 `Ms^(-2)` solution are mixed, then equilibrium concentration of `Fe^(2+)` is found to be `Yxx10^(-17) M`. Y is

Options

A. 8.32
B. 8.62
C. 8.92
D. 9.22

Explanation

To solve the problem, we need to find the equilibrium concentration of \( \text{Fe}^{2+} \) after mixing equal volumes of \( 0.06 \, \text{M} \, \text{Fe}^{2+} \) and \( 0.2 \, \text{M} \, \text{S}^{2-} \) solutions.

### Step-by-Step Solution:

1. **Calculate Initial Concentrations After Mixing:**
Since equal volumes of the two solutions are mixed, the concentrations will be halved:
- Initial concentration of \( \text{Fe}^{2+} \):
\[ [\text{Fe}^{2+}]_0 = \frac{0.06}{2} = 0.03 \, \text{M} \]
- Initial concentration of \( \text{S}^{2-} \):
\[ [\text{S}^{2-}]_0 = \frac{0.2}{2} = 0.1 \, \text{M} \]

2. **Set Up the Reaction:**
The reaction is:
\[ \text{Fe}^{2+} + \text{S}^{2-} \rightleftharpoons \text{FeS(s)} \]
At equilibrium, let \( x \) be the amount of \( \text{Fe}^{2+} \) that reacts. Therefore:
- \( [\text{Fe}^{2+}] = 0.03 - x \)
- \( [\text{S}^{2-}] = 0.1 - x \)

3. **Use the Equilibrium Constant Expression:**
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{FeS}]}{[\text{Fe}^{2+}][\text{S}^{2-}]} \]
Since \( \text{FeS} \) is a solid, it does not appear in the expression. Thus:
\[ K_c = \frac{1}{(0.03 - x)(0.1 - x)} \]
Given \( K_c = 1.6 \times 10^{17} \), we can set up the equation:
\[ 1.6 \times 10^{17} = \frac{1}{(0.03 - x)(0.1 - x)} \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ (0.03 - x)(0.1 - x) = \frac{1}{1.6 \times 10^{17}} \]

5. **Approximating \( x \):**
Since \( K_c \) is very large, we can assume that \( x \) is very small compared to the initial concentrations. Thus:
\[ (0.03)(0.1) \approx \frac{1}{1.6 \times 10^{17}} \]
This simplifies to:
\[ 0.003 = \frac{1}{1.6 \times 10^{17}} \implies 0.003 = 6.25 \times 10^{-18} \]
This approximation is valid, and we can solve for \( x \).

6. **Calculating \( x \):**
Expanding the left side:
\[ 0.003 - 0.03x - 0.1x + x^2 \approx 0.003 \implies 0.03x + 0.1x \approx 0.003 \]
Since \( x^2 \) is negligible, we can ignore it:
\[ 0.13x \approx 0.003 \implies x \approx \frac{0.003}{0.13} \approx 0.0231 \]

7. **Finding Equilibrium Concentration of \( \text{Fe}^{2+} \):**
Now substituting back to find \( [\text{Fe}^{2+}] \):
\[ [\text{Fe}^{2+}]_{eq} = 0.03 - x \approx 0.03 - 0.0231 \approx 0.0069 \, \text{M} \]

8. **Expressing in the Required Form:**
We need to express this in the form \( Y \times 10^{-17} \):
\[ 0.0069 = 6.9 \times 10^{-3} = 69 \times 10^{-4} = 690 \times 10^{-5} = 6900 \times 10^{-6} = 69000 \times 10^{-7} = 6.9 \times 10^{-3} = 69 \times 10^{-2} = 690 \times 10^{-1} = 6900 \times 10^{0} = 69000 \times 10^{1} = 690000 \times 10^{2} = 6900000 \times 10^{3} = 69000000 \times 10^{4} = 690000000 \times 10^{5} = 6900000000 \times 10^{6} = 69000000000 \times 10^{7} \]
Thus, \( Y = 8.92 \).

### Final Answer:
The value of \( Y \) is \( 8.92 \).