Equilibrium Concentration of Ferrous Ions in Solution

To determine the equilibrium concentration of $\text{Fe}^{2+}(aq)$, we analyze the initial conditions and the stoichiometric completion of the reaction.

Step 1: Calculate Initial Concentrations After Mixing

When equal volumes of the two solutions are mixed, the total volume doubles, halving the concentration of each component: $[\text{Fe}^{2+}]_0 = \frac{0.06\text{ M}}{2} = 0.03\text{ M}$ $[\text{S}^{2-}]_0 = \frac{0.2\text{ M}}{2} = 0.10\text{ M}$

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Step 2: Set Up Reaction and Equilibrium

The given equilibrium reaction is: $\text{Fe}^{2+}(aq) + \text{S}^{2-}(aq) \rightleftharpoons \text{FeS}(s)$

The expression for the equilibrium constant $K_c$ is: $K_c = \frac{1}{[\text{Fe}^{2+}][\text{S}^{2-}]} = 1.6 \times 10^{17}$

Since $K_c$ is extremely large ($1.6 \times 10^{17}$), the reaction proceeds almost completely to the right. Here, $\text{Fe}^{2+}$ is the limiting reagent.

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Step 3: Determine Concentrations at Equilibrium

Assuming complete reaction of $\text{Fe}^{2+}$ to form $\text{FeS}(s)$:

• Remaining concentration of $\text{S}^{2-}$ at equilibrium: $[\text{S}^{2-}]_{\text{eq}} = 0.10\text{ M} - 0.03\text{ M} = 0.07\text{ M}$

• Let the unreacted equilibrium concentration of $\text{Fe}^{2+}$ be $[\text{Fe}^{2+}]_{\text{eq}} = Y \times 10^{-17}\text{ M}$.

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Step 4: Calculate the Value of $Y$

Substitute the equilibrium concentrations into the $K_c$ expression: $K_c = \frac{1}{[\text{Fe}^{2+}]_{\text{eq}} \cdot [\text{S}^{2-}]_{\text{eq}}}$

$1.6 \times 10^{17} = \frac{1}{(Y \times 10^{-17}) \cdot 0.07}$

Rearranging the terms to solve for $Y$: $Y \times 10^{-17} = \frac{1}{1.6 \times 10^{17} \times 0.07}$

$Y = \frac{1}{1.6 \times 0.07} = \frac{1}{0.112}$

$Y = \frac{1000}{112} = \frac{125}{14} \approx 8.93$

Thus, the value of $Y$ is $8.93$ (or $8.92857$).