For the following reaction, the equilibrium constant Kc at 298 K is 1.6 * 10^17. Fe2+(aq) + S2-(aq)

Q: Why do we halve the initial concentrations?

When you mix equal volumes of two solutions, the total volume becomes twice the original volume of each. Since concentration is moles divided by volume, doubling the volume halves the concentration.

Q: Why isn't FeS included in the $K_c$ expression?

FeS is a pure solid. The concentration of a pure solid remains constant during a reaction, so it is incorporated into the equilibrium constant itself and does not appear in the $K_c$ expression.

Q: How do we know to assume the reaction goes to completion?

The equilibrium constant $K_c$ is $1.6 \times 10^{17}$, which is astronomically large. This means the products are overwhelmingly favored, and the forward reaction will consume almost all of the limiting reagent.

Q: Why do we ignore $y$ when calculating the equilibrium concentration of $S^{2-}$?

We found that $S^{2-}$ has a concentration of $0.07 + y$. Since $K_c$ is so large, the amount of FeS that dissolves back ($y$) is incredibly small (on the order of $10^{-17}$). Adding this tiny number to $0.07$ doesn't change it significantly, so $0.07 + y \approx 0.07$.

Q: What if $K_c$ was a small number like $10^{-3}$?

If $K_c$ were small, the reaction wouldn't go to completion. You couldn't use this approximation. Instead, you would set up an ICE table with $(0.03 - x)$ and $(0.1 - x)$ and solve the resulting quadratic equation for $x$.

Question

For the following reaction, the equilibrium constant $K_c$ at 298 K is $1.6 	imes 10^{17}$.

$	ext{Fe}^{2+}	ext{(aq)} + 	ext{S}^{2-}	ext{(aq)} ightleftharpoons 	ext{FeS(s)}$

When equal volumes of 0.06 M $	ext{Fe}^{2+}	ext{(aq)}$ and 0.2 M $	ext{S}^{2-}	ext{(aq)}$ solutions are mixed, the equilibrium concentration of $	ext{Fe}^{2+}	ext{(aq)}$ is found to be $Y 	imes 10^{-17}$ M. The value of Y is ______.

Sigma Percentile · Accepted Answer

### The Art of Mixing Solutions

Imagine you are standing in a chemistry lab with two beakers in front of you. One beaker holds a solution of iron(II) ions, $	ext{Fe}^{2+}$, and the other holds sulfide ions, $	ext{S}^{2-}$.

The problem states that we are mixing **equal volumes** of these two solutions. This is a classic trap! When you mix equal volumes, the total volume of the final mixture becomes exactly double the original volume of each individual solution.

Because concentration is defined as moles divided by volume, doubling the volume means the concentration of every solute is instantly halved.

Therefore, our initial concentrations in the mixed beaker are:

$[	ext{Fe}^{2+}] = \frac{0.06}{2} = 0.03 	ext{ M}$

$[	ext{S}^{2-}] = \frac{0.2}{2} = 0.1 	ext{ M}$

### The Power of a Massive Equilibrium Constant

Now, let's look at the chemical reaction taking place:

$	ext{Fe}^{2+}	ext{(aq)} + 	ext{S}^{2-}	ext{(aq)} ightleftharpoons 	ext{FeS(s)}$

The problem gives us the equilibrium constant, $K_c = 1.6 	imes 10^{17}$. Take a moment to appreciate how astronomically large this number is!

A massive $K_c$ tells us a very important physical reality: the forward reaction is overwhelmingly favored. The ions desperately want to combine and crash out of the solution as solid iron sulfide. For all practical purposes, this reaction will proceed almost to $100\%$ completion.

### The Limiting Reagent and The Approximation

Since the reaction goes nearly to completion, we must identify the **limiting reagent**. We have $0.03 	ext{ M}$ of $	ext{Fe}^{2+}$ and $0.1 	ext{ M}$ of $	ext{S}^{2-}$. Clearly, $	ext{Fe}^{2+}$ is present in a smaller amount, so it will be completely consumed first.

If we assume the reaction goes all the way, the $	ext{Fe}^{2+}$ concentration drops to approximately zero. The remaining $	ext{S}^{2-}$ concentration will be:

$0.1 	ext{ M} - 0.03 	ext{ M} = 0.07 	ext{ M}$

But remember, this is an equilibrium system. The concentration of $	ext{Fe}^{2+}$ cannot be exactly zero. A microscopic amount of the solid $	ext{FeS}$ will dissolve back into the solution.

Let's call this tiny equilibrium concentration of iron $y$.

$[	ext{Fe}^{2+}]_{	ext{eq}} = y$

Consequently, the equilibrium concentration of sulfide will be $0.07 + y$. However, because $K_c$ is so large, $y$ is going to be incredibly small. Adding $y$ to $0.07$ will not change it in any meaningful way.

Thus, we can safely make the approximation:

$[	ext{S}^{2-}]_{	ext{eq}} \approx 0.07 	ext{ M}$

### The Final Calculation

Now we are ready to apply the equilibrium law. The expression for $K_c$ is:

$K_c = \frac{1}{[	ext{Fe}^{2+}][	ext{S}^{2-}]}$

Notice that solid $	ext{FeS}$ is excluded from the expression because the concentration of a pure solid is constant.

Let's substitute our known values into the equation:

$1.6 	imes 10^{17} = \frac{1}{y 	imes 0.07}$

Now, we just need to isolate $y$:

$y = \frac{1}{1.6 	imes 10^{17} 	imes 0.07}$

Multiplying the terms in the denominator gives:

$y = \frac{1}{0.112 	imes 10^{17}}$

To make the calculation easier, we can rewrite this as:

$y = \frac{10}{1.12} 	imes 10^{-17}$

Solving this fraction yields:

$y = 8.928 	imes 10^{-17} 	ext{ M}$

The problem asks for the value of $Y$ where the concentration is $Y 	imes 10^{-17} 	ext{ M}$. Comparing our result, we find that $Y = 8.928$.

Rounding to two decimal places, we arrive at our final answer:

**$Y \approx 8.93$**

Enter Numerical Value:

Visualized Solution

$M i x in g t h e S o l u t i o n s$

$T h e R e a c t i o n an d E q u i l ib r i u m C o n s t an t$

$I d e n t i f y in g t h e L imi t in g R e a g e n t$

$S e tt in g u p E q u i l ib r i u m C o n ce n t r a t i o n s$

$A ppl y in g t h e E q u i l ib r i u m L a w$

$C a l c u l a t in g y$

$F in d in g Y$

$R e f l ec t i o n$

The Sigma Insight: Solubility Product and Common Ion Effect

The Art of Mixing Solutions

The Power of a Massive Equilibrium Constant

The Limiting Reagent and The Approximation

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mixing the Solutions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Reaction and Equilibrium Constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying the Limiting Reagent

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up Equilibrium Concentrations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Equilibrium Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflection