1. Write the equilibrium expression

For $Fe^{2+}(aq) + S^{2-}(aq) \rightleftharpoons FeS(s)$

the solid $FeS$ is not included in the equilibrium expression, so $K_c = \frac{1}{[Fe^{2+}][S^{2-}]}$

Given: $K_c = 1.6 \times 10^{17}$

Hence, $[Fe^{2+}][S^{2-}] = \frac{1}{1.6 \times 10^{17}}$

2. Find concentrations just after mixing

Equal volumes of:

• $0.06\,M$ $Fe^{2+}$
• $0.20\,M$ $S^{2-}$

are mixed, so each concentration becomes half: $[Fe^{2+}]_0 = \frac{0.06}{2} = 0.03\,M$ $[S^{2-}]_0 = \frac{0.20}{2} = 0.10\,M$

3. Determine the limiting ion for precipitation

Reaction is $1:1$. Initially after mixing:

• $Fe^{2+} = 0.03\,M$
• $S^{2-} = 0.10\,M$

So $Fe^{2+}$ is limiting, and almost all of it precipitates. Leftover $S^{2-}$ after complete precipitation would be approximately $0.10 - 0.03 = 0.07\,M$

Since $K_c$ is extremely large, this approximation is valid.

4. Use equilibrium condition to find $[Fe^{2+}]$

Let equilibrium concentration of $Fe^{2+}$ be $x$. Then $[S^{2-}]$ at equilibrium is approximately $0.07\,M$.

Using $[Fe^{2+}][S^{2-}] = \frac{1}{1.6 \times 10^{17}}$ we get $x(0.07) = \frac{1}{1.6 \times 10^{17}}$

So $x = \frac{1}{1.6 \times 10^{17} \times 0.07}$

$x = \frac{1}{1.12 \times 10^{16}}$

$x = 8.93 \times 10^{-17}\,M$

5. Match with the required form

Given: $[Fe^{2+}]_{eq} = Y \times 10^{-17} \, M$

Thus, $Y = 8.93 \approx 8.9$

6. Comparison with stored answer

Derived answer: $8.9$

Stored correct answer: $8.9$

They match.