Use the References to access important values if needed for this question. Consider the insoluble compound iron(II) hydroxide , Fe(OH)2 . The iron(II) ion also forms a complex with cyanide ions Write a balanced net ionic equation to show why the solubility of Fe(OH), (s) increases in the presence of cyanide ions and calculate the equilibrium constant for this reaction. For Fe(CN),+, Kf=7.7x1036. Be sure to specify states such as (aq) or (s). 4- K= Submit Answer Retry Entire Group 9 more group attempts remaining

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**Exercise: Solubility of Iron(II) Hydroxide in the Presence of Cyanide Ions**

**Objective:**
Understand and demonstrate why the solubility of iron(II) hydroxide, Fe(OH)₂, increases in the presence of cyanide ions.

---

**Instructions:**

Consider the insoluble compound iron(II) hydroxide, Fe(OH)₂. The iron(II) ion also forms a complex with cyanide ions. 

**Task:**
Write a balanced net ionic equation to show why the solubility of Fe(OH)₂(s) increases in the presence of cyanide ions and calculate the equilibrium constant for this reaction.

**Given Data:**
For the complex ion \([Fe(CN)_6]^{4-}\), the formation constant \( K_f \) is \( 7.7 \times 10^{36} \).

Be sure to specify states such as (aq) or (s).

\[
\begin{array}{c}
\text{Fe(OH)}_2(s) \longrightarrow \text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \\
\\
\text{Fe}^{2+}(aq) + 6\text{CN}^-(aq) \longrightarrow \text{[Fe(CN)}_6]^{4-}(aq)\\
\\
K = \frac{[\text{Product}]^x}{[\text{Reactant}]^y} \\
\end{array}
\]

Here is how the calculation steps and factors are visually represented:

1. Dissolution of Fe(OH)₂(s):
\[
\text{Fe(OH)}_2(s) \longrightarrow \text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) 
\]
2. Formation of the complex ion in solution:
\[
\text{Fe}^{2+}(aq) + 6\text{CN}^-(aq) \longrightarrow \text{[Fe(CN)}_6]^{4-}(aq)
\]

To find the equilibrium constant (K) for the overall reaction, multiply the solubility product constant \( K_{sp} \) of Fe(OH)₂ by the formation constant \( K_f \) of the complex ion:
\[
K = K_{sp}
Transcribed Image Text:**Exercise: Solubility of Iron(II) Hydroxide in the Presence of Cyanide Ions** **Objective:** Understand and demonstrate why the solubility of iron(II) hydroxide, Fe(OH)₂, increases in the presence of cyanide ions. --- **Instructions:** Consider the insoluble compound iron(II) hydroxide, Fe(OH)₂. The iron(II) ion also forms a complex with cyanide ions. **Task:** Write a balanced net ionic equation to show why the solubility of Fe(OH)₂(s) increases in the presence of cyanide ions and calculate the equilibrium constant for this reaction. **Given Data:** For the complex ion \([Fe(CN)_6]^{4-}\), the formation constant \( K_f \) is \( 7.7 \times 10^{36} \). Be sure to specify states such as (aq) or (s). \[ \begin{array}{c} \text{Fe(OH)}_2(s) \longrightarrow \text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \\ \\ \text{Fe}^{2+}(aq) + 6\text{CN}^-(aq) \longrightarrow \text{[Fe(CN)}_6]^{4-}(aq)\\ \\ K = \frac{[\text{Product}]^x}{[\text{Reactant}]^y} \\ \end{array} \] Here is how the calculation steps and factors are visually represented: 1. Dissolution of Fe(OH)₂(s): \[ \text{Fe(OH)}_2(s) \longrightarrow \text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \] 2. Formation of the complex ion in solution: \[ \text{Fe}^{2+}(aq) + 6\text{CN}^-(aq) \longrightarrow \text{[Fe(CN)}_6]^{4-}(aq) \] To find the equilibrium constant (K) for the overall reaction, multiply the solubility product constant \( K_{sp} \) of Fe(OH)₂ by the formation constant \( K_f \) of the complex ion: \[ K = K_{sp}
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