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K(a) for the reaction, Fe^(3+)(aq)+H(2...

`K_(a)` for the reaction,
`Fe^(3+)(aq)+H_(2)O(l)hArr Fe(OH)^(2+)(aq) +H_(3)O^(o+)(aq)` is `6.5 xx 10^(-3)`, what is the maximum `pH` value which could be used so that at least `80%` of the total iron `(III)` in a dilute solution exsists as `Fe^(3+)`?

A

`2`

B

`2.41`

C

`2.79`

D

`1.60`

Text Solution

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The correct Answer is:
To solve the problem, we need to determine the maximum pH value at which at least 80% of the total iron (III) exists as \( \text{Fe}^{3+} \) in a dilute solution. We are given the dissociation constant \( K_a \) for the hydrolysis reaction of \( \text{Fe}^{3+} \). ### Step-by-Step Solution: 1. **Write the hydrolysis reaction**: \[ \text{Fe}^{3+}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{Fe(OH)}^{2+}(aq) + \text{H}_3\text{O}^+(aq) \] 2. **Define the initial concentration**: Let the initial concentration of \( \text{Fe}^{3+} \) be \( C \). 3. **Set up the equilibrium concentrations**: At equilibrium: - Concentration of \( \text{Fe}^{3+} \) = \( C(1 - \alpha) \) - Concentration of \( \text{Fe(OH)}^{2+} \) = \( C\alpha \) - Concentration of \( \text{H}_3\text{O}^+ \) = \( C\alpha \) 4. **Use the condition for 80% of \( \text{Fe}^{3+} \)**: We need at least 80% of the total iron to exist as \( \text{Fe}^{3+} \): \[ 1 - \alpha \geq 0.8 \implies \alpha \leq 0.2 \] 5. **Express \( K_a \)**: The expression for \( K_a \) is given by: \[ K_a = \frac{[\text{Fe(OH)}^{2+}][\text{H}_3\text{O}^+]}{[\text{Fe}^{3+}]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \] 6. **Substitute the known values**: Given \( K_a = 6.5 \times 10^{-3} \) and \( \alpha = 0.2 \): \[ 6.5 \times 10^{-3} = \frac{C(0.2)^2}{1 - 0.2} = \frac{C(0.04)}{0.8} \] Rearranging gives: \[ C = \frac{6.5 \times 10^{-3} \times 0.8}{0.04} = 0.13 \, \text{mol/L} \] 7. **Calculate the concentration of \( \text{H}_3\text{O}^+ \)**: \[ [\text{H}_3\text{O}^+] = C \alpha = 0.13 \times 0.2 = 0.026 \, \text{mol/L} \] 8. **Calculate the pH**: \[ \text{pH} = -\log[\text{H}_3\text{O}^+] = -\log(0.026) \approx 1.585 \] 9. **Final answer**: The maximum pH value is approximately \( 1.6 \).
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