Problem 157 - Chemistry

Question

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)=\) \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) is \(6.0 \times 10^{-3}\). a. Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). b. Will a \(1.0 M\) solution of iron(II) nitrate have a higher or lower \(\mathrm{pH}\) than a \(1.0 \mathrm{M}\) solution of iron(III) nitrate? Explain.

Step-by-Step Solution

Answer

a. The pH of a 0.10 M solution of Fe(H2O)6^3+ is approximately 1.19. b. A 1.0 M solution of iron(II) nitrate will have a higher pH (less acidic) than a 1.0 M solution of iron(III) nitrate. This is because iron(II) ions are less acidic than iron(III) ions.

Step 1: Form the equilibrium expression from Ka

The equilibrium expression for the reaction is given by: Ka = \(\frac{[Fe(H2O)5(OH)^{2+}][H3O^+]}{[Fe(H2O)6^{3+}]}\) Step 2: Set up the reaction table

Step 2: Set up an ICE table (Initial, Change, Equilibrium) to represent concentrations

Let x be the change in concentration [H3O+] and then [Fe(H2O)5(OH)²⁺] is also x (due to stoichiometry 1:1). The initial concentration of [Fe(H2O)6^3+] is 0.10 M. Create the ICE table as follows: ``` Fe(H2O)6^3+ + H2O ⟷ Fe(H2O)5(OH)^{2+} + H3O^+ Initial (M): 0.10 0 0 Change: -x +x +x Equilibrium: 0.10-x x x ``` Step 3: Substitute values in the equilibrium expression

Step 3: Substitute equilibrium values from the ICE table into the equilibrium expression

Replace the equilibrium concentrations with the corresponding values from the ICE table: Ka = \(\frac{x \cdot x}{0.10-x}\) Step 4: Solve for x

Step 4: Solve for x

Since Ka = \(6.0 \times 10^{–3}\) and the approximations made will simplify the equation, we can write: \(6.0 \times 10^{–3}\) = \(\frac{x^2}{0.10}\) Now, solve for x: x = \(6.4 \times 10^{-2}\) Step 5: Calculate the pH

Step 5: Calculate the pH

Since x represents the concentration of H3O+ ions, the pH can be determined using the formula pH = -log[H3O+]. pH = -log(\(6.4 \times 10^{-2}\)) ≈ 1.19 The pH of a 0.10 M solution of Fe(H2O)6^3+ is 1.19. b. Will a 1.0 M solution of iron(II) nitrate have a higher or lower pH than a 1.0 M solution of iron(III) nitrate?

Step 6: Compare the chemical properties of iron(II) and iron(III) ions

Iron(II) ions (Fe²⁺) are less acidic than iron(III) ions (Fe³⁺). Iron(II) nitrate solutions contain Fe²⁺, while iron(III) nitrate solutions contain Fe³⁺ ions. Fe³⁺ ions have a higher positive charge and are more likely to attract electron pairs from the water molecule when hydrated, making the solution more acidic. Therefore, a 1.0 M solution of iron(II) nitrate will have a higher pH (less acidic) than a 1.0 M solution of iron(III) nitrate.

Key Concepts

pH Calculation

Calculating the pH of a solution involves understanding the concentration of hydrogen ions ([H₃O⁺]) in the solution. The formula for calculating pH is: \[ \text{pH} = -\log [\text{H}_3\text{O}^+] \]. In the exercise provided, we found that the concentration of [H₃O⁺] is \(6.4 \times 10^{-2}\, \text{M}\). Therefore, by applying the formula, we get a pH of approximately 1.19.
This indicates a highly acidic solution. Understanding this concept of pH calculation helps differentiate between acidic, neutral, and basic solutions.

• A pH below 7 indicates an acidic solution.
• A pH of 7 is neutral.
• A pH above 7 indicates a basic solution.

Acid-Base Chemistry

Acid-base chemistry revolves around the concept of proton transfer. In our context, Fe(H₂O)₆³⁺ acts as an acid, donating a proton (H⁺) to H₂O to form H₃O⁺ . This reaction can be better understood by recognizing the acid as a substance that increases [H₃O⁺] in solution.
The reason Fe(H₂O)₆³⁺ behaves this way relates to its ability to polarize the water molecules, pulling electrons towards itself and releasing protons. This makes solutions of iron(III) ions more acidic compared to those with iron(II) ions, which possess less of this proton-releasing ability. Acid-base reactions often involve understanding the strength of acids through their equilibrium constants, as discussed next.

Iron Complexes

Iron complexes, such as Fe(H₂O)₆³⁺ , play a vital role in many chemical reactions. These complexes involve iron cations surrounded by water molecules. The difference between iron(II) and iron(III) complexes is significant in terms of acidity.

• Iron(III) ions (Fe³⁺) , having a higher positive charge, strongly attract electron pairs, facilitating proton release and making the solution more acidic.
• Iron(II) ions (Fe²⁺) , with a lower charge, are less acidic than iron(III) ions.

By understanding these complexes, one can predict the pH of solutions and their behavior in chemical reactions. These concepts are also important in fields like biochemistry, where metal ion complexes influence enzyme activity and function.

Chemical Equilibrium

Chemical equilibrium occurs when a reaction's forward and reverse rates are equal, leading to constant concentrations of reactants and products. In our given reaction, the equilibrium constant (K_{ ext{a}}) is used to express this balance.
\[ K_a = \frac{[\text{Fe(H}_2\text{O})_5(\text{OH})^{2+}][\text{H}_3\text{O}^+]}{[\text{Fe(H}_2\text{O})_6^{3+}]} \]
The value of K_{ ext{a}} = 6.0 \times 10^{-3} indicates how far the reaction proceeds towards forming products. A higher K_ ext{a} would signify a stronger acid, meaning greater donation of protons.
This equilibrium concept ties disciplines like thermodynamics and kinetics together, providing insights into the stability and reactivity of compounds under various conditions.