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Chapter 14: Problem 157

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.

Short Answer

Expert verified
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 by step solution

01

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
02

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
03

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
04

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
05

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?
06

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.

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