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Chapter 14: Problem 157

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons{\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)}\) is \(6.0 \times 10^{-3}\). a. Calculate the \(\mathrm{pH}\) of a \(0.10-M\) 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 \(1.0-M\) 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 about 1.11. b. A 1.0-M solution of iron(II) nitrate will have a lower pH than a 1.0-M solution of iron(III) nitrate.

Step by step solution

01

Write the Ka expression for the reaction

The Ka expression for the given reaction can be written as: \[K_a = \frac{[\text{Fe}(H_2O)_5(\text{OH})^{2+}][\text{H}_3\text{O}^+]}{[\text{Fe}(H_2O)_6^{3+}]}\] Here, [Fe(H2O)5(OH)^2+], [H3O^+], and [Fe(H2O)6^3+] represent the concentrations of the corresponding species at equilibrium.
02

Determine the concentrations of the species at equilibrium

Let x represent the concentration of Fe(H2O)5(OH)^2+ and H3O^+ at equilibrium. We start with a 0.10-M solution of Fe(H2O)6^3+, so the initial concentration of Fe(H2O)6^3+ will be 0.10 M. Since the reaction has proceeded to equilibrium, the concentration of Fe(H2O)6^3+ would have decreased by x, so the concentration at equilibrium is (0.10 - x) M. Now we can express the equilibrium concentrations in terms of x: [Fe(H2O)5(OH)^2+] = x [H3O^+] = x [Fe(H2O)6^3+] = 0.10 - x
03

Use the Ka expression to find x

We can now substitute the concentrations into the Ka expression: \[6.0 \times 10^{-3} = \frac{x\cdot x}{0.10 - x}\] To solve for x, we can make the assumption that x is much smaller than 0.10, so we can approximate (0.10 - x) as 0.10. This simplifies the equation to: \[6.0 \times 10^{-3} = \frac{x^2}{0.10}\] Solve for x: \[x = \sqrt{6.0 \times 10^{-3} \times 0.10} \approx 7.75 \times 10^{-2}\] Since x represents the concentration of H3O^+ ions, the concentration of H3O^+ at equilibrium is approximately 7.75 x 10^(-2) M.
04

Calculate the pH of the solution

Now we can calculate the pH of the solution, which is the negative logarithm of the H3O^+ concentration: \[pH = -\log(7.75 \times 10^{-2}) \approx 1.11\] a. The pH of a 0.10-M solution of Fe(H2O)6^3+ is about 1.11.
05

Compare the pH of iron(II) nitrate and iron(III) nitrate solutions

Iron(II) ions have a charge of +2, while iron(III) ions have a charge of +3. This means that iron(III) ions have a higher charge density, which allows them to interact more strongly with water molecules, and subsequently cause a higher pH in solution. b. A 1.0-M solution of iron(II) nitrate will have a lower pH than a 1.0-M solution of iron(III) nitrate.

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Most popular questions from this chapter

Which of the following represent conjugate acid-base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

Trichloroacetic acid \(\left(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\) is a corrosive acid that is used to precipitate proteins. The \(\mathrm{pH}\) of a \(0.050-M\) solution of trichloroacetic acid is the same as the \(\mathrm{pH}\) of a \(0.040-\mathrm{M} \mathrm{HClO}_{4}\) solution. Calculate \(K_{\mathrm{a}}\) for trichloroacetic acid.

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}{ }^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 146.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: \(\mathrm{CO}_{2}\) blood levels increase during cardiac arrest.)

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\)

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)
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