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Chapter 15: Problem 21

Write the equilibrium constant expression for the following reaction, $$\begin{array}{r} \mathrm{Fe}(\mathrm{OH})_{3}+3 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ K=9.1 \times 10^{3} \end{array}$$ and compute the equilibrium concentration for \(\left[\mathrm{Fe}^{3+}\right]\) at \(\left.\mathrm{pH}=7 \text { (i.e., }\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7}\right)\)

Short Answer

Expert verified
\([\text{Fe}^{3+}] = 9.1 \times 10^{-18}\) M

Step by step solution

01

Write the Equilibrium Constant Expression

Recall that the equilibrium constant expression for a reaction is defined by \(K = \frac{[\text{products}]}{[\text{reactants}]}\), raising concentrations to the power of their stoichiometric coefficients. This gives \(K = \frac{[\text{Fe}^{3+}] [\text{H}_{2}O]^3}{[\text{Fe(OH)}_3][\text{H}^{+}]^3}\). Since H2O is a liquid, its concentration is not included in the equilibrium constant expression, simplifying the expression to \(K = \frac{[\text{Fe}^{3+}]}{[\text{Fe(OH)}_3][\text{H}^{+}]^3}\)
02

Substitute known values into the equation

We are given that \(K = 9.1 \times 10^{3}\) and \([\text{H}^{+}] = 1.0 \times 10^{-7}\). Since Fe(OH)3 is the precipitate (solid), we assign its concentration as 1, leading to: \[9.1 \times 10^{3} = \frac{[\text{Fe}^{3+}]}{(1)(1.0 \times 10^{-7})^3}\]
03

Solve for Fe^{3+ concentration}

Multiply both sides of the equation by \((1.0 \times 10^{-7})^3\) to solve for \([\text{Fe}^{3+}]\): \[ [\text{Fe}^{3+}] = 9.1 \times 10^{3} \times (1.0 \times 10^{-7})^3 \]

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Most popular questions from this chapter

In the gas phase, iodine reacts with cyclopentene \(\left(\mathrm{C}_{5} \mathrm{H}_{8}\right)\) by a free radical mechanism to produce cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) and hydrogen iodide. Explain how each of the following affects the amount of \(\mathrm{HI}(\mathrm{g})\) present in the equilibrium mixture in the reaction \begin{array}{r} \mathrm{I}_{2}(\mathrm{g})+\mathrm{C}_{5} \mathrm{H}_{8}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{6}(\mathrm{g})+2 \mathrm{HI}(\mathrm{g}) \\ \Delta H^{\circ}=92.5 \mathrm{kJ} \end{array} (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{C}_{5} \mathrm{H}_{6}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst; (e) adding an inert gas such as He to a constant-volume reaction mixture.

Continuous removal of one of the products of a chemical reaction has the effect of causing the reaction to go to completion. Explain this fact in terms of Le Châtelier's principle.

Nitrogen dioxide obtained as a cylinder gas is always a mixture of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) .\) A \(5.00 \mathrm{g}\) sample obtained from such a cylinder is sealed in a \(0.500 \mathrm{L}\) flask at \(298 \mathrm{K}\). What is the mole fraction of \(\mathrm{NO}_{2}\) in this mixture? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3}$$

For the reaction \(\mathrm{SO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{aq}), K=1.25 \mathrm{at}\) \(25^{\circ} \mathrm{C} .\) Will the amount of \(\mathrm{SO}_{2}(\mathrm{g})\) be greater than or less than the amount of \(\mathrm{SO}_{2}(\mathrm{aq}) ?\)

Rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}),\) is caused by the oxidation of iron by oxygen. Write the equilibrium constant expression first in terms of activities, and then in terms of concentration and pressure.
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