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Chapter 17: Problem 50

The hydroxide ion concentration in a saturated solution of iron(III) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{3}\), is \(2.2 \times 10^{-10} \mathrm{M}\). Calculate the value for the equilibrium constant.

Short Answer

Expert verified
The equilibrium constant \( K_{sp} \) is \( 7.51 \times 10^{-40} \).

Step by step solution

01

Write the Dissolution Reaction

The dissolution of iron(III) hydroxide can be expressed by the equation: \[ \mathrm{Fe(OH)_3 (s)} \rightleftharpoons \mathrm{Fe^{3+} (aq)} + 3\mathrm{OH^- (aq)} \] This shows that one formula unit of \( \mathrm{Fe(OH)_3} \) dissociates into one iron(III) ion and three hydroxide ions in solution.
02

Define the Equilibrium Expression

The expression for the solubility product constant \( K_{sp} \) is given by: \[ K_{sp} = [\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3 \] Here, \([\mathrm{Fe^{3+}}]\) and \([\mathrm{OH^-}]\) are the molar concentrations of the ions in a saturated solution.
03

Relate Ion Concentrations

Since each formula unit of \( \mathrm{Fe(OH)_3 (s)} \) produces 1 \( \mathrm{Fe^{3+}} \) ion and 3 \( \mathrm{OH^-} \) ions, if \([\mathrm{OH^-}] = 2.2 \times 10^{-10} \) M, then \([\mathrm{Fe^{3+}}] = \frac{1}{3} \times 2.2 \times 10^{-10} = x\).
04

Calculate Fe(III) Ion Concentration

Substituting the stoichiometric relationship into \([\mathrm{Fe^{3+}}] = x\), we get \(x = \frac{1}{3} \times 2.2 \times 10^{-10} = 7.33 \times 10^{-11} \) M.
05

Substitute into the Equilibrium Expression

Using the relation from the equilibrium expression \( K_{sp} = [\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3\), substitute \([\mathrm{Fe^{3+}}] = 7.33 \times 10^{-11} \) M and \([\mathrm{OH^-}] = 2.2 \times 10^{-10} \) M. \[ K_{sp} = (7.33 \times 10^{-11})(2.2 \times 10^{-10})^3 \]
06

Final Calculation for Ksp

Solve for \( K_{sp} \): \[ K_{sp} = (7.33 \times 10^{-11})(2.2 \times 10^{-10})^3 = 7.51 \times 10^{-40} \] So, the equilibrium constant \( K_{sp} \) is \( 7.51 \times 10^{-40} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product
The solubility product, represented as \( K_{sp} \), is an equilibrium constant that applies to the dissolution of sparingly soluble salts. It quantifies how much of a substance can dissolve in a given solvent to form a saturated solution.
The \( K_{sp} \) value is derived from the dissociation of ionic compounds into their respective ions. For example, for iron(III) hydroxide \( \mathrm{Fe(OH)_3} \), it dissociates into iron ions \( \mathrm{Fe^{3+}} \) and hydroxide ions \( \mathrm{OH^-} \).
  • The \( K_{sp} \) expression is written as \( [\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3 \).
  • The concentrations in the equation are those of the ions at equilibrium in a saturated solution.
The smaller the \( K_{sp} \) value, the less soluble the compound is in water. In this iron(III) hydroxide example, a very low \( K_{sp} \) signifies low solubility.
Chemical Equilibrium
Chemical equilibrium in the context of solubility is reached when the rate of dissolution of a solid equals the rate of precipitation, resulting in no net change in the amounts of reactants and products.
For a dissolution reaction like \( \mathrm{Fe(OH)_3 (s) \rightleftharpoons Fe^{3+} (aq) + 3OH^- (aq)} \), equilibrium is achieved when the ion concentrations remain constant over time.
  • The equilibrium state is dynamic, meaning the processes of dissolution and precipitation continue at equal rates.
  • At equilibrium, the concentration of dissolved ions in a saturated solution can be used to calculate \( K_{sp} \).
Equilibrium concepts help us understand how changes in conditions, such as the addition of other ions, can shift the balance and affect solubility.
Dissolution Reaction
A dissolution reaction involves the breaking apart of solid compounds into their constituent ions when dissolved in a solvent, usually water.
For iron(III) hydroxide, the dissolution is represented by the equation \( \mathrm{Fe(OH)_3 (s) \rightleftharpoons Fe^{3+} (aq) + 3OH^- (aq)} \).
This reaction demonstrates how a solid substance becomes solvated, where the water molecules surround and stabilize the ions.
  • The stoichiometry of the reaction shows how many ions each formula unit produces; here, one unit of \( \mathrm{Fe(OH)_3} \) gives one \( \mathrm{Fe^{3+}} \) and three \( \mathrm{OH^-} \).
  • Understanding this process is crucial for calculating concentrations and the solubility product.
Recognizing how a solid dissolves helps in predicting and calculating the amounts of each ion present in a saturated solution.

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Most popular questions from this chapter

Write the general equilibrium constant expression for each of the following: (a) \(3 \mathrm{~A} \rightleftarrows 2 \mathrm{C}\) (b) \(A+B \rightleftarrows 2 C\) (c) \(3 \mathrm{~A}+5 \mathrm{~B} \rightleftarrows \mathrm{C}+4 \mathrm{D}\)

The Haber Process uses a metal oxide catalyst to produce ammonia gas. Does the catalyst increase the amount of ammonia gas? Explain.

The Contact Process uses a metal catalyst to produce sulfur trioxide gas for the manufacture of sulfuric acid. Does the catalyst increase the amount of sulfur trioxide gas? Explain.

Write the equilibrium constant expression for each of the following weak acids: (a) \(\mathrm{HCHO}_{2}(a q) \rightleftarrows \mathrm{H}^{+}(a q)+\mathrm{CHO}_{2}^{-}(a q)\) (b) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \rightleftarrows \mathrm{H}^{+}(a q)+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q)\) (c) \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \rightleftarrows \mathrm{H}^{+}(a q)+\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}(a q)\)

Given the equilibrium concentrations for each gas at \(500^{\circ} \mathrm{C}\), calculate the value of \(K_{\mathrm{eq}}\) for the manufacture of ammonia. $$\begin{array}{ccc}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) & \rightleftarrows \mathrm{NH}_{3}(g) \\\0.400 \mathrm{M} & 1.20 \mathrm{M} & 0.195\mathrm{M}\end{array}$$
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