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Ch.17 - Applications of Aqueous Equilibria
Chapter 17, Problem 123

Calculate the molar solubility of AgI in: (a) Pure Water (b) 0.10 M NaCN: Kf for [Ag(CN)2]- is 3.0 x 10^20

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<b>Step 1:</b> Determine the solubility product constant (K<sub>sp</sub>) for AgI. The dissolution of AgI in water can be represented as: AgI(s) ⇌ Ag<sup>+</sup>(aq) + I<sup>-</sup>(aq). Look up the K<sub>sp</sub> value for AgI in a chemistry reference.
<b>Step 2:</b> Calculate the molar solubility of AgI in pure water. Let the molar solubility be 's'. Then, [Ag<sup>+</sup>] = s and [I<sup>-</sup>] = s. Use the expression K<sub>sp</sub> = [Ag<sup>+</sup>][I<sup>-</sup>] to solve for 's'.
<b>Step 3:</b> For part (b), consider the complexation of Ag<sup>+</sup> with CN<sup>-</sup>. The formation of the complex [Ag(CN)<sub>2</sub>]<sup>-</sup> can be represented as: Ag<sup>+</sup>(aq) + 2 CN<sup>-</sup>(aq) ⇌ [Ag(CN)<sub>2</sub>]<sup>-</sup>(aq). Use the given formation constant (K<sub>f</sub>) to express the equilibrium condition.
<b>Step 4:</b> Set up the equilibrium expressions for the dissolution of AgI and the formation of the complex ion. Assume the initial concentration of CN<sup>-</sup> is 0.10 M. Use the K<sub>sp</sub> and K<sub>f</sub> values to write the overall equilibrium expression for the system.
<b>Step 5:</b> Solve the equilibrium expression to find the molar solubility of AgI in 0.10 M NaCN. This involves substituting the expressions for the concentrations of Ag<sup>+</sup>, I<sup>-</sup>, and [Ag(CN)<sub>2</sub>]<sup>-</sup> into the equilibrium equations and solving for the solubility.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at equilibrium, expressed in moles per liter (M). It is a crucial concept in understanding how solutes interact with solvents and is influenced by factors such as temperature and the presence of other ions in solution.
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Common Ion Effect

The common ion effect describes the decrease in solubility of a salt when a common ion is added to the solution. In the case of AgI, the presence of NaCN introduces CN⁻ ions, which can shift the equilibrium and reduce the solubility of AgI due to the formation of complex ions, thereby affecting the molar solubility.
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Formation Constant (Kf)

The formation constant (Kf) quantifies the stability of a complex ion in solution. For the complex ion [Ag(CN)2]⁻, a high Kf value (3.0 x 10^20) indicates that the complex is very stable, which significantly influences the solubility of AgI in the presence of CN⁻ ions, as it drives the equilibrium towards the formation of the complex and reduces the concentration of free Ag⁺ ions.
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Related Practice
Textbook Question
Dissolution of 5.0 x 10^-3 mol of Cr(OH)3 in 1.0 L of 1.0 M NaOH gives a solution of the complex ion [Cr(OH)4]- (Kf = 8 x10^29). What fraction of the chromium in such a solution is present as uncomplexed Cr3+?
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Textbook Question
Write a balanced net ionic equation for each of the follow-ing dissolution reactions, and use the appropriate Ksp and Kf values in Appendix C to calculate the equilibrium constant for each. (a) AgI in aqueous NaCN to form [Ag(CN)2]-
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Textbook Question
Write a balanced net ionic equation for each of the follow-ing dissolution reactions, and use the appropriate Ksp and Kf values in Appendix C to calculate the equilibrium constant for each. (b) Cu(OH)2 in aqueous NH3 to form [Cu(NH3)4]2+
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Textbook Question
Calculate the molar solubility of Cr(OH)3 in 0.50 M NaOH; Kf for Cr(OH)4- is 8 x 10^29.
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Open Question
Zinc hydroxide, Zn(OH)2 (Ksp = 4.1 x 10^-17), is nearly insoluble in water but is more soluble in strong base because Zn2+ forms the soluble complex ion [Zn(OH)4]2- (Kf = 3 x 10^15). (a) What is the molar solubility of Zn(OH)2 in pure water? (You may ignore the OH- from the self-dissociation of water.) (b) What is the pH of the solution in part (a)? (c) What is the molar solubility of Zn(OH)2 in 0.10 M NaOH?
Textbook Question
Citric acid (H3Cit) can be used as a household cleaning agent to dissolve rust stains. The rust, represented as Fe(OH)3, dissolves because the citrate ion forms a soluble complex with Fe3+ (a) Using the equilibrium constants in Appendix C and Kf = 6.3 x 10^11 for Fe(Cit), calculate the equilibrium constant K for the reaction. (b) Calculate the molar solubility of Fe(OH)3 in 0.500 M solution of H3Cit.
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