Question

What mass of AgI will dissolve in \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{NH}_{3}\) ? Neglect change in conc. of \(\mathrm{NH}_{3}\). [Given : \(\left.K_{s p}(\mathrm{Agl})=1.5 \times 10^{-16} ; \mathrm{K}_{f}\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]=1.6 \times 10^{7}\right] ;\) (At. wt. \(\mathrm{Ag}=108 ; \mathrm{I}=127\) ) (a) \(4.9 \times 10^{-5} \mathrm{~g}\) (b) \(0.0056 \mathrm{~g}\) (c) \(0.035 \mathrm{~g}\) (d) \(0.011 \mathrm{~g}\)

Short answer

After calculations, the mass of AgI that will dissolve in 1.0 L of 1.0 M NH3 is approximately 0 g, which is so small that it rounds to zero, and none of the answer choices corresponds to the calculated value.

Step-by-step solution

Write the complex formation reaction

The formation of the complex ion Ag(NH3)2+ occurs as follows: AgI(s) ⇌ Ag+ (aq) + I− (aq) (Dissolution of AgI)Ag+ (aq) + 2NH3 (aq) → Ag(NH3)2+ (aq) (Formation of complex ion)

Write expressions for the equilibrium constants

Write the solubility product (Ksp) and formation constant (Kf) expressions for the dissolution and formation reactions: Ksp(AgI) = [Ag+][I−]Kf[Ag(NH3)2+] = [Ag+][NH3]^2Given: Ksp(AgI) = 1.5 × 10^−16, Kf[Ag(NH3)2+] = 1.6 × 10^7

Express Ag+ concentration through the complex ion formation

Since the concentration of ammonia is large compared to the silver ion concentration, we'll assume that most Ag+ ions form the complex. Therefore, [Ag(NH3)2+] is given by: [Ag(NH3)2+] ≈ 2[NH3][Ag+] = (2)(1.0 M)[Ag+][Ag+] = [Ag(NH3)2+] / (2)(1.0 M)

Combine the Kf and Ksp expressions

By combining the expressions and assuming [Ag+] is formed primarily from the complex ion: Kf[Ag(NH3)2+] = (Ksp(AgI)) / ([I−])Rearrange to solve for [I−], the iodide ion concentration: [I−] = (Ksp(AgI)) / (Kf[Ag(NH3)2+])Insert the values: [I−] = (1.5 × 10^−16) / (1.6 × 10^7) = 9.375 × 10^−24 M

Calculate the mass of dissolved AgI

With the concentration of iodide ions known, we can calculate the moles of AgI that dissolved: moles of AgI = [I−] × 1.0 L = 9.375 × 10^−24 molesNow, calculate the mass: mass = moles × molar mass of AgI = (9.375 × 10^−24 moles) × (108 g/mol for Ag + 127 g/mol for I) = (9.375 × 10^−24 moles) × (235 g/mol) = 2.2 × 10^−21 g

Key concepts

Complex Ion Formation

Complex ion formation is essential to understanding why some compounds are more soluble in certain solutions than in pure water. Complex ions consist of a metal ion at the center surrounded by molecules or ions, often referred to as ligands. In this context, the ligands are ammonia (\textbf{NH}\(_3\)) molecules, which coordinate to the silver ion (\textbf{Ag}\(^+\)), resulting from the dissolution of silver iodide (\textbf{AgI} in this case).

The process occurs in two steps: the initial dissociation of \textbf{AgI} into \textbf{Ag}\(^+\) and \textbf{I}\(^-\) ions, followed by the silver ion reacting with ammonia to form the complex ion, \textbf{Ag}(\textbf{NH}\(_3\))\(_2^+\). This ion is more stable due to the formation of a coordination bond between the metal ion and the ligands. This coordination leads to an increased solubility of \textbf{AgI} in the presence of ammonia because the equilibrium shifts to form more of the complex ion, thereby reducing the concentration of free \textbf{Ag}\(^+\) ions in solution.

Chemical Equilibrium

Chemical equilibrium is the point at which the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. In the dissolution of \textbf{AgI}, equilibrium is achieved between the solid phase and its dissociated ions. With the introduction of ammonia, a new equilibrium is established for the formation of the complex ion, \textbf{Ag}(\textbf{NH}\(_3\))\(_2^+\), from \textbf{Ag}\(^+\) and \textbf{NH}\(_3\).

Because of this, \textbf{NH}\(_3\) is referred to as a 'ligand' that increases the solubility of \textbf{AgI} by shifting the dissolution equilibrium towards the right. Equilibrium principles, such as Le Châtelier's principle, explain this behavior by stating that when a system at equilibrium is disturbed, it will adjust itself to minimize the disturbance and achieve a new equilibrium state. This explains why the addition of excess \textbf{NH}\(_3\) causes more \textbf{AgI} to dissolve as the equilibrium shifts to reduce the concentration of uncomplexed \textbf{Ag}\(^+\) ions.

Ksp and Kf Expressions

The solubility product constant (\textbf{Ksp}) is a special type of equilibrium constant that measures the propensity of a solid to dissolve into its constituent ions. It is defined for the dissolution of solid compounds in aqueous solutions. \textbf{Ksp} is essentially the product of the concentrations of the ions at equilibrium, each raised to the power of their coefficient in the balanced dissolution equation.

On the other hand, the formation constant (\textbf{Kf}) refers to the equilibrium constant for the formation of a complex ion from its central metal ion and attached ligands. The higher the \textbf{Kf}, the more favorably the complex ion is formed. In the step-by-step solution presented, the relationship between \textbf{Ksp} and \textbf{Kf} is exploited to determine the solubility of \textbf{AgI} in an ammonia solution. It demonstrates that a higher \textbf{Kf} value can effectively 'pull' more of the metal ion (\textbf{Ag}\(^+\)) out of the solid phase, transforming it into a complex ion, which enhances the solubility of the solid in the solution.

This balance between \textbf{Ksp} and \textbf{Kf} is crucial for predicting and understanding the solubility behavior of various ionic compounds, especially in the presence of ligands that can form complex ions.