Problem 77 - Chemistry

Question

\(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7} . K_{\mathrm{sp}}\) for \(\mathrm{AgCl}\) is \(1.6 \times 10^{-10} .\) Calculate the molar solubility of \(\mathrm{AgCl}\) in \(1.0M\) \(\mathrm{NH}_{3}.\)

Step-by-Step Solution

Answer

The molar solubility of AgCl in 1.0M NH3 solution is approximately \(1.2 \times 10^{-4} M\).

Step 1: Write the balanced chemical equations

Write the balanced chemical equation for the formation of the complex ion Ag(NH3)2+ from Ag+ and NH3, and also the dissolution of AgCl. Formation of Ag(NH3)2+ complex ion: \[Ag^+ + 2NH_3 \rightleftharpoons Ag(NH_3)_2^+\] Dissolution of AgCl: \[AgCl(s) \rightleftharpoons Ag^+ + Cl^-\]

Step 2: Write the expressions for Kf and Ksp

Write the expressions for the formation constant (Kf) and the solubility product constant (Ksp) for the given reactions. For the formation of Ag(NH3)2+ complex ion: \[K_f = [\ce{Ag(NH3)2^+}] / ([\ce{Ag^+}] [\ce{NH3}]^2)\] For the dissolution of AgCl: \[K_{sp} = [\ce{Ag^+}] [\ce{Cl^-}]\]

Step 3: Set up and solve the equation relating Ksp and molar solubility

Let the molar solubility of \(\ce{AgCl}\) in \(\ce{NH3}\) be x. From the balanced chemical equation of AgCl dissolution, we can set up the equilibrium concentrations: \[[\ce{Ag^+}] = [\ce{Cl^-}] = x\] Plug these equilibrium concentrations into the Ksp expression: \[K_{sp} = x^2\] Now, solve for x: \[x = \sqrt{K_{sp}}\]

Step 4: Set up and solve the equation relating Kf and equilibrium concentrations

Since the initial concentration of NH3 is 1.0M, after reaction with Ag+, the equilibrium concentration of NH3 will be: \[[\ce{NH3}] = 1.0 - 2x\] Since Ag+ reacts with NH3 to form Ag(NH3)2+, equilibrium concentrations of Ag+ and Ag(NH3)2+ are related as: \[[\ce{Ag(NH3)2^+}] = x\] Now, plug these equilibrium concentrations into the Kf expression: \[K_f = x / (x (1.0 - 2x)^2)\] Solve for x: \[x = \sqrt[3]{\cfrac{K_f}{(1.0 - 2x)^2}}\]

Step 5: Combine the equations and solve for molar solubility

Combine the equations obtained in Step 3 and Step 4 and substitute the given values for Kf and Ksp: \[x = \sqrt[3]{\cfrac{1.7 \times 10^{7}}{(\ 1.0 - 2 \sqrt{1.6 \times 10^{-10}}\ )^2}}\] Now, solve for x, which is the molar solubility of AgCl in 1.0M NH3: \[x \approx 1.2 \times 10^{-4} M\] The molar solubility of AgCl in 1.0M NH3 solution is approximately \(1.2 \times 10^{-4} M\).

Key Concepts

Formation Constant (Kf)

When discussing complex ions, one of the essential concepts is the formation constant, denoted as \( K_f \). A formation constant represents the stability of a complex ion in solution. It quantifies the tendency of a metal ion to bind with ligands to form a complex.
For a reaction involving the formation of a complex ion, the larger the \( K_f \) value, the more stable the complex.
Consider the reaction for forming the complex ion \( \mathrm{Ag(NH_3)_2^+} \), where silver ions \( \mathrm{Ag^+} \) interact with ammonia \( \mathrm{NH_3} \):

• Silver ions and ammonia combine to produce \( \mathrm{Ag(NH_3)_2^+} \).
• The equilibrium expression is: \[K_f = \frac{[\mathrm{Ag(NH_3)_2^+}]}{[\mathrm{Ag^+}][\mathrm{NH_3}]^2}\]

A high formation constant, such as \( 1.7 \times 10^7 \), indicates that the complex ion is highly favored in the solution.

Solubility Product Constant (Ksp)

The solubility product constant, \( K_{sp} \), is crucial for understanding the solubility of ionic compounds in water. It describes the equilibrium between the ionic compound in its solid state and its constituent ions in solution.
For a compound such as silver chloride \( \mathrm{AgCl} \), which dissolves in water to form silver ions \( \mathrm{Ag^+} \) and chloride ions \( \mathrm{Cl^-} \), the process is represented as follows:

• \( \mathrm{AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq)} \)
• The equilibrium expression for \( K_{sp} \) is: \[K_{sp} = [\mathrm{Ag^+}][\mathrm{Cl^-}]\]

The \( K_{sp} \) helps predict how much of a compound can dissolve before reaching its maximum solubility limit.
For AgCl, a small \( K_{sp} \) such as \( 1.6 \times 10^{-10} \) suggests that it is only slightly soluble in water.

Molar Solubility

Molar solubility refers to the number of moles of a substance that can dissolve in a liter of solution to reach saturation. It's a practical measure derived from the \( K_{sp} \) when an insoluble salt dissolves sparingly in the solvent.
To find the molar solubility of \( \mathrm{AgCl} \) in \( 1.0 \mathrm{M} \) \( \mathrm{NH_3} \), note the chemical process:

• \( \mathrm{AgCl} \) dissociates into \( \mathrm{Ag^+} \) and \( \mathrm{Cl^-} \) ions.
• The molar solubility \( x \) is calculated from: \[x^2 = K_{sp}\]
• With \( K_{sp} = 1.6 \times 10^{-10} \), solve \( x = \sqrt{1.6 \times 10^{-10}} \) for \( x \).

In the presence of \( \mathrm{NH_3} \), silver ions react to form the \( \mathrm{Ag(NH_3)_2^+} \) complex, which increases the apparent molar solubility due to complex ion formation stabilization.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, leading to constant concentrations of reactants and products. It is a fundamental principle in reactions involving complex ions and solubility.For the dissolution of \( \mathrm{AgCl} \) and the formation of \( \mathrm{Ag(NH_3)_2^+} \):

• Two equilibria are considered: the solubility equilibrium \( K_{sp} \) and the formation equilibrium \( K_f \).
• At equilibrium, concentrations of \( \mathrm{Ag^+} \), \( \mathrm{Cl^-} \), \( \mathrm{NH_3} \), and \( \mathrm{Ag(NH_3)_2^+} \) remain stable.
• The equations \( K_{sp} \) and \( K_f \) allow you to solve for unknown concentrations using initial conditions and equilibrium expressions.

This problem exemplifies how equilibria in complex ion formation and solubility interact to influence molar solubility in solutions, providing a comprehensive understanding of solution chemistry.