Question

Use values of \(K_{s p}\) for AgI and \(K_{f}\) for \(A g(C N)_{2}^{-}\) to (a) calculate the molar solubility of Agl in pure water, (b) calculate the equilibrium constant for the reaction \(\operatorname{AgI}(s)+2 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{I}^{-}(a q), \quad(\mathbf{c})\) determine the molar solubility of AgI in a 0.100 \(\mathrm{MNaCN}\) solution.

Short answer

\(S_{water} = \sqrt{K_{sp}}\) \(K = \frac{K_{sp}}{K_{f}}\) \(K = \frac{[Ag(CN)_{2}^{-}][I^{-}]}{[Ag^{+}][CN^{-}]^{2}}\) Molar solubility of AgI in 0.100 M NaCN solution: ICE table concentrations: [Ag\(^\oplus\)] = S, [CN\(^{-}\)] = 0.100 - 2S, [Ag(CN)\(_2^-\)] = [I\(^-\)] = S. Plug in these values into the equilibrium expression, and solve for S representing the molar solubility of AgI in the 0.100 M NaCN solution.

Step-by-step solution

Molar Solubility of AgI in Pure Water

Using the given Ksp for AgI, we can set up an expression relating the concentration of Ag+ and I- to the solubility product constant: Ksp = [Ag+][I-] Since the molar solubility of AgI in water is S, the concentrations of Ag+ and I- are also S. The equation becomes: Ksp = (S)(S) = S^2 Now solve for S.

Equilibrium Constant for the Reaction

Using the given Kf for Ag(CN)2-, we can set up an expression relating the concentration of Ag+, CN-, and Ag(CN)2- to the formation constant: Kf = [Ag(CN)2-]/([Ag+][CN-]^2) Now, we can use the previously determined value for S to calculate the equilibrium constant for the reaction: K = Ksp/Kf

Molar Solubility of AgI in 0.100 M NaCN Solution

In a 0.100 M NaCN solution, consider the initial concentrations of Ag+, CN-, and I-. Initially, Ag+ and I- are at 0, and CN- is at 0.100 M. We know that for every Ag+ that dissolves, one I- is also released. However, we also consider the reaction with CN-: AgI(s) + 2 CN-(aq) ⇌ Ag(CN)2-(aq) + I-(aq) The reaction consumes Ag+ and 2 CN- ions to produce 1 Ag(CN)2- ion and a I- ion. To find the molar solubility of AgI in a 0.100 M NaCN solution, we can create an ICE table for the reaction: Initial: 0 0.100 0 0 Change: +S -2S +S +S Equilibrium: S 0.100-2S S S Using the equilibrium constant calculated in step 2: K = ([Ag(CN)2-][I-])/([Ag+][CN-]^2) Replace the concentrations with the values in the equilibrium row and solve for S, which represents the molar solubility of AgI in a 0.100 M NaCN solution.

Key concepts

Solubility Product Constant

In chemistry, the solubility product constant, often denoted as Ksp, is a special type of equilibrium constant that measures the solubility of a sparingly soluble ionic compound. It quantifies how much of a compound can dissolve in a solution to reach a state of dynamic equilibrium, where the rate of dissolution equals the rate of precipitation.

The Ksp value is specific to each ionic compound at a given temperature and is typically represented by the concentrations of the ions in their stoichiometric ratios. For example, for a generic salt AB that dissociates into A+ and B- ions, the Ksp expression is:
Ksp = [A+][B-]
This relation indicates that as the product of the concentrations of the ions increases, so does the extent of their solubility in the solution, up to the limit defined by the Ksp value. For compounds with multiple ions, such as A2B3, the expression would account for the stoichiometry: Ksp = [A+]^2[B-]^3.

Ksp

Ksp, or the solubility product constant, can be a tool to predict whether a precipitation reaction will occur. A solution is considered saturated when it contains the maximum concentration of ions that can be present without forming a precipitate under set conditions. When the product of the concentrations of the ions exceeds the Ksp, the solution becomes supersaturated, and precipitation occurs to restore the balance.

In practice, to determine the molar solubility of a substance like AgI (silver iodide) in water, you would set up an equilibrium equation using Ksp:Ksp = [Ag+][I-]Since AgI dissociates into Ag+ and I- in a 1:1 molar ratio, their concentrations at equilibrium are equal, and the Ksp expression simplifies to the square of the molar solubility (S): Ksp = S^2.

Equilibrium Constant

The equilibrium constant, denoted as K, describes the balance between the reactants and products in a reversible chemical reaction at equilibrium. It's determined by the temperature and is independent of the initial concentrations of reactants and products. In the context of solubility, it helps to ascertain the extent of a reaction and predict which way the equilibrium will shift in response to changes in conditions, such as concentration or temperature.

For a generic reaction where aA + bB ⇌ cC + dD, the equilibrium constant expression is:
K = [C]^c[D]^d / [A]^a[B]^b
For example, the equilibrium constant for AgI dissolving in a solution with CN- ions involves both Ksp and Kf, the formation constant for the complex ion Ag(CN)2-. The overall equilibrium constant for the reaction is calculated by dividing Ksp by Kf: K = Ksp / Kf.

Complex Ion Formation

Complex ion formation refers to the process where simple ions combine to form a larger, coordinated complex. These complexes are unique as they can have very high formation constants (Kf), which indicates they are highly stable in solution. In the context of silver iodide (AgI), when the CN- ions are introduced, a complex ion Ag(CN)2- is formed:

Ag+ + 2CN- ⇌ Ag(CN)2-

This reaction is important in the study of solubility because the creation of a complex ion can significantly increase the solubility of an otherwise insoluble salt. For example, the interaction between AgI and CN- ions in solution changes the equilibrium, resulting in a higher solubility of AgI due to the removal of Ag+ ions from the solution, which shifts the equilibrium towards more dissolution of AgI to form the stable complex ion.

ICE Table

An ICE table, which stands for Initial, Change, and Equilibrium, is a tabular method used to keep track of the changes in concentrations or partial pressures of species in a chemical reaction. It's extremely useful for solving equilibrium problems.

The ICE table organizes the reaction's initial amounts, the changes that occur as the system moves towards equilibrium, and the final equilibrium concentrations. For example, let's consider the equilibrium process of AgI dissolving in the presence of CN- using an ICE table:

Initial Concentrations:

0 (Ag+), 0 (I-), 0.100 M (CN-)

Change in Concentrations:

+S (Ag+), +S (I-), -2S (CN-)

Equilibrium Concentrations:

S (Ag+), S (I-), 0.100 M - 2S (CN-)
Using the expressions for Ksp and the equilibrium constant of the complex ion formation, we can solve for S, which is the molar solubility of AgI in a CN- containing solution. The ICE table simplifies the complex calculations involved in finding the equilibria of chemical reactions.