Problem 81 - General Chemistry: Principles and Modern Applications

Question

The solubility of \(\mathrm{CdCO}_{3}(\mathrm{s})\) in \(1.00 \mathrm{M} \mathrm{KI}(\mathrm{aq})\) is \(1.2 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Given that \(K_{\mathrm{sp}}\) of \(\mathrm{CdCO}_{3}\) is \(5.2 \times 10^{-12},\) what is \(K_{\mathrm{f}}\) for \(\left[\mathrm{CdI}_{4}\right]^{2-} ?\)

Step-by-Step Solution

Answer

The formation constant (\(K_{f}\)) for [\CdI4]^{2-} is \((1.2 \times 10^{-3})(4)^4 = 1.22 \times 10^{1}\).

Step 1: Given

The solubility of CdCO3, \(s\) is given as \(1.2 \times 10^{-3}\) mol/L. Also, the \(K_{sp}\) of CdCO3 is given, which is \(5.2 \times 10^{-12}\). We are to find \(K_{f}\) for [CdI4]2-.

Step 2: Express solubility equilibrium

The solubility equilibrium and constant expression for \(CdCO3{s})\ are: \CdCO3 \(s\) \( \rightleftharpoons \) Cd2+ \(aq\) + CO3 2- \(aq\), \( K_{sp} \) = \[ [Cd2+][CO3 2-] \] . The solubility is the same as the concentration of Cd2+ ions, so the \(K_{sp}\) expression can be written as \(5.2 \times 10^{-12}\ = s^2\).

Step 3: Solve for s

Solving the equation for s, you get \(s = sqrt{5.2 \times 10^{-12}}\). Given that the solubility of \(CdCO3{s}\) in \(1.00 M (KI(aq))\) is \(1.2 \times 10^{-3} Mol/L\), this must be the concentration of \(Cd2+\) ions in solution.

Step 4: Express formation equilibrium

The formation equilibrium and constant expression for [CdI4]2- are: \( Cd2+ (aq) + 4I- (aq) \rightleftharpoons [CdI4]^{2-} (aq)\), \(K_f\) = \[ [\CdI4]^{2-}/[Cd2+] [I-]^4] \]. Considering 4I- ions come from 1KI molecule, the concentration of I- ions is 4 M.

Step 5: Solve for \(K_{f}\)

Rearranging the expression for \(K_{f}\) to be expressed in terms of given quantites gives \(K_{f} = [\CdI4]^{2-}/[Cd2+] [I-]^4 = [\s] [I-]^4. Substituting the known values, \(K_{f} = (1.2 \times 10^{-3}) (4)^4 \), yielding the answer.

Key Concepts

Ksp (Solubility Product Constant)

Understanding the concept of the solubility product constant, or Ksp, is essential for students tackling solubility equilibrium problems. Ksp is a special type of equilibrium constant used for sparingly soluble salts. It helps us predict whether a salt will precipitate out of solution and to what extent it will dissolve.

The Ksp value gives us a quantitative measure of the solubility of a compound. It is defined as the product of the concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced equation. Let's simplify this with an example:
For a generic salt AB having the dissolution reaction: \[ AB_{(s)} \rightleftharpoons A^{+}_{(aq)} + B^{-}_{(aq)} \],the Ksp would be: \[ K_{sp} = [A^{+}][B^{-}] \].This Ksp value doesn't change unless the temperature changes. It is unique for each compound and is known under standard conditions. A low Ksp value indicates a less soluble salt, while a higher value means greater solubility. When solving solubility problems, it is helpful to start by writing out the dissolution reaction and the expression for Ksp to clearly understand the relation between the solubility and the ion concentrations.

Complex Ion Formation

Complex ion formation is a reaction where simple ions combine to form a charged species consisting of a central metal atom or ion bonded to one or more ligands. These ligands are molecules or ions that can donate at least one pair of electrons to the metal center. The formation of a complex ion in a solution can significantly affect the solubility of certain compounds.

A classic example involves the reaction of metal ions in solution with ammonia, halide ions, or other Lewis bases to form coordination compounds. For instance, if a metal ion, M, forms a complex with ligand L in the reaction: \[ M^{n+} + xL \rightleftharpoons ML_x^{(n-x)+} \],the formation constant (Kf) can be applied, which is defined as: \[ K_f = \frac{[ML_x^{(n-x)+}]}{[M^{n+}][L]^x} \].In equilibrium calculations, the stability of the complex ion is an essential factor. A higher Kf indicates a more stable complex ion, which can also lead to a higher solubility of the metal ion in the solution, due to the Le Chatelier's principle. Consequently, when ligands are added to a solution of metal ions, the equilibrium may shift to form more complex ions, thereby altering the overall solubility of salts.

Equilibrium Constant (Kf)

The formation of a complex ion is governed by the equilibrium constant (Kf), often referred to as the formation constant or stability constant. Kf specifically quantifies the formation of complex ions in solution. It is a crucial part of solubility equilibrium problems involving complex ion formation because it indicates the favorability of complex formation.

For a general complex formation reaction: \[ M^{n+} + xL \rightleftharpoons ML_x^{(n-x)+} \],the equilibrium constant (Kf) can be represented by: \[ K_f = \frac{[ML_x^{(n-x)+}]}{[M^{n+}][L]^x} \].The magnitude of the Kf is quite informative. A high Kf suggests a strong attraction between the metal ion and the ligands, leading to a highly stable complex ion. Such complexes tend to pull more metal ions out of the solid phase into the solution as the system strives to reach equilibrium, effectively increasing the solubility of the metal ion. The concept of Kf is essential in understanding how metal ions behave in various environmental and biological contexts and is closely associated with the principles of Le Chatelier's principle and the solubility product (Ksp).