Problem 109 - Chemistry The Central Science

Question

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14} .\) (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?\) \((\mathbf{b} ) \)The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

Step-by-Step Solution

Answer

(a) The molar solubility of Cd(OH)₂ without NaBr solution is \(s = 9.47 \times 10^{-6} \mathrm{mol} / \mathrm{L}\). (b) The initial concentration of NaBr needed to increase the molar solubility of Cd(OH)₂ to \(1.0 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) is \(x = 6.68 \times 10^{-2} \mathrm{mol} / \mathrm{L}\).

Step 1: (a) Finding the molar solubility of Cd(OH)₂ without NaBr solution

First, let's write the solubility equilibrium equation for Cd(OH)₂: Cd(OH)₂(s) ⇌ Cd²⁺(aq) + 2OH⁻(aq) Now, let's denote the molar solubility of Cd(OH)₂ as 's'. Therefore, the concentration of Cd²⁺ will be 's' and the concentration of OH⁻ will be '2s' in the solution. The solubility product constant (Kₛₚ) is given by: Kₛₚ = [Cd²⁺][OH⁻]² We have the value of Kₛₚ as 2.5 × 10⁻¹⁴. Substituting the values, we get: \(2.5 \times 10^{-14} = s \times (2s)^2\) Now, solve this equation for 's' to find the molar solubility of Cd(OH)₂.

Step 2: (b) Finding the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)₂

To find the initial concentration of NaBr, we need to use the complex formation constant (Kf) for the complex ion CdBr₄²⁻. The complex formation reaction can be written as: Cd²⁺(aq) + 4Br⁻(aq) ⇌ CdBr₄²⁻(aq) Now, let's denote the initial concentration of NaBr required as 'x'. Since NaBr is a strong electrolyte, it will completely dissociate in water and will provide 'x' moles of Br⁻ per liter. The final concentration of Br⁻ should be 4 times the increased molar solubility. Now, we can write the expression for the complex formation constant (Kf) as: Kf = [CdBr₄²⁻] / ([Cd²⁺][Br⁻]⁴) We have the value of Kf as 5 × 10³ and are given that the increased molar solubility of Cd(OH)₂ (concentration of Cd²⁺) is 1.0 × 10⁻³ mol/L, which can be written in terms of 'x': Kf = (1.0 × 10⁻³) / ((1.0 × 10⁻³)(x⁴)) Now, solve this equation for 'x' to find the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)₂.

Key Concepts

Ksp

The concept of the solubility product constant, often denoted as \(K_{sp}\), is crucial for understanding solubility equilibrium in ionic compounds. \(K_{sp}\) refers to the equilibrium constant for a solid substance dissolving in an aqueous solution. When an ionic compound dissolves, it dissociates into its constituent ions. For \[ \text{Cd(OH)}_2 \] it dissociates as follows:

• \[ \text{Cd(OH)}_2(s) \rightleftharpoons \text{Cd}^{2+}(aq) + 2\text{OH}^{-}(aq) \]

Here, the product of the concentrations of the ions, each raised to the power of its coefficient in the balanced equation, gives you the \(K_{sp}\). For \[ \text{Cd(OH)}_2, \]\(K_{sp} = [\text{Cd}^{2+}][\text{OH}^{-}]^2\).This value, \(2.5 \times 10^{-14}\), is very tiny, indicating that \[ \text{Cd(OH)}_2 \] is not very soluble. The primary use of this constant is to help calculate the molar solubility of a compound, as shown in step 1 of the solution process.

Complex ion formation

Complex ion formation is a process where metal ions form a complex with certain ligands. In this context, \(\text{Cd}^{2+}\) ions combine with \(\text{Br}^-\) ions to form the complex ion \(\text{CdBr}_4^{2-}\). This greatly affects the solubility of the original compound because the formation of the complex ion reduces the concentration of free metal ions in solution, effectively increasing the solubility of the salt.In the context of \(\text{Cd(OH)}_2\), when this compound is exposed to an excess of \(\text{Br}^-\) ions from NaBr, it can form a stable complex given by the equation:

• \[ \text{Cd}^{2+}(aq) + 4\text{Br}^- (aq) \rightleftharpoons \text{CdBr}_4^{2-} (aq) \]

The formation of this complex ion \(\text{CdBr}_4^{2-}\) hence increases the solubility of \(\text{Cd(OH)}_2\) by shifting the equilibrium of the dissolution reaction. This is an essential concept when considering how different ions and chemicals might affect the solubility of compounds in various scenarios.

Molar solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution until the solution becomes saturated. It's an important concept allowing us to quantify how much of a compound can dissolve in solution, given its \(K_{sp}\). For \(\text{Cd(OH)}_2\), the molar solubility is the concentration of \(\text{Cd}^{2+}\) ions, 's', when equilibrium is reached:

• \[2.5 \times 10^{-14} = s \times (2s)^2\] simplifies to \(s = \sqrt[3]{\frac{2.5 \times 10^{-14}}{4}}\)

This calculates the molar solubility without any complex formation.However, the presence of complex ions can change this solubility drastically. As in the case above, adding \(\text{NaBr}\) leads to complex ion formation, thereby increasing the molar solubility significantly to \(1.0 \times 10^{-3} \text{ mol/L}\). Knowing how to calculate and adjust molar solubility is useful in various chemical applications and is key to handling reactions in the lab.

Complex ion stability constant (Kf)

The stability constant of a complex ion, known as \(K_f\), reflects the stability of the complex ion in solution. Specifically, it measures the propensity of the complex to form from the central metal ion and its ligands. A larger \(K_f\) value indicates a more stable complex.For the complex ion \(\text{CdBr}_4^{2-}\), the equation:

• \[ K_f = \frac{[\text{CdBr}_4^{2-}]}{[\text{Cd}^{2+}][\text{Br}^-]^4} \]

Utilizes the given \(K_f\) value of \(5 \times 10^3\), showing how increased concentrations of \(\text{Br}^-\) (provided by NaBr) push the formation of more \(\text{CdBr}_4^{2-}\), thus enhancing solubility.Understanding \(K_f\) is essential for predicting and manipulating the solubility of salts in complex solutions. When designing experiments or in industrial processes, knowing the stability constants allows for precise control of compound dissolution and reactions.