Problem 61 - Chemistry The Central Science In Si Units

Question

Calculate the solubility of \(\mathrm{Mn}(\mathrm{OH})_{2}\) in grams per liter when buffered at \(\mathrm{pH}\) (a) \(7.0,(\mathbf{b}) 9.5,(\mathbf{c}) 11.8\).

Step-by-Step Solution

Answer

Solubility of Mn(OH)₂ increases with higher pH values: lower at pH 7, higher at pH 11.8 due to [OH⁻] concentration.

Step 1: Write the Dissolution Equation

The dissolution of manganese(II) hydroxide in water can be represented by the equilibrium equation: \[ \text{Mn(OH)}_2 (s) \rightleftharpoons \text{Mn}^{2+} (aq) + 2 \text{OH}^- (aq) \].

Step 2: Express the Solubility Product, Ksp

The solubility product (\( K_{sp} \)) for Mn(OH)₂ can be expressed as: \[ K_{sp} = [\text{Mn}^{2+}][\text{OH}^-]^2 \]. Assume the solubility of Mn(OH)₂ is \( s \), then \([\text{Mn}^{2+}] = s \) and \([\text{OH}^-] = 2s \).

Step 3: pH and pOH Relationship

We know that \( \text{pH} + \text{pOH} = 14 \). Calculate \( \text{pOH} \) from the given \( \text{pH} \) values to find the concentration of \( \text{OH}^- \) ions, \([\text{OH}^-] = 10^{-\text{pOH}} \).

Step 4: Solubility Adjustment for pH 7.0

At \( \text{pH} = 7.0 \), \( \text{pOH} = 14 - 7 = 7 \). Therefore, \([\text{OH}^-] = 10^{-7} \). Using \( K_{sp} = s \times (10^{-7})^2 \), calculate \( s \).

Step 5: Solubility Adjustment for pH 9.5

For \( \text{pH} = 9.5 \), \( \text{pOH} = 14 - 9.5 = 4.5 \). Thus, \([\text{OH}^-] = 10^{-4.5} \). Substitute into the equation: \( K_{sp} = s \times (10^{-4.5})^2 \), and solve for \( s \).

Step 6: Solubility Adjustment for pH 11.8

With \( \text{pH} = 11.8 \), \( \text{pOH} = 14 - 11.8 = 2.2 \). So, \([\text{OH}^-] = 10^{-2.2} \). Use the equation: \( K_{sp} = s \times (10^{-2.2})^2 \), and calculate \( s \).

Step 7: Conversion to Grams per Liter

Convert the solubilities from molarity (mol/L) to grams/L by multiplying \( s \) by the molar mass of \( \text{Mn(OH)}_2 \). The molar mass is \( 54.94 + 2(16.00 + 1.01) = 88.95 \) g/mol. The solubility in grams per liter is \( s \times 88.95 \).

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, abbreviated as Ksp, is a value that represents the solubility of a sparingly soluble compound in water. It quantifies the extent to which a solid will dissolve in water to form a saturated solution.
In simple terms, Ksp is used to tell us how much of the solid will dissolve. A higher Ksp value indicates greater solubility, while a lower Ksp value signifies limited solubility.
For any given compound like manganese(II) hydroxide (Mn(OH)8), we write a dissolution equation to represent it breaking down into its ions:

• Mn(OH)8(s) 1 Mn2+ (aq) + 2 OH7 (aq)

The Ksp expression is then formulated based on these ions. For Mn(OH)8, the expression is:

• \[ K_{sp} = [\text{Mn}^{2+}][\text{OH}^-]^2 \]

Here, the concentration of OH^- is squared because there are two OH^- ions for each formula unit of Mn(OH)8. Knowing the Ksp allows us to calculate how the solubility adjusts under different conditions, like changes in pH, which are fundamental in solubility calculations.

Acid-Base Chemistry

Acid-base chemistry plays a significant role in understanding solubility, especially when dealing with substances that produce or consume H+ and OH- ions. The pH of a solution is a measure of its acidity or basicity.
The pH scale ranges from 0 to 14, where:

• A pH less than 7 indicates an acidic solution.
• A pH greater than 7 means the solution is basic.
• A pH of 7 is neutral.

The relationship between pH and pOH is expressed as:

• \[ \text{pH} + \text{pOH} = 14 \]

This relationship helps us calculate the concentration of hydroxide ions [OH^-], especially in solutions where an acid or base affects solubility.
The concentration of [OH^-] is determined as \(10^{-\text{pOH}}\). As pH changes, it alters the concentrations of the ions in the solution, subsequently affecting the solubility of compounds like Mn(OH)8. Thus, understanding acid-base chemistry is crucial in adjusting solubility for different pH levels.

Equilibrium Reactions

Equilibrium reactions occur when a reversible chemical reaction reaches a state where the forward and reverse reaction rates are equal. This means no net change in the concentrations of reactants and products.
In the context of solubility, equilibrium reactions describe the balance between a solid and its dissolved ions in solution:

• Mn(OH)8(s) \(\rightleftharpoons\) Mn2+ (aq) + 2 OH7 (aq)

At equilibrium, the equilibrium expression, or Ksp in this case, quantifies the product of the concentrations of the ions raised to their stoichiometric coefficients.
This helps in understanding how the balance shifts when conditions such as concentration, temperature, or pH change. If external conditions change, Le Chatelier's principle predicts how the equilibrium will shift to accommodate this change, often impacting solubility.
Evaluating equilibrium reactions in solubility is essential for calculating exact solubility values, particularly when additional factors, like common ions, can shift the equilibrium.