Problem 56 - Chemistry: The Central Science

Question

Calculate the molar solubility of \(\mathrm{Fe}(\mathrm{OH})_{2}\) when buffered at \(\mathrm{pH}\) (a) \(8.0\), (b) \(10.0\), (c) \(12.0\)

Step-by-Step Solution

Answer

The molar solubility of \(\mathrm{Fe}(\mathrm{OH})_{2}\) when buffered at different pH values is: (a) pH = 8.0: \(1.13 \times 10^{-5}\, \mathrm{M}\) (b) pH = 10.0: \(2.64 \times 10^{-7}\, \mathrm{M}\) (c) pH = 12.0: \(7.07 \times 10^{-10}\, \mathrm{M}\)

Step 1: Write the dissolution equation for \(\mathrm{Fe}(\mathrm{OH})_{2}\)

The dissolution of \(\mathrm{Fe}(\mathrm{OH})_{2}\) in water is given by the following balanced chemical equation: \[ \mathrm{Fe}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Fe}^{2+}(aq) + 2\mathrm{OH}^-(aq) \]

Step 2: Use given pH values to find the \(\mathrm{OH}^-\) concentrations

To find the concentration of \(\mathrm{OH}^-\) ions at the given pH values, we will use the relationship between pH, pOH and the concentration of \(\mathrm{OH}^-\) ions: \[ \mathrm{pOH} = -\log[\mathrm{OH}^-] \] Since \(\mathrm{pH} + \mathrm{pOH} = 14\), we can find the pOH for each case, and then the concentration of \(\mathrm{OH}^-\) ions: (a) For pH = 8.0: \[ \mathrm{pOH} = 14 - 8 = 6 \Rightarrow [\mathrm{OH}^-] = 10^{-6} \,\mathrm{M}\] (b) For pH = 10.0: \[ \mathrm{pOH} = 14 - 10 = 4 \Rightarrow [\mathrm{OH}^-] = 10^{-4} \,\mathrm{M}\] (c) For pH = 12.0: \[ \mathrm{pOH} = 14 - 12 = 2 \Rightarrow [\mathrm{OH}^-] = 10^{-2} \,\mathrm{M}\]

Step 3: Use the Ksp of \(\mathrm{Fe}(\mathrm{OH})_{2}\) to find molar solubility

The solubility product constant (Ksp) for \(\mathrm{Fe}(\mathrm{OH})_{2}\) is \(\approx 1.78 \times 10^{-15}\). Using the dissolution equation, we set up the following Ksp expression: \[ K_{sp} = [\mathrm{Fe}^{2+}][\mathrm{OH}^-]^2 \] We let the molar solubility of \(\mathrm{Fe}(\mathrm{OH})_{2}\) be \(S\). Therefore, \([\mathrm{Fe}^{2+}] = S\) and \([\mathrm{OH}^-] = 2S + [\mathrm{OH}^-]_{buffer}\), where \([\mathrm{OH}^-]_{buffer}\) is the \(\mathrm{OH}^-\) concentration due to the buffer at a given pH. Now, we can calculate the molar solubility (\(S\)) at each pH value: (a) For pH = 8.0: \[ K_{sp} = S(2S + 10^{-6})^2 \Rightarrow S = 1.13 \times 10^{-5} \,\mathrm{M} \] (b) For pH = 10.0: \[ K_{sp} = S(2S + 10^{-4})^2 \Rightarrow S = 2.64 \times 10^{-7} \,\mathrm{M} \] (c) For pH = 12.0: \[ K_{sp} = S(2S + 10^{-2})^2 \Rightarrow S = 7.07 \times 10^{-10} \,\mathrm{M} \] Therefore, the molar solubility of \(\mathrm{Fe}(\mathrm{OH})_{2}\) at: (a) pH = 8.0 is \(1.13 \times 10^{-5}\, \mathrm{M}\) (b) pH = 10.0 is \(2.64 \times 10^{-7}\, \mathrm{M}\) (c) pH = 12.0 is \(7.07 \times 10^{-10}\, \mathrm{M}\)

Key Concepts

Solubility Product Constant

The solubility product constant, Ksp, is a fundamental concept in chemistry that describes how a compound dissolves in a solution. It's especially pertinent when dealing with sparingly soluble salts like \textbf{Fe(OH)2}. When a salt's solubility is low, it dissociates into its constituent ions to a limited extent, and an equilibrium is established between the undissolved solid and the ions in solution.

The Ksp is the product of these ions' molar concentrations, each raised to the power of its stoichiometric coefficient in the dissolution equation. For instance, \textbf{Fe(OH)2} dissociates into one Fe2+ ion and two OH- ions, its Ksp expression becomes \[ K_{sp} = [\mathrm{Fe}^{2+}][\mathrm{OH}^-]^2 \] Understanding this constant is crucial for predicting the extent of a salt's solubility under varying conditions, such as different pH levels as seen in the given exercise.

Dissolution Equation

The dissolution equation illustrates how a compound, like \textbf{Fe(OH)2}, separates into its ionic components in a solution. Writing this equation is the first step to understanding the solubility behavior of the compound. The dissolution equation for \textbf{Fe(OH)2} is represented as \[ \mathrm{Fe(OH)_2(s)} \rightleftharpoons \mathrm{Fe^{2+}(aq)} + 2\mathrm{OH^{-}(aq)} \] It indicates that solid \textbf{Fe(OH)2} dissociates into one Fe2+ ion and two OH- ions. For each molecule of \textbf{Fe(OH)2} that dissolves, we obtain one Fe2+ ion and double the amount of OH- ions in the solution, which is critical for calculating solubility especially in the presence of additional sources of ions like a buffer solution.

pH and pOH Relationship

The relationship between pH and pOH is essential to comprehend the acidity or basicity of a solution. pH is a measure of the hydrogen ion concentration, whereas pOH quantifies the hydroxide ion concentration. In a neutral solution at 25°C, pH and pOH always add up to 14 due to the ion product constant for water \(K_w = [H^+][OH^-] = 1 \times 10^{-14}\text{ at 25}°C\text{M}\).

The formula \[ \mathrm{pOH} = -\log[\mathrm{OH}^-] \] allows us to convert between pOH and the concentration of OH- ions. If the pH is known, we can find the pOH by subtracting the pH from 14 and then find the hydroxide ion concentration, which is crucial for solubility calculations of compounds affected by pH, like \textbf{Fe(OH)2}.

For example, at a pH of 8.0, corresponding pOH would be 6, suggesting a hydroxide ion concentration of \[ 10^{-6} \text{M} \]. This interconnection plays a significant role in determining the solubility of hydroxides in different pH conditions.

Concentration Calculations

Concentration calculations are vital for determining how much of a substance is present in a given volume of solution, which is directly related to its molarity (M). Molarity is defined as moles of solute per liter of solution. To find the molar solubility of a compound like \textbf{Fe(OH)2}, we combine its dissolution equation, the Ksp value, and the pH or pOH of the solution.

In the context of our exercise, we account for the \textbf{Fe(OH)2} that dissolves and the hydroxide ions from the buffer. By expressing the molar solubility of \textbf{Fe(OH)2} as \(S\) and the concentration of hydroxide ions from the buffer as \( [\mathrm{OH}^-]_{buffer} \), we can insert these values into the Ksp expression to ascertain the solubility under different pH conditions. The process involves setting up and solving a quadratic equation, as demonstrated in the solution steps for different pH levels, allowing us to understand how solubility changes in response to acidity or basicity of the environment.