Problem 69 - Chemistry The Molecular Science

Question

Iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2},\) has a solubility in water of \(6.0 \times 10^{-1} \mathrm{mg} / \mathrm{L}\) at a given temperature. (a) From this solubility, calculate the \(K_{\mathrm{sp}}\) of iron(II) hydroxide. Explain why the calculated \(K_{\mathrm{sp}}\) differs from the experimental value of \(8.0 \times 10^{-16}\) (b) Calculate the hydroxide concentration needed to precipitate \(\mathrm{Fe}^{2+}\) ions such that no more than \(1.0 \mu \mathrm{g} \mathrm{Fe}^{2+}\) per liter remains in the solution.

Step-by-Step Solution

Answer

(a) Calculated \(K_{sp}\) is \(1.20 \times 10^{-15}\), differing due to experimental variables. (b) Maximum \([\text{OH}^-] = 6.69 \times 10^{-4} \text{ mol/L}\).

Step 1: Determine Molar Solubility from Given Solubility

Convert the given solubility from mg/L to mol/L. The molar mass of Fe(OH)₂ is approximately 89.9 g/mol. Thus, \[\text{Solubility in mol/L} = \frac{6.0 \times 10^{-1} \text{ mg/L}}{1000 \text{ mg/g} \times 89.9 \text{ g/mol}} = 6.68 \times 10^{-6} \text{ mol/L}.\]

Step 2: Write Dissolution Equation and Expression for K_sp

The dissolution of Fe(OH)₂ in water is given by:\[\text{Fe(OH)}_2 (s) \rightleftharpoons \text{Fe}^{2+} (aq) + 2 \text{OH}^- (aq)\]The solubility product constant expression \(K_{sp}\) is:\[K_{sp} = [\text{Fe}^{2+}][\text{OH}^-]^2\]

Step 3: Calculate Ion Concentrations at Equilibrium

The molar solubility is 6.68 \times 10^{-6} mol/L, which is also the concentration of \(\text{Fe}^{2+}\) ions. The concentration of \(\text{OH}^-\) ions will be twice that, as each formula unit of Fe(OH)₂ produces two OH⁻ ions.\[[\text{OH}^-] = 2 \times 6.68 \times 10^{-6} = 1.34 \times 10^{-5} \text{ mol/L}\]

Step 4: Calculate K_sp

Substitute the ion concentrations into the expression for \(K_{sp}\):\[K_{sp} = (6.68 \times 10^{-6})(1.34 \times 10^{-5})^2\]Calculate the \(K_{sp}\):\[K_{sp} = 1.20 \times 10^{-15}\]

Step 5: Compare to Experimental Value

The calculated \(K_{sp}\) value is higher than the experimental \(K_{sp}\) value of \(8.0 \times 10^{-16}\). This discrepancy can arise from experimental errors, differences in temperature, supersaturation effects, or other environmental factors.

Step 6: Calculate Required Hydroxide Concentration for Precipitation

For no more than 1.0 µg of Fe²⁺ per liter, convert µg to mol:\[1.0 \text{µg/L} \times \frac{1 \text{ mg}}{1000 \text{µg}} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{1}{55.85 \text{ g/mol}} = 1.79 \times 10^{-8} \text{ mol/L}\]Set up the inequality using \(K_{sp}\):\[1.79 \times 10^{-8} \times [\text{OH}^-]^2 \leq 8.0 \times 10^{-16}\]Solve for \([\text{OH}^-]\):\[[\text{OH}^-]^2 \leq \frac{8.0 \times 10^{-16}}{1.79 \times 10^{-8}} = 4.47 \times 10^{-8}\]\[[\text{OH}^-] \leq 6.69 \times 10^{-4} \text{ mol/L}\]

Key Concepts

Solubility Product

The solubility product, abbreviated as \(K_{sp}\), is a crucial concept in chemistry that describes the solubility of sparingly soluble ionic compounds. It is especially relevant in calculating the solubility of compounds like iron(II) hydroxide, \(\text{Fe(OH)}_2\), in water.
To understand \(K_{sp}\), think of it as a mathematical representation of the solubility equilibrium of a dissolved ionic compound. It is calculated using the concentrations of the ions involved, based on the dissolution reaction. For \(\text{Fe(OH)}_2\), the reaction can be represented as:
\[\text{Fe(OH)}_2 (s) \rightleftharpoons \text{Fe}^{2+} (aq) + 2 \text{OH}^- (aq)\]
The solubility product expression for this reaction is:
\[K_{sp} = [\text{Fe}^{2+}][\text{OH}^-]^2\]
This equation tells us that the product of the ion concentrations at equilibrium raised to the power of their coefficients (from the balanced equation) equals \(K_{sp}\). Thus, \(K_{sp}\) is directly related to how much of the compound can dissolve in solution before the solution becomes saturated and precipitation occurs.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of ions or molecules at the point where the rates of the forward and reverse reactions are equal in a system. This concept is paramount in determining how much of a compound can dissolve in a solution.
When \(\text{Fe(OH)}_2\) dissolves, it reaches an equilibrium state where the concentration of \(\text{Fe}^{2+}\) ions is equal to the molar solubility of the compound. The concentration of \(\text{OH}^-\) ions, however, is twice the molar solubility because each unit of \(\text{Fe(OH)}_2\) releases two \(\text{OH}^-\) ions into the solution.
For instance, if the molar solubility is \(6.68 \times 10^{-6} \text{ mol/L}\), then:

• \([\text{Fe}^{2+}] = 6.68 \times 10^{-6} \text{ mol/L}\)
• \([\text{OH}^-] = 2 \times 6.68 \times 10^{-6} = 1.34 \times 10^{-5} \text{ mol/L}\)

This information is crucial for plugging into the \(K_{sp}\) expression to find the solubility product.

Precipitation Reaction

Precipitation reactions occur when two solutions combine to form an insoluble solid, known as a precipitate. This happens when the product of the concentrations of the resulting ions exceeds the \(K_{sp}\) of the compound.
In the exercise, we explore precipitating iron ions by adding hydroxide ions. The goal is to add enough \(\text{OH}^-\) to reduce the concentration of \(\text{Fe}^{2+}\) ions in solution to less than \(1.0 \mu \text{g/L}\).
To achieve this, we set up the inequality based on \(K_{sp}\), where the goal is to keep \([\text{Fe}^{2+}][\text{OH}^-]^2\) less than or equal to the known \(K_{sp}\). By solving the inequality:

• Convert the \(1.0 \mu\text{g/L}\) of \(\text{Fe}^{2+}\) to molarity.
• Use this value to determine the hydroxide ion concentration needed.

The calculation confirms the amount of \(\text{OH}^-\) needed to ensure that precipitation occurs, thereby limiting the \(\text{Fe}^{2+}\) in the solution to the desired level.