Problem 41 - Chemistry

Question

Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\) \(\left(K_{\mathrm{sp}}=4 \times 10^{-38}\right)\) in each of the following. a. water b. a solution buffered at pH \(=5.0\) c. a solution buffered at pH\(=11.0\)

Step-by-Step Solution

Answer

The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in different solutions is as follows: a. water - \(2.22 \times 10^{-10}\) mol/L b. solution buffered at pH = 5.0 - \(4 \times 10^{-11}\) mol/L c. solution buffered at pH = 11.0 - \(4 \times 10^{-29}\) mol/L

Step 1: Write the balanced dissolution equation of \(\mathrm{Fe}(\mathrm{OH})_{3}\)

The dissolution of iron (III) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water can be represented as: \[\mathrm{Fe(OH)_3 (s)} \rightleftharpoons \mathrm{Fe^{3+} (aq)} + 3\mathrm{OH^{-} (aq)}\] This balanced equation shows that for every mole of \(\mathrm{Fe}(\mathrm{OH})_{3}\) that dissolves, one mole of \(\mathrm{Fe^{3+}}\) ions and 3 moles of \(\mathrm{OH^{-}}\) ions are produced.

Step 2: Write the expression for the solubility product constant, \(K_{sp}\)

For the balanced dissolution equation, the solubility product constant is given by: \[K_{\text{sp}} = [\mathrm{Fe^{3+}}][\mathrm{OH^{-}}]^3\]

Step 3: Express the solubility (mol/L) in terms of \(x\)

Let \(x\) be the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in moles per liter. Then, \[[\mathrm{Fe^{3+}}] = x\] \[[\mathrm{OH^{-}}] = 3x\]

Step 4: Solve the expression for the solubility in water (a.)

Substitute the values of \([\mathrm{Fe^{3+}}]\) and \([\mathrm{OH^{-}}]\) in the \(K_{sp}\) expression: \[K_{\text{sp}} = x(3x)^3 = 27x^4\] Given \(K_{sp} = 4 \times 10^{-38}\), we can find the solubility in water: \[x^4 = \frac{4 \times 10^{-38}}{27}\] Solve for x: \[x = \sqrt[4]{\frac{4 \times 10^{-38}}{27}}\] \[x \approx 2.22 \times 10^{-10}\ \text{mol/L}\]

Step 5: Solve the expression for the solubility in a solution buffered at pH = 5.0 (b.)

First, calculate the hydroxide ion concentration, \([\mathrm{OH^{-}}]\), using the pH value: \[\text{pH} + \text{pOH} = 14\] \[5 + \text{pOH} = 14\] \[\text{pOH} = 9\] \[ [\mathrm{OH^{-}}] = 10^{-9}\] Now, substitute the \([\mathrm{OH^{-}}]\) value in the \(K_{sp}\) expression: \[K_{sp} = [\mathrm{Fe^{3+}}][(10^{-9})]^3\] \[4 \times 10^{-38} = [\mathrm{Fe^{3+}}] \times 10^{-27}\] Solve for \([\mathrm{Fe^{3+}}]\): \[[\mathrm{Fe^{3+}}] \approx 4 \times 10^{-11}\ \text{mol/L}\]

Step 6: Solve the expression for the solubility in a solution buffered at pH = 11.0 (c.)

Using the pH value, calculate the hydroxide ion concentration, \([\mathrm{OH^{-}}]\): \[\text{pH} + \text{pOH} = 14\] \[11 + \text{pOH} = 14\] \[\text{pOH} = 3\] \[ [\mathrm{OH^{-}}] = 10^{-3}\] Now, substitute the \([\mathrm{OH^{-}}]\) value in the \(K_{sp}\) expression: \[K_{sp} = [\mathrm{Fe^{3+}}][(10^{-3})]^3\] \[4 \times 10^{-38} = [\mathrm{Fe^{3+}}] \times 10^{-9}\] Solve for \([\mathrm{Fe^{3+}}]\): \[[\mathrm{Fe^{3+}}] \approx 4 \times 10^{-29}\ \text{mol/L}\] #Conclusion# To summarize, the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in different solutions is as follows: a. water - \(2.22 \times 10^{-10}\) mol/L b. solution buffered at pH = 5.0 - \(4 \times 10^{-11}\) mol/L c. solution buffered at pH = 11.0 - \(4 \times 10^{-29}\) mol/L

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, or \(K_{sp}\), is an essential concept in understanding how much of a compound can dissolve in water. It is a specific type of equilibrium constant used for solid substances dissolving in water to form a saturated solution. When a solid dissolves, it breaks apart into its respective ions in the solution.

The formula for \(K_{sp}\) involves the concentration of these ions raised to the power of their coefficients in the balanced equation. For example, for the dissolution of Fe(OH)\[ \mathrm{Fe(OH)_3 (s)} \rightleftharpoons \mathrm{Fe^{3+} (aq)} + 3\mathrm{OH^{-} (aq)} \]the \(K_{sp}\) expression is:\[ K_{\text{sp}} = [\mathrm{Fe^{3+}}][\mathrm{OH^{-}}]^3 \]

This means we multiply the concentration of the Fe³⁺ ions by the cube of the concentration of OH⁻ ions. Low \(K_{sp}\) values indicate low solubility. The example given \(K_{sp} = 4 \times 10^{-38}\) indicates that Fe(OH)₃ is only sparingly soluble in water.

Hydroxide Ion Concentration

The hydroxide ion concentration, denoted as \([\mathrm{OH^{-}}]\), is a critical factor in determining the solubility of compounds like Fe(OH)₃ in water. This concentration can change the solubility of a substance, influenced by the pH of the solution.

In different pH-buffered solutions, the hydroxide ion concentration varies significantly. For a solution at a neutral pH of 7, the concentration of \([\mathrm{OH^{-}}]\) is small because pure water dissociates to provide equal amounts of hydrogen and hydroxide ions.

For instance:

• At pH = 5.0, the \([\mathrm{OH^{-}}]\) is \(10^{-9}\) M.
• At pH = 11.0, the \([\mathrm{OH^{-}}]\) is \(10^{-3}\) M.

These variations cause the solubility of Fe(OH)₃ to change; higher \([\mathrm{OH^{-}}]\) concentrations mean lower solubility because the solution is closer to saturation with hydroxide ions.

pH and pOH

The concepts of pH and pOH are interconnected and crucial for understanding the behavior of ions in a solution, particularly in acid-base chemistry. They are scales used to describe the acidity and alkalinity of a solution, respectively.

While pOH measures hydroxide ion concentration, calculated similarly:\[ \text{pOH} = -\log[\mathrm{OH^-}] \]
In any aqueous solution, pH and pOH are related by the equation:\[ \text{pH} + \text{pOH} = 14 \]

Thus, knowing either pH or pOH allows you to determine the other.

\(\text{pOH} = 9\)

\([\mathrm{OH^{-}}] = 10^{-9}\) M

Conversely, a solution at pH = 11 has:

• \(\text{pOH} = 3\)
• \([\mathrm{OH^{-}}] = 10^{-3}\) M

Understanding these relationships helps predict how changes in pH will affect a solution's chemical dynamics and solubility.