Question

Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\left(K_{\mathrm{sp}}=4 \times\right.\) \(10^{-33}\) ) in each of the following. a. water b. a solution buffered at \(\mathrm{pH}=5.0\) c. a solution buffered at \(\mathrm{pH}=11.0\)

Short answer

The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is approximately: a. \(3.33 \times 10^{-9}\) moles per liter in water. b. \(4.0 \times 10^{-6}\) moles per liter in a solution buffered at \(\mathrm{pH}=5.0\). c. \(4.0 \times 10^{-24}\) moles per liter in a solution buffered at \(\mathrm{pH}=11.0\).

Step-by-step solution

Write the balanced chemical equation for the dissolution of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water

The balanced chemical equation for the dissolution of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water is: \(\mathrm{Fe}(\mathrm{OH})_{3(s)} \rightleftharpoons \mathrm{Fe}^{3+}_{(aq)} + 3\mathrm{OH}^-_{(aq)}\)

Write the expression for the solubility product

The \(K_{sp}\) expression for the dissolution of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is: \(K_{sp} = [\mathrm{Fe}^{3+}] [(\mathrm{OH}^-)^3]\)

Determine the concentrations of ions in terms of solubility

Let the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water be \(s\) mol/L. Then \([\mathrm{Fe}^{3+}] = s\) mol/L and \([\mathrm{OH}^-] = 3s\) mol/L

Plug the concentrations into the \(K_{sp}\) expression and solve for the solubility

Now we need to substitute the given \(K_{sp}\) value and the expressions for concentrations of ions in the \(K_{sp}\) expression, which is: \(4 \times 10^{-33} = s(3s)^3\) Now we need to solve this equation for the solubility \(s\): \(4 \times 10^{-33} = 27s^4\) \(s^4 = \frac{4 \times 10^{-33}}{27}\) \(s = \sqrt[4]{\frac{4 \times 10^{-33}}{27}} \approx 3.33 \times 10^{-9}\) mol/L The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water is approximately \(3.33 \times 10^{-9}\) moles per liter. b. Solution buffered at \(\mathrm{pH}=5.0\)

Calculate the hydroxide ion concentration

To find the hydroxide ion concentration, we first need to calculate the hydronium ion concentration using the given pH: \(\mathrm{pH} = -\log[H^+]\) \(5.0 = -\log[H^+]\) \(H^+ = 10^{-5}\) M Now, we can use the ion product of water, \(K_w = [\mathrm{H}^+][\mathrm{OH}^-]\), to find the concentration of hydroxide ions given that \(K_w = 1.0 \times 10^{-14}\) for water at 25°C, and the concentration of hydronium ions has been calculated: \([\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}^+]} = \frac{1.0 \times 10^{-14}}{10^{-5}} = 1.0 \times 10^{-9}\) M

Calculate the solubility

Now, we can go back to our \(K_{sp}\) expression for \(\mathrm{Fe}(\mathrm{OH})_{3}\) in the presence of a solution buffered at \(\mathrm{pH}=5.0\): \(K_{sp} = [\mathrm{Fe}^{3+}] [\mathrm{OH}^-]^3\) Given that \([\mathrm{OH}^-]\) is now \(1.0 \times 10^{-9}\) M, and letting the solubility in this case be \(s_b\): \(4 \times 10^{-33} = [\mathrm{Fe}^{3+}] (1.0 \times 10^{-9})^3\) \(4 \times 10^{-33} = s_b\ (1.0 \times 10^{-27} )\) \(s_b = \frac{4 \times 10^{-33}}{1.0 \times 10^{-27}} = 4.0 \times 10^{-6}\) mol/L The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in a solution buffered at \(\mathrm{pH}=5.0\) is \(4.0 \times 10^{-6}\) moles per liter. c. Solution buffered at \(\mathrm{pH}=11.0\)

Calculate the hydroxide ion concentration

As in part (b), we start by calculating the hydronium ion concentration from the given pH and then using the ion product of water to find the hydroxide ion concentration: \(\mathrm{pH} = -\log[H^+]\) \(11.0 = -\log[H^+]\) \(H^+ = 10^{-11}\) M \([\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}^+]} = \frac{1.0 \times 10^{-14}}{10^{-11}} = 1.0\times 10^{-3}\) M

Calculate the solubility

Now, we use the \(K_{sp}\) expression for \(\mathrm{Fe}(\mathrm{OH})_{3}\) in the presence of a solution buffered at \(\mathrm{pH}=11.0\): \(K_{sp} = [\mathrm{Fe}^{3+}] [\mathrm{OH}^-]^3\) Given that \([\mathrm{OH}^-]\) is now \(1.0 \times 10^{-3}\) M, and letting the solubility in this case be \(s_c\): \(4 \times 10^{-33} = [\mathrm{Fe}^{3+}] (1.0 \times 10^{-3})^3\) \(4 \times 10^{-33} = s_c\ (1.0 \times 10^{-9})\) \(s_c = \frac{4 \times 10^{-33}}{1.0 \times 10^{-9}} = 4.0 \times 10^{-24}\) mol/L The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in a solution buffered at \(\mathrm{pH}=11.0\) is \(4.0 \times 10^{-24}\) moles per liter.

Key concepts

Solubility Product Constant

In chemistry, the solubility product constant, often denoted as \(K_{sp}\), is a crucial value that helps predict the solubility of a sparingly soluble compound. It represents the concentration of ions in a saturated solution of an ionic compound at equilibrium. For example, when \(\mathrm{Fe(OH)}_3\) dissolves in water, it dissociates into \(\mathrm{Fe}^{3+}\) and \(\mathrm{OH}^-\) ions, as shown in the reaction:

• \(\mathrm{Fe(OH)}_3 (s) \rightleftharpoons \mathrm{Fe}^{3+}_{(aq)} + 3\mathrm{OH}^-_{(aq)}\)

The expression for the solubility product constant of \(\mathrm{Fe(OH)}_3\) is:

• \(K_{sp} = [\mathrm{Fe}^{3+}] [\mathrm{OH}^-]^3\)

Here, \([\mathrm{Fe}^{3+}]\) and \([\mathrm{OH}^-]\) are the molar concentrations of the dissociated ions. By knowing \(K_{sp}\), you can calculate the solubility of the compound in moles per liter (M) by simply solving the equation for the concentration of the ions.For instance, when calculating the solubility of \(\mathrm{Fe(OH)}_3\) in pure water using its \(K_{sp}\) of \(4 \times 10^{-33}\), you set up the equation based on the stoichiometry of the reaction and solve for the solubility \(s\) as demonstrated in the step-by-step solution.

pH and Solubility

The pH of a solution plays a significant role in determining the solubility of certain compounds, especially hydroxides. The pH scale ranges from 0 to 14, indicating how acidic or basic a solution is. For hydroxides like \(\mathrm{Fe(OH)}_3\), changes in pH affect the concentration of the \(\mathrm{OH}^-\) ions, which in turn affects solubility.

• **Low pH (Acidic):** More \(\mathrm{H}^+\) ions are present, and these react with \(\mathrm{OH}^-\) ions to form water, which effectively decreases the concentration of \(\mathrm{OH}^-\) ions. As a result, the solubility of \(\mathrm{Fe(OH)}_3\) increases as it dissolves more to re-establish equilibrium.
• **High pH (Basic):** A higher concentration of \(\mathrm{OH}^-\) ions are present in the solution. This already "saturates" the solution with \(\mathrm{OH}^-\) ions, and thus the solubility of \(\mathrm{Fe(OH)}_3\) decreases as it requires less \(\mathrm{Fe(OH)}_3\) to dissolve to maintain equilibrium.

When the solution is buffered at different pH values, such as at pH 5.0 and pH 11.0 in this exercise, we can see the solubility values of \(4.0 \times 10^{-6}\) mol/L and \(4.0 \times 10^{-24}\) mol/L respectively, illustrating how drastically solubility can be affected by pH changes.

Buffer Solutions

Buffer solutions are essential in maintaining a stable pH in a chemical environment. They consist of a weak acid and its conjugate base or a weak base and its conjugate acid. This balance allows the solution to resist changes in pH when small amounts of acid or base are added.In the context of solubility, buffers are particularly useful because they can maintain a specific pH, which, as mentioned in the previous section, impacts the solubility of compounds like \(\mathrm{Fe(OH)}_3\).
Here's how buffer solutions work:

• **Acidic Buffer (low pH):** Contains a weak acid and its salt. It helps to maintain the pH by providing \(\mathrm{H}^+\) ions, which lowers \([\mathrm{OH}^-]\), increasing the solubility of \(\mathrm{Fe(OH)}_3\).
• **Basic Buffer (high pH):** Contains a weak base and its salt. It maintains a higher \(\mathrm{OH}^-\) concentration, thus decreasing the solubility of \(\mathrm{Fe(OH)}_3\).

These properties of buffers prove crucial in numerous chemical and biological processes, where maintaining a specific solubility profile and pH is necessary for desired reactions or to support cellular functions.