Problem 102 - General Chemistry: Principles and Modern Applications

Question

Will \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) precipitate from a buffer solution that is \(0.45 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.35 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\) and also \(0.275 \mathrm{M}\) in \(\mathrm{Al}^{3+}(\mathrm{aq}) ?\) For \(\mathrm{Al}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=\) \(1.3 \times 10^{-33} ;\) for \(\mathrm{CH}_{3} \mathrm{COOH}, K_{\mathrm{a}}=1.8 \times 10^{-5}\).

Step-by-Step Solution

Answer

Yes, taking into account the solubility equilibrium and the conditions present in the solution, Al(OH)3 will precipitate from the buffer solution.

Step 1: Write the equations for the dissociation of Al(OH)3 and CH3COOH

The dissociation of Aluminium hydroxide is given as: \n \[Al(OH)_3(s) \rightleftharpoons Al^{3+}(aq) + 3OH^-(aq)\] \n And the dissociation of Acetic acid is represented as: \n \[CH3COOH(aq) \rightleftharpoons H^+(aq) + CH3COO^-(aq)\]

Step 2: Express the equilibrium constants

The established equilibrium can then be described using the solubility product, \(K_{sp}\), of the Aluminium hydroxide and the acid dissociation constant, \(K_a\), for the acetic acid. \n For Aluminium hydroxide, the \(K_{sp}\) expression given as: \n \[K_{sp} = [Al^{3+}][OH^-]^3 = 1.3x10^{-33}\] \n For Acetic acid, \(K_a\) can be expressed as: \n \[K_a = \frac{[H^+][CH3COO^-]}{[CH3COOH]} = 1.8x10^{-5}\]. This equation can be rearranged for [H^+] given that the Sodium acetate concentration is the same as the Acetate ion concentration: \n \[[H^+] = \frac{K_a [CH3COOH]}{[CH3COO^-]} \]

Step 3: Calculate [OH^-] from the buffer system (waters ion product)

The concentration of the hydroxide ions can be determined using the ion product of water, \(K_w = [H^+][OH^-] = 1x10^{-14}\) at 25°C. Rearrange this formula to give [OH^-], which gives us: \n \[[OH^-] = \frac{K_w}{[H^+]}\]

Step 4: Substitute back into the Ksp equation

After finding the [OH^-] from the buffer system, return to the \(K_{sp}\) expression and plug in the given [Al^{3+}] and the calculated [OH^-]. If the product is larger than the \(K_{sp}\), then the solution is supersaturated and Al(OH)3 will precipitate.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, is a crucial concept in understanding the solubility of sparingly soluble salts in water. It helps us predict whether a precipitate will form when two ionic species are mixed in a solution. Let's simplify this concept for better comprehension.

Ksp is specifically the product of the concentrations of the ions in a saturated solution, each raised to the power of their respective coefficients in the balanced equation. For example, the solubility product for aluminum hydroxide, \(Al(OH)_3\), can be represented as: \[K_{sp} = [Al^{3+}] [OH^-]^3\].

Visualize a dynamic balance between the solid salt that dissolves in water and the dissolved ions that recombine to form the solid. At equilibrium, the rate at which the salt dissolves equals the rate at which the solid forms, and this equilibrium has a specific Ksp value.

Each sparingly soluble salt has its unique Ksp reflecting its solubility level. The lower the Ksp, the less soluble the compound. For instance, with \(Al(OH)_3\)'s Ksp given as \(1.3 \times 10^{-33}\), it indicates an extremely low solubility, meaning \(Al(OH)_3\) tends not to dissolve well in water.

When assessing whether a precipitate will form, compare the product of the ionic concentrations to the Ksp. If the product is greater than Ksp, precipitation will occur since the solution is supersaturated with ions.

Acid Dissociation Constant (Ka)

The acid dissociation constant, Ka, is a measure of the strength of an acid in solution. It quantifies the acid's ability to donate protons to the surrounding water molecules, consequently forming hydronium ions (\(H_3O^+\)) or hydrogen ions (\(H^+\)) and the conjugate base. Grasping the concept of Ka allows us to evaluate the extent of acid dissociation in water.

The generic equation for the dissociation of an acid, \(HA\), into its conjugate base, \(A^-\), and hydrogen ions is: \[HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)\].Building on this, the expression for Ka becomes: \[K_a = \frac{[H^+][A^-]}{[HA]}\].Put simply, Ka is the ratio of the concentration of the products (ions) to that of the undissociated acid.

A higher Ka value indicates a stronger acid because it implies a greater degree of ionization in solution. Conversely, a weaker acid has a lower Ka value and less ionization. For acetic acid \(CH_3COOH\), with a Ka of \(1.8 \times 10^{-5}\), it's considered a weak acid as it only partially ionizes in the water.

Buffer Solution

Buffer solutions are ingenious systems that resist changes in pH when small amounts of acid or base are added. Their secret lies in containing both an acid and its conjugate base, or a base and its conjugate acid, in appreciable concentrations. This duo acts as a chemical sponge, absorbing excess hydrogen (\(H^+\)) or hydroxide (\(OH^-\)) ions and thus stabilizing the solution's pH.

In our case, a buffer solution consisting of acetic acid \(CH_3COOH\) and its conjugate base, sodium acetate \(NaCH_3COO\), helps to maintain a fairly constant pH despite adding aluminum ions (\(Al^{3+}\)).

The functioning of a buffer can be illustrated through the Henderson-Hasselbalch equation, though this equation is often unnecessary when working with strong acid/strong base pairings in a buffer. Instead, understanding how the buffer maintains pH by neutralizing the added acid or base is the key concept.

As a practical application, if \(Al^{3+}\) ions are introduced into the buffer, and \(Al(OH)_3\) begins to precipitate, the buffer resists a rapid pH change by adjusting the ratio of \(CH_3COOH\) to \(CH_3COO^-\), therefore, demonstrating its buffering capability and highlighting its importance in many biological and industrial processes.