Problem 18 - Analytical Chemistry

Question

A monoprotic organic acid with a \(K_{a}\) of \(6.7 \times 10^{-4}\) is \(3.5 \%\) ionized when \(100 \mathrm{~g}\) of it is dissolved in \(1 \mathrm{~L}\). What is the formula weight of the acid?

Step-by-Step Solution

Answer

The formula weight of the acid is approximately 2346 g/mol.

Step 1: Calculate moles of ionized acid

The degree of ionization is given as 3.5%, which means that 3.5% of the acid molecules are ionized. Let the initial concentration of the solution be denoted by \([HA]_0\). At equilibrium, the concentration of ionized acid \([H^+]\) is 0.035 \([HA]_0\).

Step 2: Set up ionization equilibrium

According to the ionization equation of the monoprotic acid, \(HA \rightleftharpoons H^+ + A^-\), the equilibrium concentrations are \([H^+] = 0.035 \cdot [HA]_0\), \([A^-] = 0.035 \cdot [HA]_0\), and \([HA] = [HA]_0 - 0.035 \cdot [HA]_0 = 0.965 \cdot [HA]_0\).

Step 3: Use the ionization constant formula

The expression for the ionization constant \(K_a\) is \(K_{a} = \frac{[H^+][A^-]}{[HA]}\). Substitute the equilibrium concentrations into this formula: \(6.7 \times 10^{-4} = \frac{(0.035 \cdot [HA]_0)^2}{0.965 \cdot [HA]_0}\).

Step 4: Solve for initial concentration

Rearrange the equation: \([HA]_0 = \frac{(0.035)^2}{0.965} \times 6.7 \times 10^{-4}\). Calculate \([HA]_0\) and find the initial molarity of the acid. \([HA]_0 \approx 4.26 \times 10^{-2} \ mol/L\).

Step 5: Calculate formula weight

To find the molecular weight, divide the mass given by the number of moles. Since the mass is 100 g, and the solution is 1 L, the number of moles is \(0.0426 \ mol\). Molecular weight = \(\frac{100}{0.0426} \approx 2346 \ g/mol\).

Key Concepts

Ionization constant

The ionization constant, denoted as \(K_a\), is a measure of the strength of an acid in solution. It indicates how well an acid can donate its protons to the solvent, typically water. For a monoprotic acid, which donates a single proton, the ionization constant is derived from its dissociation reaction in water:

• \(HA \rightleftharpoons H^+ + A^-\)

In this equation, \(HA\) represents the undissociated acid, \(H^+\) the proton, and \(A^-\) the conjugate base. The expression for the ionization constant \(K_a\) is given by:

• \[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Here, \([H^+]\), \([A^-]\), and \([HA]\) are the molar concentrations of the proton, conjugate base, and undissociated acid at equilibrium. A higher \(K_a\) value indicates a stronger acid, as more of the acid dissociates in water.

Molecular weight calculation

Calculating the molecular weight of a substance involves dividing the mass of the substance by the number of moles. In this exercise, 100 grams of the acid are dissolved, and we've found the concentration to be approximately \(0.0426 \ mol/L\). Hence, the number of moles in 1 liter (i.e., the volume of our solution) is 0.0426.

• Molecular weight = \(\frac{\text{mass}}{\text{moles}}\)
• \(= \frac{100 \ g}{0.0426 \ mol} \approx 2346 \ g/mol\)

This high molecular weight suggests a large, possibly complex organic acid. Understanding molecular weight is vital as it is a fundamental property that affects many aspects of chemical behavior and reactivity.

Equilibrium concentration

Equilibrium concentration is the concentration of reactants and products in a chemical reaction that remains constant over time. For a weak acid like the one in this exercise, only a small fraction ionizes, and this is quantified by the degree of ionization, which is 3.5%.

• At equilibrium: \([H^+] = 0.035 \cdot [HA]_0\)
• \([A^-] = 0.035 \cdot [HA]_0\)
• \([HA] = [HA]_0 - 0.035 \cdot [HA]_0 = 0.965 \cdot [HA]_0\)

These expressions show that the equilibrium concentration of the proton and conjugate base are both equal, and the remaining concentration of the acid is the initial concentration minus the ionized part. This balance of concentrations is crucial for calculating properties like \(K_a\), which in turn help determine the behavior and strengths of acids.

Monoprotic acid

Monoprotic acids are acids that can donate one proton (hydrogen ion) per molecule to an aqueous solution. Examples include hydrochloric acid (HCl) and acetic acid (CH₃COOH). In aqueous solution, they typically follow a dissociation pattern where the acid molecule separates to form a proton and a conjugate base:

• \(HA \rightleftharpoons H^+ + A^-\)

The characteristic feature of monoprotic acids is their simple 1:1 ionization pattern.

• Simultaneously producing one proton \([H^+]\)
• One conjugate base \([A^-]\)

This simplicity makes them an excellent model for understanding acid-base chemistry, as they simplify equilibrium calculations involving ionization and \(K_a\). Moreover, the analysis of their molecular weight and equilibrium concentrations is straightforward owing to this 1:1 stoichiometry.