Question

A \(0.100-\mathrm{g}\) sample of the weak acid HA (molar mass \(=\) \(100.0 \mathrm{g} / \mathrm{mol}\) ) is dissolved in \(500.0 \mathrm{g}\) water. The freezing point of the resulting solution is \(-0.0056^{\circ} \mathrm{C}\). Calculate the value of \(K_{\mathrm{a}}\) for this acid. Assume molality equals molarity in this solution.

Short answer

The value of the acid dissociation constant (\(K_{a}\)) for the weak acid HA is approximately \(1.125\times 10^{-3}\).

Step-by-step solution

Calculate the molality of the solution

First, we need to find the molality of the solution (i.e., the number of moles of solute per kilogram of solvent). We can determine the molality using the molar mass of the acid and the mass of the solute and solvent. Given mass of acid (HA): \(0.100\,\text{g}\) Molar mass of acid (HA): \(100.0\,\text{g/mol}\) Mass of solution: \(0.100\,\text{g}\) + \(500.0\,\text{g}\) = \(500.100\,\text{g}\) Moles of HA: \(\frac{0.100\,\text{g}}{100.0\,\text{g/mol}}=0.001\,\mathrm{mol}\) Mass of water: \(500.0\,\text{g}\) = \(0.500\,\mathrm{kg}\) Molality, \(m=\frac{\mathrm{moles\,of\,solute}}{\mathrm{mass\,of\,solvent\,(kg)}}=\frac{0.001\,\mathrm{mol}}{0.500\,\mathrm{kg}}=0.002\,\mathrm{mol/kg}\)

Determine the ions in the solution

Next, we'll use the freezing point depression information to determine the number of ions that the acid produces when dissolved. The freezing point depression is given by the equation: \(ΔT_{f}=K_{f} \times m \times i\) ΔTₓ=f is the freezing point depression, Kₓ is the freezing point depression constant for water (1.86 °C/molal), m is the molality, and i is the van't Hoff factor (the number of particles into which the solute dissociates when dissolved). We need to solve for i. First, let's plug in the known values: \(-0.0056^{\circ} \mathrm{C} = 1.86\, °C \mathrm{molal/kg} \times 0.002\,\mathrm{mol/kg} \times i\)

Calculate the van't Hoff factor (i)

Now, let's solve for i: \(i=\frac{-0.0056^{\circ} \mathrm{C}}{1.86\, °C \mathrm{molal/kg} \times 0.002\,\mathrm{mol/kg}} = 1.5\) Since the value of i is close to 2, the acid HA dissociates into two ions, H⁺ and A⁻.

Calculate the dissociation

Now, let's calculate the dissociation of HA into ions: HA \(\rightleftharpoons\) H⁺ + A⁻ Initial concentration of HA: 0.002 M (as molality and molarity are nearly equal) \[ \begin{array}{c|c|c|c} & HA & H+ & A^- \\ \hline Initial & 0.002 & 0 & 0 \\ Change & -x & +x & +x \\ Equilibrium & 0.002-x & x & x \end{array} \]

Apply the equilibrium constant expression

Now, we'll use the acid dissociation constant expression: \(K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\) As the value of i is close to 2, we can assume that the equilibrium concentration of H⁺ and A⁻ is equal (x). Thus, \(K_{a} = \frac{x^2}{0.002 - x}\) Since i is close to 2, we know that a significant amount of the weak acid HA has dissociated into H⁺ and A⁻. So, we can make an approximation: \(x<<0.002\) Therefore, \(K_{a} \approx \frac{x^2}{0.002}\)

Calculate the concentration of H⁺

At this point, we have enough information to determine the concentration of H⁺ ions in the solution. Since we know they are equal to the concentration of A⁻ ions, we can calculate x using the van't Hoff factor from step 3: \(x = \frac{1.5}{2}[HA]\) \(x = 0.0015\,\text{M}\)

Calculate the value of \(K_{a}\)

Finally, we can use the calculated value of x to find the value of \(K_{a}\): \(K_{a} = \frac{x^2}{0.002} = \frac{(0.0015\,M)^2}{0.002\,M} = \frac{2.25 \times 10^{-6} \,M^2}{0.002\,M}= 1.125 \times 10^{-3}\) Thus, the value of \(K_{a}\) for the weak acid HA is approximately \(1.125\times 10^{-3}\).

Key concepts

Weak Acid

A weak acid is a type of acid that only partially dissociates into its ions in aqueous solution. Unlike strong acids, which completely dissociate, weak acids establish an equilibrium between the undissociated molecules and the ions formed in solution.

This partial dissociation is expressed through the acid dissociation constant, also known as the equilibrium constant for the dissociation of an acid. When a weak acid, represented as HA, dissolves in water, it forms H⁺ (hydrogen ions) and A⁻ (conjugate base).

In chemical notation, this equilibrium can be expressed as:
- HA ⇌ H⁺ + A⁻

The strength of a weak acid is measured by its acid dissociation constant, Ka. The smaller the Ka value, the weaker the acid, indicating less dissociation. In our exercise, HA is a weak acid with a calculated Ka of approximately 1.125 × 10⁻³, suggesting a moderate tendency to donate protons.

Freezing Point Depression

Freezing point depression is a colligative property, meaning it depends on the number of particles in a solution rather than their identity. When a solute is dissolved in a solvent, the freezing point of the solution becomes lower than that of the pure solvent.

This phenomenon occurs because the solute particles interfere with the formation of the solid lattice structure of the solvent, requiring a lower temperature to achieve freezing.

The change in freezing point, ∆Tₓ=f, is calculated using the formula:
- ∆Tₓ=f = Kₓf × m × i

Where:

• ΔTₓ=f is the freezing point depression, which is the difference between the freezing point of the pure solvent and the solution.
• Kₓf is the freezing point depression constant (for water, it is 1.86 °C/molal).
• m is the molality of the solution.
• i is the van't Hoff factor, representing the number of particles the solute produces in solution.

In this exercise, the freezing point of the solution is –0.0056°C, indicating the presence and effect of dissolved ions on the freezing point.

van't Hoff Factor

The van't Hoff factor, denoted as i, is a crucial concept for understanding how solutes affect the colligative properties of solutions. It represents the number of particles formed in solution from one formula unit of solute.

In this context, a higher van't Hoff factor signifies a greater number of species in solution, which can lead to a more pronounced effect on properties like boiling point elevation and freezing point depression.

For example:
- A nonelectrolyte like glucose would have an i value of 1, because it does not dissociate into ions. - Sodium chloride (NaCl), a common electrolyte, dissociates into 2 ions (Na⁺ and Cl⁻), giving it an i value of 2. - The weak acid HA in our solution has an experimental i value of 1.5, indicating partial dissociation into ions.

This calculation reflects the degree of dissociation and helps estimate the impact on the solution's properties.

Equilibrium Constant Expression

An equilibrium constant expression is used to quantify the concentration of reactants and products in a reversible chemical reaction at equilibrium. For acids, this constant is known as the acid dissociation constant, or Ka.

The expression for Ka involves the concentrations of the ions produced and the undissociated acid. For the weak acid HA, the dissociation reaction can be expressed as:
- HA ⇌ H⁺ + A⁻

Here, the equilibrium constant expression is given by:
- Ka = [H⁺] × [A⁻] / [HA]

Where:

• [H⁺] and [A⁻] are the molar concentrations of hydrogen ions and conjugate base in solution.
• [HA] is the molar concentration of the remaining undissociated acid.

This equation helps determine the extent of dissociation and the strength of the acid.

In the given exercise, approximation methods were used to simplify calculations, assuming that the change in concentration of HA during dissociation is minimal compared to its initial concentration.