Question

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125\(M \mathrm{NaOH}\) , requiring 16.00 \(\mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding 2.00 \(\mathrm{mL}\) . NaOH was 6.912 . Calculate \(K_{\mathrm{a}}\) for the weak acid.

Short answer

The acid dissociation constant for the weak acid is approximately \(1.72 \times 10^{-5}\).

Step-by-step solution

Determine the moles of NaOH added and the initial moles of the weak acid

To determine the number of moles of NaOH that reacted, we can use the volume and molarity given: Moles of NaOH = Molarity × Volume Moles of NaOH = 0.125 M × 16.00 mL × \(\frac{1\,\mathrm{L}}{1000\,\mathrm{mL}}\) Moles of NaOH = 0.00200 moles Since the weak acid is monoprotic, the moles of weak acid will equal the moles of NaOH at the equivalence point. Moles of weak acid = Moles of NaOH = 0.00200 moles

Identify the balanced chemical equation for the reaction

The reaction between the weak acid (HA) and NaOH can be represented by the following balanced equation: HA + OH⁻ → A⁻ + H₂O

Calculate the concentration of the weak acid and its conjugate base when 2.00 mL NaOH is added

The number of moles of NaOH added when 2.00 mL is used can be calculated as follows: Moles of NaOH added at 2.00 mL = Molarity × Volume Moles of NaOH added at 2.00 mL = 0.125 M × 2.00 mL × \(\frac{1\,\mathrm{L}}{1000\,\mathrm{mL}}\) Moles of NaOH added at 2.00 mL = 0.00025 moles Using the balanced chemical equation, the moles of weak acid that reacted are equal to the moles of NaOH added at 2.00 mL: Moles of HA reacted = 0.00025 moles The remaining moles of the weak acid (HA), and the moles of the conjugate base (A⁻) formed can be calculated as: Moles of HA remaining = Initial moles of HA - Moles of HA reacted Moles of HA remaining = 0.00200 moles - 0.00025 moles = 0.00175 moles Moles of A⁻ formed = Moles of HA reacted = 0.00025 moles Now we will determine the total volume of the solution at this point: Total volume = Initial volume of weak acid + Volume of NaOH added Total volume = 16.00 mL + 2.00 mL = 18.00 mL To calculate the concentration of HA and A⁻ at this point, we will divide the moles by the total volume: Concentration of HA = \(\frac{0.00175\,\mathrm{moles}}{0.018\,\mathrm{L}}\) = 0.09722 M Concentration of A⁻ = \(\frac{0.00025\,\mathrm{moles}}{0.018\,\mathrm{L}}\) = 0.01389 M

Use an ICE table to express the initial, change, and equilibrium concentrations

We will create an ICE table for the weak acid ionization: | | HA | H₂O (l) | H₃O⁺ | A⁻ | |------- |------------|----------|-----------|-----------| | I | 0.09722 M | -- | 0 M | 0.01389 M | | C | -x | -- | +x | +x | | E | 0.09722-x M | -- | x M | 0.01389+x M |

Solve the Ka expression for the weak acid, using the equilibrium concentrations from the ICE table

We know the pH after adding 2.00 mL of NaOH is 6.912, so we can find the concentration of H₃O⁺ ions at this point: pH = -log[H₃O⁺] 6.912 = -log[H₃O⁺] Solve for [H₃O⁺]: [H₃O⁺] = 10^(-6.912) = 1.2 × 10^(-7) M Now, we can use the equilibrium concentrations from the ICE table to find Ka for the weak acid: Default variables: x = [H₃O⁺] = 1.2 × 10^(-7) M 0.09722 - x ≈ 0.09722 M (since x is very small) Using the ionization equation for the weak acid: Ka = \(\frac{[A⁻][H₃O⁺]}{[HA]}\) Ka = \(\frac{(0.01389 + x)(1.2 × 10^{-7})}{0.09722}\) Solve for Ka: Ka = 1.72 × 10^(-5) Thus, the acid dissociation constant for the weak acid is approximately 1.72 × 10^(-5).

Key concepts

Weak Acid

A weak acid is an acid that partially dissociates in water. This means only a fraction of its molecules release hydrogen ions (H⁺) into the solution. Unlike strong acids, which fully dissociate, weak acids establish an equilibrium between the undissociated molecules and the released ions.

Characteristics of weak acids:

• They have a higher pH compared to strong acids of the same concentration.
• The degree of ionization is usually less than 5%.
• Examples include acetic acid, formic acid, and citric acid.

This partial dissociation makes the calculation of the acid dissociation constant, \(K_a\), an important task since it quantifies the strength of a weak acid. It helps us understand how well an acid releases its protons in a given solution.

In the case of a Monoprotic weak acid, like the example in the titration exercise, it gives off one proton per molecule. During titration, it reacts with a base, such as \(\mathrm{NaOH}\), through a neutralization reaction to form water and a conjugate base.

Equivalence Point

In acid-base titration, the equivalence point is a critical milestone. It is the point where the total number of moles of hydrogen ions (\(\text{H}^+\)) from the acid equals the number of moles of hydroxide ions (\(\text{OH}^-\)) from the base added. This balance marks the complete neutralization of the acid by the base.

Features of the equivalence point include:

• The solution's pH may not always be neutral (pH 7) especially in weak acid-strong base titrations.
• The presence of only the conjugate base of the weak acid, which can affect the pH.
• Indicator color change often signifies the equivalence point, depending on the titration type.

In our weak acid titration example, after adding 16.00 mL of \(0.125 \,\mathrm{M} \text{NaOH}\), the equivalence point was achieved. The number of moles of acid equaled the number of moles of \(\text{NaOH}\), which allowed for calculating reactions quantities and concentrations.

Acid Dissociation Constant (Ka)

The acid dissociation constant, \(K_a\), provides a measure of an acid's ionization strength in a solution. It reflects how well an acid can donate its protons to the solvent. For weak acids, \(K_a\) values are usually small, indicating less dissociation and weaker acidic nature.

Calculating \(K_a\) involves using equilibrium concentrations from the acid's dissociation. In this example, calculations were based on measurements taken after adding a known volume of \(\text{NaOH}\) to a weak acid solution and finding the resulting pH.

To find \(K_a\) in this exercise:

• The concentration of hydronium ions (\([\text{H}_3\text{O}^+]\)) at the measured pH was first determined.
• An ICE table was used to organize and relate the initial concentrations, changes in concentrations, and equilibrium concentrations.
• \(K_a\) was derived from the equilibrium expression, \(K_a = \frac{[\text{A}^-][\text{H}_3\text{O}^+]}{[\text{HA}]}\), utilizing these equilibrium concentrations.

Ultimately, \(K_a\) provides insights into both the extent of the ionization of the weak acid and its tendency to release protons upon dissolving in water.