Question

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) with 0.100 \(\mathrm{M} \mathrm{NaOH}\) .

Short answer

The pH values at various stages of the titration of 0.100 M propanoic acid with 0.100 M NaOH are: 1. Before the addition of NaOH: pH = 2.44 2. After the addition of 12.5 mL of NaOH (halfway to the equivalence point): pH = 4.88 3. At the equivalence point (after the addition of 25 mL of NaOH): pH = 8.99

Step-by-step solution

Calculate the initial moles of propanoic acid and sodium hydroxide

To find the initial moles of propanoic acid (HC3H3O2) and sodium hydroxide (NaOH), we can use the given volumes and concentrations. Moles of propanoic acid: moles = (volume in L) × (concentration in mol/L) Moles of HC3H3O2 = (0.025 L) × (0.1 mol/L) = 0.0025 moles

Set up and solve the equilibrium expression.

The reaction between propanoic acid and sodium hydroxide looks like this: HC3H3O2 + OH- ↔ C3H3O2- + H2O Because the Ka for propanoic acid is given, we can set up the equilibrium expression to solve for [OH-]: Ka = \[ \frac{[C3H3O2^-]_{eq}[H^+]_{eq}}{[HC3H3O2]_{eq}} = 1.3 \times 10^{-5} \] \[ [H^+]_{eq} = \frac{Ka[HC3H3O2]_{eq}}{[C3H3O2^-]_{eq}} \] We need to subsequently calculate the moles of hydroxide ions added to the solution at each stage of titration. In this case, let's calculate the pH at the following stages: 1. Before the addition of NaOH 2. After the addition of 12.5 mL of NaOH (halfway to the equivalence point) 3. At the equivalence point (after the addition of 25 mL of NaOH)

Find the pH at each stage

1. Before the addition of NaOH: Since no hydroxide ions have been added yet, we calculate the pH of the propanoic acid solution using the given Ka value. \[ [H^+]_{eq} = \sqrt{Ka \times [HC3H3O2]} = \sqrt{1.3 \times 10^{-5} \times 0.1} \] \[ [H^+]_{eq} = 3.60 \times 10^{-3} \] pH = -log([H+]) = 2.44 2. After the addition of 12.5 mL of NaOH (halfway to the equivalence point): The number of moles of NaOH added is: Moles of NaOH = (0.0125 L) × (0.1 mol/L) = 0.00125 moles Since both the base and acid are present at this stage, a buffering effect occurs: \[ [OH^-] = [\dfrac{moles \: of \: OH^- \\ added}{Total \: volume \: after \: base \: added}] \\ \] \[ [OH^-] = \dfrac{0.00125}{0.0375}= 3.33 \times 10^{-2} \\ \] Then use the buffer equation: \[ pH = pKa + log \dfrac{[C3H3O2^-]}{[HC3H3O2]} \] \[ pH = -log(K_a) + log \dfrac{0.00125}{0.00250} \] \[ pH = 4.88 \] 3. At the equivalence point (after the addition of 25 mL of NaOH): At the equivalence point, equal amounts of acid and base have been added. Therefore, the number of moles of NaOH added is equal to the initial moles of propanoic acid: Moles of NaOH = 0.0025 moles \[ [OH^-] = [\dfrac{moles \: of \: OH^- \\ added}{Total \: volume \: after \: base \: added}] \\ \] \[ [OH^-] = (0.0025 \: moles) \div (0.05 \: L) = 5 \times 10^{-2} \: M \] Using the Kb value for the conjugate base, C3H3O2-, we can calculate the pH: \[ K_b = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{1.3 \times 10^{-5}} = 7.69 \times 10^{-10} \] \[ K_b = \dfrac{[OH^-][HC3H3O2]}{[C3H3O2^-]} \] \[ [OH^-] = \sqrt{K_b \times [C3H3O2^-]} = \sqrt{7.69 \times 10^{-10} \times 5 \times 10^{-2}} \] \[ [OH^-] = 9.75 \times 10^{-6} \] First, calculate the pOH and then pH: pOH = -log([OH-]) = 5.01 pH = 14 - pOH = 14 - 5.01 = 8.99 The final pH values at the various stages of titration are: 1. Before the addition of NaOH: 2.44 2. After the addition of 12.5 mL of NaOH (halfway to the equivalence point): 4.88 3. At the equivalence point (after the addition of 25 mL of NaOH): 8.99

Key concepts

Propanoic Acid

Propanoic acid is an organic compound that belongs to a group of carboxylic acids. Its chemical formula is \( \text{HC}_3\text{H}_5\text{O}_2 \), and it is characterized by a three-carbon chain attached to a carboxyl group. This acid is commonly used in organic chemistry due to its simple structure.
In aqueous solutions, propanoic acid partially dissociates to form propanoate ions (\( \text{C}_3\text{H}_5\text{O}_2^- \)) and hydrogen ions (\( \text{H}^+ \)). The acid dissociation constant \( K_a \) for propanoic acid is \( 1.3 \times 10^{-5} \), which indicates that it is a weak acid. Understanding its dissociation is crucial for calculating pH, especially during titrations.

• The acidic nature is due to the carboxyl group \( \text{-COOH} \), which releases \( \text{H}^+ \) ions.
• The strength and behavior of propanoic acid in solution can be manipulated by altering the concentration or through a titration process.

This property makes propanoic acid an excellent candidate for studies on acid-base equilibrium and buffer solutions.

pH Calculation

Calculating the pH of a solution is crucial in understanding the acidity or basicity of that solution. pH is the negative logarithm of the hydrogen ion concentration, given by \( \text{pH} = -\log[\text{H}^+] \).
When dealing with weak acids like propanoic acid, it's necessary to use the acid dissociation constant \( K_a \) in the calculations.

• Initial pH Calculation: For propanoic acid, the initial pH can be calculated without any added base by using \( K_a \):
  \( [\text{H}^+] = \sqrt{K_a \times [\text{HC}_3\text{H}_5\text{O}_2]} \).
• During Titration: The concentration of hydrogen ions changes, requiring calculations at different titration stages to determine the pH.
• Significance of Equivalence Point: At the equivalence point, using the \( K_b \) of the conjugate base helps to find the pH since equal moles of acid and base have reacted.

Each stage in titration (pre-addition, midpoint, and equivalence) offers valuable insights into how solutions respond to chemical changes.

Acid-Base Equilibrium

In the context of titration, understanding acid-base equilibrium is essential. It describes how acids and bases interact with each other in a solution.
Equilibrium in this setting refers to the balance established when the forward and reverse reactions between the acid-base pairs occur at the same rate.

• Equilibrium Expression: For propanoic acid, it can be set up as:
  \( K_a = \frac{[\text{C}_3\text{H}_5\text{O}_2^-][\text{H}^+]}{[\text{HC}_3\text{H}_5\text{O}_2]} \)
• Role of \( K_a \): This constant is pivotal in predicting how much of the acid will dissociate into its ions in a given solution.
• Influence of Titration: Adding a strong base like NaOH shifts the equilibrium, affecting the concentrations of each species involved.

This equilibrium principle allows for precise calculations of pH during the progression of a titration.

Buffer Solution

A buffer solution is a combination of weak acid and its conjugate base, which resists changes in pH upon the addition of small amounts of an acid or a base.
When propanoic acid is titrated with a strong base like NaOH, a buffer solution is temporarily formed before reaching the equivalence point.

• Henderson-Hasselbalch Equation: This equation allows calculation of pH in buffer solutions:
  \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \)
• Buffer Capacity: The effectiveness of a buffer is called its buffer capacity, determined by the concentrations of the acid and its conjugate base.
• Significance: Buffers are essential in many biological and chemical systems, as they help maintain a stable pH, which is crucial for reactions and processes.

During titration, before reaching the equivalence point, the pH change is gradual due to the buffering action of the propanoic acid and its conjugate base pair.