Question

A \(35.0\)-mL sample of \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is titrated with \(0.150 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of base have been added: (a) \(0 \mathrm{~mL}\), (b) \(17.5 \mathrm{~mL}\) (c) \(34.5 \mathrm{~mL}\) (d) \(35.0 \mathrm{~mL}\) (e) \(35.5 \mathrm{~mL}\) (f) \(50.0 \mathrm{~mL}\)..

Short answer

The calculated pH values after adding different volumes of NaOH are: (a) pH ≈ 2.87, (b) pH ≈ 4.76, (c) pH ≈ 6.37, (d) pH ≈ 8.61, (e) pH ≈ 12.04, and (f) pH ≈ 13.21.

Step-by-step solution

Identify the given values

The initial concentration of acetic acid (\(CH_3COOH\)) is 0.150 M and it is titrated with a 0.150 M sodium hydroxide (NaOH) solution. The volume of acetic acid is 35.0 mL. We need to find the pH of the solution for different volumes of the NaOH added.

Calculate the initial moles of acetic acid and its conjugate base

First, calculate the initial moles of acetic acid (\(CH_3COOH\)) using its concentration and volume: moles of acetic acid = concentration × volume moles of acetic acid = 0.150 mol/L × 0.035 L = 0.00525 mol At the beginning of the titration, the number of moles of its conjugate base (\(CH_3COO^-\)) is zero.

Calculate the moles of acetic acid and its conjugate base after each volume of NaOH is added

For each volume of NaOH added, calculate the moles of NaOH and subtract it from the initial moles of acetic acid, while adding it to the moles of its conjugate base: moles of NaOH = concentration × volume of NaOH added (a) 0 mL NaOH: Moles of acetic acid = 0.00525 mol, Moles of conjugate base = 0 mol (b) 17.5 mL NaOH: Moles of acetic acid = 0.00525 - (0.150 × 0.0175) mol, Moles of conjugate base = (0.150 × 0.0175) mol (c) 34.5 mL NaOH: Moles of acetic acid = 0.00525 - (0.150 × 0.0345) mol, Moles of conjugate base = (0.150 × 0.0345) mol (d) 35.0 mL NaOH: Moles of acetic acid = 0.00525 - (0.150 × 0.0350) mol, Moles of conjugate base = (0.150 × 0.0350) mol (e) 35.5 mL NaOH: Moles of acetic acid = 0.00525 - (0.150 × 0.0355) mol, Moles of conjugate base = (0.150 × 0.0355) mol (f) 50.0 mL NaOH: Moles of acetic acid = 0.00525 - (0.150 × 0.0500) mol, Moles of conjugate base = (0.150 × 0.0500) mol

Calculate pH using the Henderson-Hasselbalch equation

Using the Henderson-Hasselbalch equation, pH = pKa + log ([conjugate base]/[acid]), and knowing that the pKa of acetic acid is 4.76, calculate the pH for each volume of NaOH added. (a) pH = 4.76 + log(0/0.00525) => undefined, since log(0) cannot be calculated. In this case, as no base has been added and the system contains only weak acid, pH can be calculated using the pH = -log(\([\ce{H3O+}]\)) equation knowing that \(\ce{CH3COOH}\) equilibrium constant \(Ka = 1.8 \times 10^{-5}\) and solving the quadratic equation. (b) - (f) Calculate the pH using the Henderson-Hasselbalch equation and the moles of acetic acid and its conjugate base found in step 3. Remember to convert moles back to concentrations by dividing the moles by total volume (volume of acetic acid + volume of added NaOH), then use this concentrations in the equation.

Report the calculated pH values

After calculating the pH values for each volume of NaOH added using the Henderson-Hasselbalch equation (or other methods for the initial pH), report the pH values: (a) pH ≈ 2.87 (b) pH ≈ 4.76 (c) pH ≈ 6.37 (d) pH ≈ 8.61 (e) pH ≈ 12.04 (f) pH ≈ 13.21

Key concepts

Acetic Acid

Acetic acid, often found in household vinegar, is a weak organic acid with the chemical formula \( \text{CH}_3\text{COOH} \). It dissociates partially in water to produce hydrogen ions and acetate ions \( \text{CH}_3\text{COO}^- \).
This partial ionization is what categorizes acetic acid as a weak acid, unlike stronger acids that fully dissociate in solution. The equilibrium between the undissociated acid and the ions it forms can be represented by the acid dissociation constant, or \( K_a \).
For acetic acid, \( K_a = 1.8 \times 10^{-5} \), highlighting its weak nature. This property is crucial during a titration process, as it influences how the pH of the solution changes as a base, such as sodium hydroxide (\( \text{NaOH} \)), is added.

• In the initial stage of a titration involving acetic acid, no sodium hydroxide is present, and the solution consists entirely of the acetic acid. This is also when the lowest pH value is observed.
• As more base is added, acetic acid gradually gets converted to its conjugate base, acetate, until it reaches the equivalence point, where theoretically, all acetic acid has been neutralized.
• Beyond the equivalence point, the solution becomes more basic, and the pH rises significantly.

Understanding these basics of acetic acid helps clarify the titration process, where you monitor changes in pH to determine the endpoint of the reaction.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is a formula used to estimate the pH of a buffer solution. In the context of an acetic acid titration, it helps us determine the pH at various points as base is added.
It is derived from the acid dissociation equilibrium expression:\[pH = pK_a + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]Here, \([\text{A}^-]\) is the concentration of the conjugate base (acetate), and \([\text{HA}]\) is the concentration of the weak acid (acetic acid). \(pK_a\) is the negative logarithm of the \(K_a\) of the acid.

• In our titration example, we use the Henderson-Hasselbalch equation once some amount of sodium hydroxide is added, creating a buffer solution of acetic acid and its conjugate base.
• The equation is particularly useful from the halfway point of the titration to the equivalence point because it accounts for the presence of both the acid and its conjugate base.
• At the halfway point, where \([\text{A}^-] = [\text{HA}]\), the \(\log\) term becomes zero, and thus \(pH = pK_a\), providing an easy way to confirm the pKa.

Using the Henderson-Hasselbalch equation allows us to understand how the ratios of reactants and products affect pH, making it a powerful tool in the analysis of buffer systems and titrations.

pH Calculation

Calculating pH is an essential skill in chemistry that tells us how acidic or basic a solution is. For titrations, specifically with acetic acid and NaOH, understanding how to properly calculate pH at different stages is crucial.
This begins with recognizing what stage the titration is at: before, at, or after the equivalence point.
**Before Adding NaOH:**

• When no base is added, the pH is calculated using only the \(K_a\) of acetic acid and its initial concentration.
• The formula \(\text{pH} = -\log [\text{H}^+]\) is used after solving for \([\text{H}^+]\) from the equilibrium expression \(K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}\).

**During Addition (Buffered Region):**

• Use the Henderson-Hasselbalch equation to find \(pH\), given the concentrations of acetic acid and its conjugate base.
• Adjust these concentrations based on the amount of \(\text{NaOH}\) added.

**Equivalence and Beyond:**

• At equivalence, all the acetic acid is converted, so the \([\text{OH}^-]\) concentration from excess \(\text{NaOH}\) dictates \(pH\).
• Beyond equivalence, only the excess base affects pH, calculated using \([\text{OH}^-]\) and the formula \(\text{pH} = 14 + \log [\text{OH}^-]\).

The ability to calculate pH accurately at different titration stages helps confirm the process' accuracy and supports further understanding of acid-base reactions.