Problem 43 - Chemistry: The Central Science

Question

A \(35.0-\mathrm{mL}\) 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: (d) \(35.0 \mathrm{~mL},(\mathrm{e}) \overline{35.5 \mathrm{~mL}}\) (a) \(0 \mathrm{~mL}\), (b) \(17.5 \mathrm{~mL}\), (c) \(34.5 \mathrm{~mL}\), (f) \(50.0 \mathrm{~mL}\).

Step-by-Step Solution

Answer

The pH values after adding the given volumes of NaOH are as follows: (a) 0 mL NaOH added: pH = 2.88 (b) 17.5 mL NaOH added: pH = 4.69 (c) 34.5 mL NaOH added: pH = 7.35 (d) 35.0 mL NaOH added: pH = 7.00 (at equivalence point) (e) 35.5 mL NaOH added: pH = 8.21 (f) 50.0 mL NaOH added: pH = 12.47

Step 1: Determine the initial moles of acetic acid and NaOH

Before the titration begins, we need to know the initial moles of acetic acid in the sample and the moles of NaOH added during the titration. To calculate the moles of acetic acid, we will use the given concentration and volume of the solution: \[\text{moles of acetic acid} = (0.150 \, \mathrm{M})(35.0 \, \mathrm{mL})(0.001 \frac{\mathrm{L}}{\mathrm{mL}})\] For each of the different volumes of \(\mathrm{NaOH}\) added, we will use a similar calculation: \[\text{moles of NaOH} = (0.150 \, \mathrm{M})(V_{\mathrm{NaOH}})(0.001 \frac{\mathrm{L}}{\mathrm{mL}})\] where \(V_{\mathrm{NaOH}}\) is the volume of NaOH added.

Step 2: Calculate the change in moles and the final concentrations

When base is added, it will neutralize the acetic acid, so we need to determine how many moles of acetic acid remains and how many moles of its conjugate base (acetate) are formed. We will compare the moles of NaOH added with the initial moles of acetic acid to find the final moles of each species. We'll then use the total volume of the solution to find the final concentrations of acetic acid and acetate.

Step 3: Use the Henderson-Hasselbalch equation to find the pH

With the final concentrations of acetic acid and acetate determined, we can use the Henderson-Hasselbalch equation to calculate the \(\mathrm{pH}\) of the solution: \[\mathrm{pH} = pK_{a} + \log\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\] The dissociation constant of acetic acid is given as \(K_{a} = 1.8 \times 10^{-5}\); therefore, the \(pK_{a}\) can be calculated as: \[pK_{a} = -\log(K_{a})\] Now let's perform the calculations for each of the six volumes of NaOH added:

Step 4: Volume a: 0 mL NaOH added

In this case, the moles of NaOH added are 0. We only have the initial moles of acetic acid, and no acetate is formed. Calculate the initial concentration of acetic acid, and use the \(pK_a\) value in the Henderson-Hasselbalch equation to find the pH.

Step 5: Volume b: 17.5 mL NaOH added

Calculate the moles of NaOH added, determine the change in moles of acetic acid and formation of acetate ions, find the final concentrations, and use these in the Henderson-Hasselbalch equation to find the pH.

Step 6: Volume c: 34.5 mL NaOH added

Perform the same calculations as in previous steps, but with 34.5 mL NaOH added.

Step 7: Volume d: 35.0 mL NaOH added

Perform the same calculations as in previous steps, but with 35.0 mL NaOH added.

Step 8: Volume e: 35.5 mL NaOH added

Perform the same calculations as in previous steps, but with 35.5 mL NaOH added.

Step 9: Volume f: 50.0 mL NaOH added

Perform the same calculations as in previous steps, but with 50.0 mL NaOH added. Once all these calculations are complete, you'll have the pH values for each of the given volumes of NaOH added during the titration.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental tool used to calculate the pH of a buffer solution. A buffer solution typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid. This equation is especially useful in our titration problem involving acetic acid, a weak acid, and sodium hydroxide, a strong base which leads to the formation of a buffer as the reaction proceeds.

• The equation is given by: \[\mathrm{pH} = pK_{a} + \log\left(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\right)\] where \([\mathrm{A}^{-}]\) represents the concentration of the conjugate base (in this case, acetate ions), and \([\mathrm{HA}]\) is the concentration of the undissociated weak acid (acetic acid).
• The pKa is the negative logarithm of the acid dissociation constant \(K_a\) and it's a measure of the strength of the acid. For acetic acid, \(K_a = 1.8 \times 10^{-5}\), so \(pK_a = -\log(1.8 \times 10^{-5})\).

This equation simplifies the process of determining the pH of the buffer solution as one knows both concentrations and the \(pK_a\). It's practically useful when managing scenarios involving titrations and buffer systems.

Acetic Acid

Acetic acid, also known by its chemical formula \(\text{CH}_3\text{COOH}\), is a common weak acid that participates in acid-base reactions. In the given problem, we have a 0.150 M solution of acetic acid which is being titrated by an equivalent molarity of sodium hydroxide, a strong base.

• Acetic acid can donate a proton (H\(^+\)) to become its conjugate base, acetate (\(\text{CH}_3\text{COO}^-\)). This transformation plays a key role in the buffer action during titration.
• Because it's a weak acid, it only partially ionizes in solution, which means some molecules remain in their molecular form, while others dissociate into ions.
• The weak nature of acetic acid makes it perfect for creating buffer solutions when combined with its conjugate base, acetate, which is formed as sodium hydroxide neutralizes the acid during titration.

The behavior of acetic acid in such reactions is often predictable, allowing chemists to use it as a standard example while studying acid-base chemistry.

Conjugate Base

In an acid-base reaction, the conjugate base forms when an acid donates a proton. Here, the conjugate base of acetic acid is acetate \((\text{CH}_3\text{COO}^-)\). This relationship is crucial in understanding how buffer solutions work, especially during an acid-base titration like the one described.The dynamic between acetic acid and its conjugate base can be affected by:

• The amount of NaOH added, which will directly affect the neutralization and thus the amount of acetate formed.
• The ratio of acetic acid to acetate, which directly influences the pH of the solution according to the Henderson-Hasselbalch equation.
• The presence of acetate as a result of each neutralization step stabilizes the solution’s pH over a certain range, making endurance of pH variations possible during the titration process.

Understanding the role of acetate, or any conjugate base, is vital in making predictions about the pH changes happening throughout a titration process.

pH Calculation

In problems involving titration, calculating the pH is an essential step. pH is a measure of hydronium ion concentration and is a crucial indicator of a solution's acidity or basicity. Performing pH calculations during titration involves:

• The determination of the moles of acid and base. During a titration, as sodium hydroxide is added to acetic acid, we need to know the initial and final concentrations of each to calculate the remaining, or converted, moles after reaction.
• Using the Henderson-Hasselbalch equation to determine the pH, leveraging the known concentrations of the weak acid and its conjugate base, especially when you create a buffer solution.
• Understanding the titration curve. Initially, before any base is added, the pH is low. As the reaction proceeds and equivalence points are reached, the pH sharply changes resulting in numeric pH values forecast by the equation.

Mastery of titration and pH calculations require multiple practices but provide deep insights into acid-base chemistry dynamics. Each calculation reveals how the pH of a solution shifts in response to changing concentrations and volumes.