Problem 45 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

The pH of the solution after various volumes of \(0.150 \ M \ NaOH\) is added to a \(35.0 \ mL\) sample of \(0.150 \ M \ CH_3COOH\) can be calculated using the acid dissociation constant and stoichiometry. The pH values are as follows: (a) after \(0 \ mL\) of NaOH, pH = \(2.86\); (b) after \(17.5 \ mL\) of NaOH, pH = \(4.74\); (c) after \(34.5 \ mL\) of NaOH, pH = \(7.95); (d) after \(35.0 \ mL\) of NaOH, pH = \(8.82\); (e) after \(35.5 \ mL\) of NaOH, pH = \(9.23\); (f) after \(50.0 \ mL\) of NaOH, pH = \(10.21\).

Step 1: The balanced chemical equation for the reaction of acetic acid (CH3COOH) and sodium hydroxide (NaOH) is: \( CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \) #Step 2: Calculate the initial moles of acetic acid and moles of NaOH added#

The initial moles of acetic acid can be calculated using the given concentration and volume of the solution: Initial moles of \(CH_3COOH =\) Concentration × Volume Initial moles of \(CH_3COOH = 0.150 M \times 0.035 L = 0.00525 mol\) At each step, calculate the moles of NaOH added using the given volumes: Moles of \(NaOH_{added} =\) Concentration × Volume We will do this for each volume given in the problem (0 mL, 17.5 mL, 34.5 mL, 35.0 mL, 35.5 mL, and 50.0 mL). #Step 3: Determine the moles of the remaining acetic acid, moles of the produced acetate ion, and moles of the remaining NaOH#

Step 2: For each volume of added NaOH, calculate the moles of the remaining acetic acid, the moles of the produced acetate ion, and the moles of the remaining NaOH using stoichiometry and an ice table: Moles of \(CH_3COOH_{remaining} =\) Initial moles of \(CH_3COOH -\) Moles of \(NaOH_{added}\) (if NaOH is limiting reagent) Moles of \(CH_3COO^-\) = Moles of \(NaOH_{added}\) (if NaOH is limiting reagent) Moles of remaining \(NaOH =\) Moles of \(NaOH_{added} -\) Initial moles of \(CH_3COOH\) (if acetic acid is limiting reagent) Calculate these values for each given volume of added NaOH. #Step 4: Calculate the pH of the solution using the ion concentrations and the acid dissociation constant of acetic acid#

In order to calculate the pH of the solution, we need to use the dissociation constant of acetic acid, which is indicated as \(K_a\): \( K_a = 1.8 \times 10^{-5} \) For each given volume, we will either have: 1) Excess acetic acid: Use the Henderson-Hasselbalch equation to calculate the pH when there is acetic acid and acetate ion: \(pH = pK_a + \log \frac{[CH_3COO^-]}{[CH_3COOH]}\) 2) Excess NaOH: Calculate the concentration of the remaining NaOH in the solution, and then use the negative logarithm to determine the pH of the solution: \[pH = -\log([OH^-])\] Perform these calculations for each volume to find the pH of the solution at each stage of the titration.

Key Concepts

Acetic Acid

Acetic acid, with the chemical formula \( \text{CH}_3\text{COOH} \), is a weak acid that is commonly found in vinegar. Because it is a weak acid, it only partially dissociates in water. This means not all acetic acid molecules donate their protons to water to form hydronium ions \( \text{H}_3\text{O}^+ \) and acetate ions \( \text{CH}_3\text{COO}^- \).

• Formula: \( \text{CH}_3\text{COOH} \)
• Weak acid: partially dissociates in solution
• Commonly found in vinegar

In the context of acid-base titration, we use the properties of acetic acid to analyze its behavior when neutralized by a strong base, like sodium hydroxide \( \text{NaOH} \). By reacting acetic acid with \( \text{NaOH} \), we can create a buffer solution consisting of acetic acid and its conjugate base, acetate.

pH Calculation

The pH of a solution is a measure of its acidity or basicity. It's calculated based on the concentration of hydrogen ions \(\text{H}^+ \) in the solution. For a strong acid or base, this calculation can be straightforward, but for a weak acid like acetic acid, the process involves understanding its partial dissociation.When calculating the pH during a titration, we look at:

• The initial pH of the acetic acid solution before any base is added.
• The change in pH as a base is added, depending on which reagent is in excess and whether a buffer is present.

In a buffer solution with both acetic acid and acetate ion present, the Henderson-Hasselbalch equation is useful for calculating the pH. If the solution reaches the equivalence point, further base addition will switch the focus to calculating the pH based on excess \( \text{NaOH} \).

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a powerful tool used to estimate the pH of a buffer solution, which consists of a weak acid and its conjugate base. This equation relates the pH, the pKa of the weak acid, and the ratio of the concentrations of the conjugate base to the acid:\[\text{pH} = \text{pK}_a + \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right)\]Here, \( [\text{CH}_3\text{COOH}] \) is the concentration of acetic acid, and \( [\text{CH}_3\text{COO}^-] \) is the concentration of the acetate ion. The value of \( \text{pK}_a \) is derived from the acid dissociation constant \( \text{K}_a \) of acetic acid.This equation is particularly useful during titrations when acetic acid and acetate ion both are present, creating a buffer system. It provides a straightforward way to understand how the pH behaves when small amounts of acid or base are added.

Stoichiometry

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. In titration, stoichiometry helps us determine how much base is needed to react completely with a given amount of acid. It involves using the balanced chemical equation to relate quantities of reactants to each other.

• Balanced equation: \( \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \)
• Calculating moles: Use concentration and volume (e.g., initial moles of acetic acid \( = 0.150 \text{ M} \times 0.035 \text{ L} = 0.00525 \text{ mol} \)).
• Identify limiting reagent: Helps determine the end of reaction completion.

By calculating moles at each step, stoichiometry helps us understand how much of each reactant remains and how much product is formed as the titration progresses.

Acid Dissociation Constant

The acid dissociation constant, \( \text{K}_a \), is a crucial concept when dealing with weak acids like acetic acid. It quantifies the strength of an acid in solution, specifically, how well an acid molecule disassociates into its ions. For acetic acid, the \( \text{K}_a \) value is \( 1.8 \times 10^{-5} \).This low \( \text{K}_a \) value indicates that acetic acid does not fully ionize in water, reflecting its nature as a weak acid.

• The dissociation equation: \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \)
• Weaker acids have smaller \( \text{K}_a \). Larger \( \text{K}_a \) would imply a stronger acid, more complete dissociation.

Understanding \( \text{K}_a \) helps in predicting how the acid behaves in different chemical environments, such as during a titration or when part of a buffer solution.