Chemical Titration Analysis: Ammonia and Vinegar Acidity

Acids and bases are utilized in volumetric analysis to determine the concentration of unknown substances. This document details two common acid-base titrations: determining ammonia concentration and calculating vinegar acidity.

Part 1: Ammonia Concentration Determination

The goal is to determine the amount of pure ammonia (NH₃) in a solution using an acid titration (valorization).

Titration Data and Dilution

• Initial Sample: 20 mL of concentrated ammonia solution.
• Dilution: The 20 mL sample was diluted to a final volume of 500 mL.
• Aliquot Titrated: 25 mL of the diluted solution was used.
• Titrant: 22.7 mL of 0.5 M HCl was required to reach the endpoint.

Calculation of Moles

The reaction is NH₃ + HCl → NH₄Cl (1:1 stoichiometry).

1. Calculate moles of HCl spent:

$$n_{\text{HCl}} = C \times V = 0.5 \text{ M} \times 0.0227 \text{ L} = 0.01135 \text{ mol}$$

2. Moles of NH₃ in the 25 mL aliquot (based on 1:1 ratio):

$$n_{\text{NH}_3, 25\text{mL}} = 0.01135 \text{ mol}$$

3. Total moles of NH₃ in the 500 mL diluted solution:

$$n_{\text{NH}_3, 500\text{mL}} = 0.01135 \text{ mol} \times \left(\frac{500 \text{ mL}}{25 \text{ mL}}\right) = 0.227 \text{ mol}$$

Mass and Final Concentration

The molar mass of NH₃ is approximately 17 g/mol.

4. Calculate the total mass of NH₃ in 500 mL:

$$m_{\text{NH}_3} = n \times \text{MM} = 0.227 \text{ mol} \times 17 \text{ g/mol} = 3.859 \text{ g}$$

5. Calculate the concentration in grams per 100 mL of diluted solution:

$$\text{Concentration} = 3.859 \text{ g} \times \left(\frac{100 \text{ mL}}{500 \text{ mL}}\right) = 0.7718 \text{ g/100 mL}$$

Part 2: Determining Acidity in Vinegar

Vinegar acidity (total volatile acids) is typically expressed in grams of acetic acid (CH₃COOH) per 100 mL of vinegar. This measurement is often referred to as the degree of acidity.

Titration Reaction and Materials

Acetic acid is the primary organic component in vinegar (formula CH₃COOH). The titration uses sodium hydroxide (NaOH):

$$\text{CH}_3\text{COOH}_{\text{(aq)}} + \text{NaOH}_{\text{(aq)}} \rightarrow \text{CH}_3\text{COONa}_{\text{(aq)}} + \text{H}_2\text{O}$$

Materials Required: Test tube, flask, burette, 100 mL volumetric flask, Erlenmeyer flask, and phenolphthalein indicator.

Hazard: Sodium hydroxide is corrosive. Appropriate safety measures must be taken.

Procedure for Dilution and Titration

1. Take 10 mL of vinegar and transfer it to a volumetric flask.
2. Add water to dilute the sample up to 100 mL.
3. Transfer 25 mL of this diluted solution into an Erlenmeyer flask.
4. Add a few drops of phenolphthalein indicator.
5. Prepare approximately 100 mL of a 0.5 M NaOH solution.

Calculating NaOH Concentration

If the mass taken was approximately 1.98 g (MM NaOH ≈ 40 g/mol), the moles are $n = 1.98/40 = 0.0495 \text{ mol}$. The actual concentration prepared in 100 mL (0.1 L) is:

$$C_{\text{NaOH}} = 0.0495 \text{ mol} / 0.1 \text{ L} = 0.495 \text{ M} \approx 0.49 \text{ M}$$

6. Pour the NaOH solution into the burette, ensuring it is primed and free of bubbles. Record the initial volume (e.g., 0.00 mL).
7. Begin adding the base dropwise until the endpoint is reached (the solution turns a persistent light pink).

Observed Titration Volume: $V_{\text{NaOH}} = 4.25 \text{ mL}$.

Numerical Calculations for Acidity

1. Moles of NaOH Spent

$$n_{\text{NaOH}} = 0.49 \text{ M} \times 4.25 \times 10^{-3} \text{ L} = 0.0020825 \text{ mol}$$

(We use $2.08 \times 10^{-3} \text{ mol}$ for subsequent steps, correcting the error in the original text's calculation.)

2. Moles of Acetic Acid in Original Sample

Since the stoichiometry is 1:1, the moles of acid in the 25 mL aliquot is $2.08 \times 10^{-3} \text{ mol}$.

Total moles of acid in the 100 mL diluted solution:

$$n_{\text{Acid, 100mL}} = 2.08 \times 10^{-3} \text{ mol} \times 4 = 8.32 \times 10^{-3} \text{ mol}$$

This quantity ($8.32 \times 10^{-3} \text{ mol}$) represents the moles of acid present in the original 10 mL vinegar sample.

3. Mass and Acidity Degree

Molar Mass of CH₃COOH (MM) = 60 g/mol.

Mass of acid in the 10 mL sample:

$$m_{\text{Acid}} = 8.32 \times 10^{-3} \text{ mol} \times 60 \text{ g/mol} \approx 0.499 \text{ g}$$

Acidity is measured in degrees, defined as the mass of acid in 100 mL of vinegar. Since 0.499 g was found in 10 mL:

$$\text{Acidity} = \frac{0.499 \text{ g}}{10 \text{ mL}} \times 100 \text{ mL} = 4.99 \text{ degrees}$$

The acidity of the vinegar is approximately 4.9 degrees.