Problem 84 - Chemistry The Central Science

Question

The distinctive odor of vinegar is due to aceticacid, \(\mathrm{CH}_{3} \mathrm{COOH},\) which reacts with sodium hydroxide according to: $$\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow_{\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCH}_{3} \mathrm{COO}(a q)}$$ If 3.45 \(\mathrm{mL}\) of vinegar needs 42.5 \(\mathrm{mL}\) of 0.115 \(\mathrm{M} \mathrm{NaOH}\) to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 -qt sample of this vinegar?

Step-by-Step Solution

Answer

In the given titration, 3.45 mL of vinegar containing acetic acid reacts with 42.5 mL of 0.115 M NaOH to reach equivalence. The balanced chemical equation for the reaction is \(CH_3COOH(aq) + NaOH(aq) \longrightarrow H_2O(l) + NaCH_3COO(aq)\). After calculating the moles of NaOH and acetic acid based on the titration data, we find the concentration of acetic acid in the vinegar sample to be 1.417 M. In a 1.00-qt (0.946353 L) sample of this vinegar, there are approximately 1.341 moles of acetic acid. Therefore, the mass of acetic acid in a 1.00-qt sample is roughly 80.51 g.

Step 1: Moles of NaOH

Calculate the moles of sodium hydroxide used in the titration using the given volume and concentration: Moles = Volume (L) × Concentration (M) Note that volume should be converted from mL to L: 42.5 mL = 42.5 / 1000 L = 0.0425 L Moles of NaOH = 0.0425 L × 0.115 M Moles of NaOH = 0.0048875 moles

Step 2: Moles of Acetic Acid

Since the stoichiometry of the reaction is 1:1, the moles of acetic acid are equal to the moles of NaOH. Moles of Acetic Acid = Moles of NaOH Moles of Acetic Acid = 0.0048875 moles

Step 3: Concentration of Acetic Acid

Find the concentration of acetic acid in the 3.45 mL vinegar sample by dividing the moles of acetic acid by the volume (in L): Concentration = Moles / Volume (L) Convert 3.45 mL to L: 3.45 mL = 3.45 / 1000 L = 0.00345 L Concentration of Acetic Acid = 0.0048875 moles / 0.00345 L Concentration of Acetic Acid = 1.417 M

Step 4: Volume of 1.00-qt Sample

Convert the volume of the 1.00-qt vinegar sample to liters: 1 qt = 0.946353 L

Step 5: Moles of Acetic Acid in 1.00-qt Sample

Calculate the moles of acetic acid in the 1.00-qt vinegar sample using the concentration: Moles = Volume (L) × Concentration (M) Moles = 0.946353 L × 1.417 M Moles = 1.341 moles

Step 6: Mass of Acetic Acid in 1.00-qt Sample

Calculate the mass of acetic acid in the 1.00-qt vinegar sample by multiplying the moles of acetic acid by the molar mass of acetic acid: Mass = Moles × Molar Mass Molar Mass of Acetic Acid = 60.05 g/mol Mass of Acetic Acid = 1.341 moles × 60.05 g/mol Mass of Acetic Acid = 80.51 g Therefore, there are approximately 80.51 grams of acetic acid in a 1.00-qt sample of this vinegar.

Key Concepts

Acetic Acid

Acetic acid, known scientifically as \(\mathrm{CH}_3 \mathrm{COOH}\), is the component responsible for vinegar's distinctive smell. It’s a common household chemical, and is used not just in cooking but also in a variety of industrial processes. Understanding its chemical behaviors helps in exploring its reactive nature in solutions.
In a titration, acetic acid acts as an "acid" which means it can donate protons to bases. The reaction between acetic acid and sodium hydroxide is a classic example of an acid-base reaction where the acid loses a proton to form water and a salt.
This reaction is significant in many fields, particularly in determining the concentration of acetic acid in solutions such as vinegar, allowing us to know how strong or weak the vinegar is in terms of its acidity.
Its molecular composition allows it to easily dissociate in water, making it effective in neutralizing bases.

Sodium Hydroxide

Sodium hydroxide, \(\mathrm{NaOH}\), is a powerful base often used in chemical titrations. In the context of our acetic acid and sodium hydroxide titration, it plays a crucial role in determining the acidity of a solution.
NaOH is known for its ability to easily dissociate in water, providing hydroxide ions \(\mathrm{OH}^-\), which are essential for neutralizing acids such as acetic acid. This base is not only used in laboratories but also in industrial settings for soap making, among other processes.
During a titration involving \(\mathrm{NaOH}\) and acetic acid, the amount of \(\mathrm{NaOH}\) used directly correlates to the acetic acid content in the sample. The point at which all acetic acid has reacted is known as the equivalence point, marking the completion of the reaction where the amount of acid exactly neutralizes the base.

Chemical Stoichiometry

Chemical stoichiometry is the calculation of reactants and products in chemical reactions, which plays a vital role in understanding the titration of acetic acid and sodium hydroxide. Stoichiometry allows chemists to predict the quantities required and expected in chemical reactions.
For the reaction of acetic acid and sodium hydroxide, the stoichiometric relationship is 1:1. This means one mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of water and one mole of sodium acetate, \(\mathrm{NaCH}_3 \mathrm{COO}\) .
Using stoichiometry, the moles of sodium hydroxide used in a titration can directly determine the moles of acetic acid in the sample. This relation simplifies the process of calculating the unknown concentrations or quantities needed for completing the reaction efficiently.

Molar Concentration

Molar concentration, often referred to as molarity (M), is a way of expressing the concentration of a substance in a liquid. It is defined as the number of moles of a solute (in this case, acetic acid) present in one liter of solution.
To calculate the molarity, you divide the number of moles of solute by the liters of solution. For example, in our exercise, finding the concentration of acetic acid involves calculating from its moles and the sample volume.

• First, convert the sample volume from milliliters to liters.
• Then, calculate the moles from the previous stoichiometric calculations.
• Finally, use the formula: \(\text{Molarity} = \frac{\text{Moles of Solute}}{\text{Liters of Solution}}\)

Molar concentration is a fundamental concept in chemistry, enabling chemists to work with precise quantities and predict the outcomes of chemical reactions more accurately. It plays a vital role in laboratory settings, industrial processes, and even in daily life applications where solution concentrations are crucial.