Problem 84 - Chemistry The Central Science

Question

The distinctive odor of vinegar is due to acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) which reacts with sodium hydroxide in the following fashion: $$\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow$$ 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

There are 80.56 grams of acetic acid in a 1.00-quart sample of this vinegar.

Step 1: Write down the balanced chemical equation

We are given the balanced chemical reaction equation: $$CH_3COOH_{(aq)} + NaOH_{(aq)} \longrightarrow CH_3COONa_{(aq)} + H_2O_{(l)}$$

Step 2: Calculate the amount of moles of NaOH used in the titration

We are given that 42.5 mL of 0.115 M NaOH is needed to reach the equivalence point in a titration. To determine the number of moles of NaOH used, we can multiply the volume of the NaOH solution by its molarity (concentration). Make sure to convert the volume from mL to L (divide by 1000): $$n_{NaOH} = 42.5 \times 10^{-3}\,L \times 0.115\,M = 0.0048875\,mol$$

Step 3: Determine the moles of acetic acid present in the vinegar

Since the stoichiometry of the balanced chemical equation tells us that one mole of acetic acid reacts with one mole of NaOH, the number of moles of acetic acid present in the vinegar is equal to the number of moles of NaOH used in the titration. Thus, we have: $$n_{CH_3COOH} = n_{NaOH} = 0.0048875\,mol$$

Step 4: Determine the molarity of acetic acid in the vinegar

The volume of the vinegar titrated is given as 3.45 mL. To find the concentration (molarity) of acetic acid in the vinegar, we can use the equation: $$M = \frac{n}{V}$$ where \(M\) is the molarity, \(n\) is the amount of moles, and \(V\) is the volume. Make sure to convert the volume from mL to L (divide by 1000): $$M_{CH_3COOH} = \frac{0.0048875\,mol}{3.45 \times 10^{-3}\,L} = 1.4174\,M$$

Step 5: Determine the grams of acetic acid in a 1.00-quart sample

We have found the molarity of acetic acid in the vinegar sample to be 1.4174 M. Now, we need to determine the grams of acetic acid in a 1.00-quart sample. We need to convert the volume from quarts to liters (1 qt = 0.946353 L): $$V_{sample} = 1.00\,qt \times 0.946353\,\frac{L}{qt} = 0.946353\,L$$ Next, we can convert the molarity of acetic acid into the moles of acetic acid present in a 1.00-quart sample: $$n_{CH_3COOH_{sample}} = 1.4174\,M \times 0.946353\,L = 1.3416\,mol$$ Lastly, we can convert the moles of acetic acid to grams by multiplying the moles by the molar mass of acetic acid (\(MM_{CH_3COOH} = 60.05\,\frac{g}{mol}\)): $$m_{CH_3COOH_{sample}} = 1.3416\,mol \times 60.05\,\frac{g}{mol} = 80.56\,g$$ Therefore, there are 80.56 grams of acetic acid in a 1.00-quart sample of this vinegar.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand how substances react with each other. It involves using a balanced chemical equation to find the relationships between the amounts of reactants and products.
For the reaction between acetic acid and sodium hydroxide, the balanced equation is: \[ \mathrm{CH}_3\mathrm{COOH}_{(aq)} + \mathrm{NaOH}_{(aq)} \longrightarrow \mathrm{CH}_3\mathrm{COONa}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \] This equation reveals that one mole of acetic acid reacts with one mole of NaOH to produce one mole of sodium acetate and water.
Understanding stoichiometry is crucial as it tells us the exact proportions needed to react the substances fully. Without it, we wouldn't know how much of each substance to use, and we might have leftover reactants or insufficient products. This **1:1 ratio** is key in solving problems related to acid-base titration.

By using this stoichiometric relationship, we can directly relate the moles of NaOH used in a titration to the moles of acetic acid present.

In the exercise, we determined that the moles of NaOH used were 0.0048875 mol. Thanks to stoichiometry, we knew that this also meant 0.0048875 mol of acetic acid was present in the vinegar.

Molarity Calculation

Molarity, often symbolized as \(M\), is a way to express the concentration of a solution. It is defined as moles of solute per liter of solution.
Understanding and calculating molarity are necessary skills for any chemist or student engaged in chemical experiments like titration.
In our exercise, we needed to find out the molarity of acetic acid in the vinegar. Here's how we do it. First, remember the formula for molarity: \[ M = \frac{n}{V} \] where \(n\) is the number of moles and \(V\) is the volume in liters.
With the moles of acetic acid calculated as 0.0048875 mol and the volume of vinegar used as 3.45 mL (or 0.00345 L when converted), the molarity \(M\) of the acetic acid is: \[ M_{\mathrm{CH}_3\mathrm{COOH}} = \frac{0.0048875}{0.00345} = 1.4174 \, M \] This means every liter of vinegar solution contains 1.4174 moles of acetic acid, helping us to understand its concentration level. This calculation was essential to determine the mass of acetic acid in a larger volume later in the problem.

Chemical Reactions

Chemical reactions involve transforming one or more substances, called reactants, into new substances, known as products. The characteristics of chemical reactions are expressed in chemical equations, which show the substances involved and the quantities required or produced.
In the vinegar titration exercise, the chemical reaction is presented as: \[ \mathrm{CH}_3\mathrm{COOH}_{(aq)} + \mathrm{NaOH}_{(aq)} \longrightarrow \mathrm{CH}_3\mathrm{COONa}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \] This is a classic example of an acid-base reaction, where acetic acid (the acid) reacts with sodium hydroxide (the base). The products of this reaction are sodium acetate (a salt) and water.
Key concepts from this reaction include the **neutralization** process. Neutralization involves the reaction of an acid and a base to form water and a salt. This reaction typically occurs in aqueous (water-based) solutions and is fundamental in processes such as titration, where precise measurement of concentrations is needed.
Knowing how chemical reactions work is essential for predicting outcomes in experiments and industrial processes, thus ensuring safety and efficiency. Through understanding this specific reaction, we've determined the necessary conditions to measure the amount of acetic acid in vinegar accurately.