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Chapter 4: Problem 77

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of \(1.049 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate the molarity of a solution of acetic acid made by dissolving \(20.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) in enough water to make \(250.0 \mathrm{~mL}\) of solution.

Short Answer

Expert verified
The molarity of the solution is approximately 1.396 M.

Step by step solution

01

Calculate the Mass of Acetic Acid

First, find the mass of the glacial acetic acid. Use the density formula:\[\text{Mass} = \text{Density} \times \text{Volume}\]Given the density \(1.049 \text{ g/mL}\) and the volume \(20.00 \text{ mL}\), the mass of acetic acid is:\[\text{Mass} = 1.049 \text{ g/mL} \times 20.00 \text{ mL} = 20.98 \text{ g}\]
02

Calculate the Moles of Acetic Acid

Next, convert the mass of acetic acid to moles. The molecular weight of acetic acid \((\text{CH}_3\text{COOH})\) is approximately \(60.05 \text{ g/mol}\). Use the formula:\[\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\]Thus, the moles of acetic acid are:\[\text{Moles} = \frac{20.98 \text{ g}}{60.05 \text{ g/mol}} \approx 0.349 \text{ mol}\]
03

Calculate the Molarity of the Solution

Finally, calculate the molarity of the solution. Molarity \((M)\) is the number of moles of solute per liter of solution. The volume of the solution is \(250.0 \text{ mL} = 0.2500 \text{ L}\). \[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]\[\text{Molarity} = \frac{0.349 \text{ mol}}{0.2500 \text{ L}} = 1.396 \text{ M}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density
Density is a fundamental concept in chemistry that defines how much mass a substance has in a given volume. It's expressed in terms of mass per unit volume, commonly grams per milliliter (g/mL) for liquids.
For example, glacial acetic acid has a density of 1.049 g/mL. This means each milliliter of acetic acid weighs 1.049 grams. Understanding density helps in calculating the mass of a substance when you know the volume, or vice versa.
  • The formula to calculate the mass from density and volume is: \(\text{Mass} = \text{Density} \times \text{Volume}\).
  • Using this equation, you can determine the mass of the liquid needed for further calculations.
This is particularly useful when preparing solutions, ensuring that the proportions are accurate for the desired concentration.
Chemical Calculations
Chemical calculations are crucial for accurately preparing solutions and predicting reactions. These calculations often involve converting between grams, moles, and molarity.
The steps typically include:
  • Calculating the mass of a substance using its density and volume.
  • Converting the mass to moles using its molar mass, which is the mass of one mole of a substance.
  • Determining the molarity of the solution, which is the concentration of a solute in a solution.
The molecular weight of acetic acid, for instance, is about 60.05 g/mol, an essential value for converting between grams and moles.
Successfully carrying out chemical calculations requires attention to detailed measurements and ensures the accuracy of experimental results.
Acetic Acid
Acetic acid is an organic compound with the chemical formula \(\text{CH}_3\text{COOH}\). It's a simple carboxylic acid and a key component of vinegar. When pure, it's referred to as glacial acetic acid due to its ice-like appearance at lower temperatures.
Important characteristics include:
  • It is a clear, colorless liquid.
  • It has a distinctive sour taste and pungent smell.
  • Its density is 1.049 g/mL at 25掳C, making it slightly denser than water.
Understanding acetic acid is vital because it's widely used in chemical synthesis, food production, and as a solvent in laboratory settings.
In solutions chemistry, it acts as both a solvent and a solute depending on the preparation and application.
Solutions Chemistry
Solutions chemistry involves understanding how substances interact when mixed. A solution is composed of a solute and a solvent. Molarity is a central concept here, representing the concentration of the solute in a given volume of solution.
In the context of the original exercise:
  • Acetic acid acts as the solute.
  • Water is the solvent, used to dissolve the acetic acid.
  • The solution's final volume is 250.0 mL or 0.250 L.
Molarity (\(M\)) is defined as the moles of solute per liter of solution, calculated using: \(\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\).
Mastery of solutions chemistry allows for precise formulation and analysis of chemical solutions and is essential for research, industrial processes, and educational purposes.

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Most popular questions from this chapter

The concept of chemical equilibrium is very important. Which one of the following statements is the most correct way to think about equilibrium? (a) If a system is at equilibrium, nothing is happening. (b) If a system is at equilibrium, the rate of the forward reaction is equal to the rate of the back reaction. (c) If a system is at equilibrium, the product concentration is changing over time. [Section 4.1]

Citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) is a triprotic acid. It occurs naturally in citrus fruits like lemons and has applications in food flavouring and preservatives. A solution containing an unknown concentration of the acid is titrated with KOH. It requires \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{KOH}\) solution to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the citric acid solution.

An aqueous solution contains \(1.2 \mathrm{~m} M\) of total ions. (a) If the solution is \(\mathrm{NaCl}(a q),\) what is the concentration of chloride ion? \((\mathbf{b})\) If the solution is \(\mathrm{FeCl}_{3}(a q),\) what is the concentration of chloride ion?

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for 10.0 min at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) \((\mathbf{c})\) Is this manufacturer in compliance with regulations?

Would you expect that an anion would be physically closer to the oxygen or to the hydrogens of water molecules that surround it in solution?
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