A typical solution used in general chemistry laboratories is 3.0 \(M\) HCl. Describe, in detail, the composition of \(2.0 \mathrm{L}\) of a 3.0-M HCI solution. How would 2.0 L of a \(3.0-M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) solution differ from the same quantity of the HCl solution?

Molarity is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution. Molarity is commonly denoted as "M". In this problem, we are given two solutions with 3.0 M concentration: one with HCl as the solute and one with HC₂H₃O₂.

To find the composition of 2.0 L of a 3.0 M HCl solution, first, we need to calculate the moles of HCl using the molarity formula: Moles of solute = Molarity × Volume of solution In our case: Moles of HCl = 3.0 M × 2.0 L = 6.0 moles of HCl The composition of the 2.0 L HCl solution: - 6.0 moles of HCl as solute - Approximately 2.0 L of water as the solvent (assuming HCl is dissolved in water)

Now we'll analyze the 2.0 L of a 3.0 M HC₂H₃O₂ solution. Moles of HC₂H₃O₂ = 3.0 M × 2.0 L = 6.0 moles of HC₂H₃O₂ The composition of the 2.0 L HC₂H₃O₂ solution: - 6.0 moles of HC₂H₃O₂ as solute - Approximately 2.0 L of water as the solvent (assuming HC₂H₃O₂ is dissolved in water)

When comparing the two solutions, we can see that both solutions have the same volume (2.0 L) and molarity (3.0 M). The main difference between these two solutions is the solute: - In the 2.0 L of 3.0 M HCl solution, the solute is Hydrochloric acid (HCl) which is a strong acid. - In the 2.0 L of 3.0 M HC₂H₃O₂ solution, the solute is Acetic acid (HC₂H₃O₂) which is a weak acid. The difference in the strength of the acids determines their degree of ionization in the solution, which in turn determines the pH values and other chemical properties of the solutions. In general, the HCl solution will have a lower pH (more acidic) and will have more dissolved ions due to its complete ionization, while the HC₂H₃O₂ solution will have a higher pH (less acidic) and will have fewer dissolved ions due to its partial ionization.