Problem 62 - General Chemistry: Principles and Modern Applications

Question

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

Step-by-Step Solution

Answer

The molarity of the original \( \mathrm{HCl} \) is \( 0.0441 \mathrm{ M} \)

Step 1: Calculate moles of \(\mathrm{CaCO}_3\)

First, calculate the moles of \(\mathrm{CaCO}_3\) from the given mass. The molar mass of \(\mathrm{CaCO}_3\) is 40.08 + 12.01 + (3 * 16.00) = 100.09 g/mol. So, there are \(0.1000 \mathrm{g} / 100.09 \mathrm{ g/mol} = 0.001 \mathrm{ mol}\) of \(\mathrm{CaCO}_3\)

Step 2: Calculate moles of \(\mathrm{HCl}\) reacted with \(\mathrm{CaCO}_3\)

From the balanced chemical equation, \(\mathrm{CaCO}_3\) reacts with \(\mathrm{HCl}\) in a 1:2 ratio. Therefore, the moles of \(\mathrm{HCl}\) reacted with \(\mathrm{CaCO}_3\) are \(2 * 0.001 \mathrm{moles} = 0.002 \mathrm{ mol}\)

Step 3: Calculate moles of excess \(\mathrm{HCl}\) titrated

To find the moles of excess \(\mathrm{HCl}\), use the volume and molarity of \(\mathrm{Ba(OH)}_2\) used in titration. The moles of \(\mathrm{Ba(OH)}_2\) are \(43.82 \mathrm{mL} * 0.01185 \mathrm{M} = 0.000519 \mathrm{ mol}\). From the balanced chemical reaction, \(\mathrm{Ba(OH)}_2\) reacts with \(\mathrm{HCl}\) in a 1:2 ratio, so the moles of \(\mathrm{HCl}\) are \(2 * 0.000519 \mathrm{ mol} = 0.001038 \mathrm{ mol}\)

Step 4: Calculate total moles of \(\mathrm{HCl}\) and molarity

The total moles of \(\mathrm{HCl}\) are the sum of the \(\mathrm{HCl}\) that reacted with \(\mathrm{CaCO}_3\) and the excess \(\mathrm{HCl}\). The total moles = \(0.002 \mathrm{ mol} + 0.001038 \mathrm{ mol} = 0.003038 \mathrm{ mol}\). The molarity is the total moles divided by the total volume in liters, which is \((25.00 \mathrm{mL} + 43.82 \mathrm{mL}) / 1000 = 0.06882 \mathrm{L}\). So, the molarity of the original \(\mathrm{HCl}\) is \(0.003038 \mathrm{ mol} / 0.06882 \mathrm{ L} = 0.0441 \mathrm{ M}\)

Key Concepts

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It relies on the principle of the conservation of mass where the mass of reactants equals the mass of products. To solve stoichiometry problems, balanced chemical equations are fundamental, as they show the mole relationships between reactants and products.

In the provided exercise, stoichiometry helps us understand how \(\mathrm{CaCO}_{3}\) and \(\mathrm{HCl}\) react. The balanced equation \(\mathrm{CaCO}_{3} + 2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_{2} + \mathrm{H}_{2}O + \mathrm{CO}_{2}\)\ shows that one mole of \(\mathrm{CaCO}_{3}\) reacts with two moles of \(\mathrm{HCl}\). By identifying the ratio, we understand that every mole of \(\mathrm{CaCO}{3}\) requires two moles of \(\mathrm{HCl}\) to fully react.

These relationships are pivotal in calculating how much \(\mathrm{HCl}\) was consumed during the reaction with \(\mathrm{CaCO}{3}\). In essence, mastering stoichiometry equips students with tools to calculate quantities involved in chemical reactions.

Titration

Titration is a technique used to determine the concentration of a solution by reacting it with a solution of known concentration. In this exercise, titration was employed to find the excess moles of \(\mathrm{HCl}\) remaining after reacting part of it with \(\mathrm{CaCO}_{3}\).

Key Steps in the Titration Process:

• Preparing a standard solution of \(\mathrm{Ba(OH)}_{2}\) with a known molarity of 0.01185 M.
• Adding this solution to the \(\mathrm{HCl}\) sample until the reaction completes as indicated by a change (endpoint).
• Using the stoichiometric relationship from the balanced equation to calculate the moles of \(\mathrm{HCl}\) that reacted with \(\mathrm{Ba(OH)}_{2}\).

During the titration in this practical scenario, \(43.82 \mathrm{mL}\) of \(\mathrm{Ba(OH)}_{2}\) was used, and the relationship between \(\mathrm{HCl}\) and \(\mathrm{Ba(OH)}_{2}\) is crucial since it tells us that two moles of \(\mathrm{HCl}\) react with one mole of \(\mathrm{Ba(OH)}_{2}\).

Understanding titration enables students to determine unknown concentrations, providing critical insight needed for real-world laboratory applications.

Molarity Calculation

Molarity is a way to express the concentration of a solute in a solution, specifically in moles per liter (M). In the given problem, calculating the molarity of the original \(\mathrm{HCl}\) solution was the main goal.

Steps to Determine Molarity:

• Calculate the total moles of solute present. This combines both the moles of \(\mathrm{HCl}\) initially reacting with \(\mathrm{CaCO}_{3}\) and those remaining excess, which were found using titration.
• Determine the total volume of solution in liters by converting the combined volume from milliliters (\[(25.00 + 43.82)/1000\]).
• Divide the total moles by the total volume in liters.

In this problem, combining the moles of \(\mathrm{HCl}\) gave us 0.003038 moles. With a total solution volume of 0.06882 liters, the molarity calculation was simple: \[\text{Molarity} = \frac{0.003038 \text{ mol}}{0.06882 \text{ L}} = 0.0441 \text{ M}\].

Being skilled in calculating molarity is essential in many fields within chemistry. This exercise exemplifies practical molarity calculations, an essential skill for students mastering chemical solutions.