Problem 113 - Chemistry: An Atoms-Focused Approach

Question

According to the label on a bottle of concentrated hydrochloric acid, the contents are \(36.0 \%\) HCl by mass and have a density of \(1.18 \mathrm{g} / \mathrm{mL}\) a. What is the molarity of this concentrated HCl? b. What volume of it would you need to prepare \(0.250 \mathrm{L}\) of \(2.00 \mathrm{MHCl} ?\) c. What mass of sodium hydrogen carbonate would be needed to neutralize the spill if a bottle containing 1.75 L of this concentrated HCl dropped on a lab floor and broke open?

Step-by-Step Solution

Answer

Answer: To neutralize a spill of 1.75 L concentrated HCl, approximately 1.74 kg of sodium hydrogen carbonate (NaHCO3) is required.

Step 1: Calculate the mass of HCl in 1 L solution

We can convert the percentage to mass and then calculate the mass of HCl in 1000 mL (1 L). Mass of HCl: \(\frac{36.0}{100} \times (1000 \cdot 1.18) = 432\,\mathrm{g}\)

Step 2: Calculate the moles of HCl

Use the molar mass of HCl (36.5 g/mol) to find the moles of HCl in 1 L solution: Moles of HCl = \(\frac{432\,\mathrm{g}}{36.5\,\mathrm{g/mol}} = 11.84\,\mathrm{moles}\)

Step 3: Calculate molarity

Molarity is calculated as moles of solute per liter of solution. Therefore, the molarity of the concentrated HCl is: Molarity = \(11.84\,\mathrm{M}\) #b. Calculate the volume of concentrated HCl required to prepare 0.250 L of 2.00 M HCl#

Step 1: Calculate the moles of HCl required

In order to prepare 0.250 L of 2.00 M HCl, the number of HCl moles required is: Moles of HCl = 2.00 M \(\times\) 0.250 L = 0.50 moles

Step 2: Calculate the volume of concentrated HCl

Now, using the molarity of concentrated HCl found in part (a): Volume of concentrated HCl = \(\frac{0.50\,\mathrm{moles}}{11.84\,\mathrm{M}} = 0.0422\,\mathrm{L}\) (or 42.2 mL) #c. Calculate the mass of sodium hydrogen carbonate required to neutralize the spill#

Step 1: Calculate moles of HCl in 1.75 L spill

First, find the moles of HCl in the 1.75 L spill: Moles of HCl = 11.84 M \(\times\) 1.75 L = 20.72 moles

Step 2: Calculate moles of NaHCO3 required for neutralizing

The reaction between HCl and NaHCO3 is: HCl + NaHCO3 \(\rightarrow\) NaCl + CO2 + H2O. From the balanced equation, we can see that one mole of HCl reacts with one mole of NaHCO3. Therefore, 20.72 moles of NaHCO3 are required.

Step 3: Calculate the mass of NaHCO3

Using the molar mass of NaHCO3 (84.01 g/mol), calculate the required mass: Mass of NaHCO3 = 20.72 moles \(\times\) 84.01 g/mol = 1739.3 g (or approximately 1.74 kg)

Key Concepts

Concentration by Mass Percentage

When we talk about concentration by mass percentage, we are referring to how much of a solute is present in a solution compared to the total mass of the solution. It's a simple way to express concentration. In the formula, this percentage is denoted as \[\% ext{ by mass} = \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100\]For example, if we have a solution that is labeled as 36% hydrochloric acid (HCl), it means that there are 36 grams of HCl in every 100 grams of this solution. This percentage helps in calculating the exact mass of the solute in any given mass of the solution. Knowing the mass percentage is crucial for converting mass to moles, which is often needed for further calculations in chemistry.

Density of Solutions

Density is a measure of how much mass is contained in a given volume of a substance. For solutions, density is an essential property as it often aids in converting between volume and mass. The formula for calculating density is:\[\text{Density} = \frac{\text{mass}}{\text{volume}}\]In the exercise, the density of the hydrochloric acid solution is given as 1.18 g/mL. This indicates that one milliliter of the solution weighs 1.18 grams. With density, we can determine the mass of solution required for calculations and vice versa. For instance, knowing the density helps to calculate the mass of 1000 mL of the 36% HCl solution, simplifying the step to determine how much solute is in that volume.

Neutralization Reaction

Neutralization reactions happen when an acid reacts with a base, producing salt and water. A common scenario is the reaction between hydrochloric acid (HCl) and sodium hydrogen carbonate (NaHCO3), as presented in the exercise. The balanced chemical equation for their reaction is:\[\text{HCl} + \text{NaHCO}_3 \rightarrow \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\]In this reaction, one mole of HCl reacts with one mole of NaHCO3. It's important to understand that this balanced equation is crucial for determining the proportions of each reactant required to fully neutralize the other. The stoichiometry of a neutralization reaction is often used to calculate the exact amounts needed, which is essential for laboratory safety and accuracy when preparing solutions.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculations based on balanced chemical equations to find out how much of each substance is involved in a reaction. Key steps in stoichiometry include:

• Balancing the chemical equation to ensure matter conservation.
• Using molar ratios derived from the balanced equation to relate the amounts of different substances.
• Converting between moles, mass, and volume depending on the information and demand of the problem.

In the given exercise, stoichiometry was used to find the molarity of the HCl solution and the mass of sodium hydrogen carbonate needed for neutralization. By knowing how to use molar ratios, students can perform precise calculations that are essential in both laboratory and industrial chemistry applications.