Problem 123 - Chemistry The Science in Context

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 concentrated HCl? b. What volume of it would you need to prepare \(0.250 \mathrm{L}\) of \(2.00 \mathrm{M} \mathrm{HCl}^{2}\) c. What mass of sodium hydrogen carbonate would be needed to neutralize the spill if a bottle containing 1.75 L of concentrated HCl dropped on a lab floor and broke open?

Step-by-Step Solution

Answer

Additionally, how much mass of sodium hydrogen carbonate is needed to neutralize the spill if a 1.75 L bottle of concentrated HCl breaks open? Answer: The molarity of concentrated HCl is 11.64 M. To prepare 0.250 L of 2.00 M HCl, 43.0 mL of concentrated HCl is needed. To neutralize the spill of a 1.75 L bottle of concentrated HCl, 1711 g of sodium hydrogen carbonate is required.

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

Density of the solution is given as 1.18 g/mL. To find the mass of HCl in 1L of solution, use the mass percentage of HCl (36.0%). Mass of HCl = mass percentage of HCl × total mass of solution Since 1000 mL = 1L, find the mass of 1L of solution. Total mass of solution = density × volume = 1.18 g/mL × 1000 mL = 1180 g Mass of HCl = 36.0% × 1180 g = (0.36)(1180) = 424.8 g

Step 2: Calculate the moles of HCl

Now, find the moles of HCl using its molar mass (HCl = 1 + 35.5 = 36.5 g/mol). Moles of HCl = mass of HCl / molar mass of HCl = 424.8 g / 36.5 g/mol = 11.64 mol

Step 3: Determine the molarity of concentrated HCl

Molarity = moles of solute / volume of solution in liters Molarity = 11.64 mol / 1 L = 11.64 M a. The molarity of concentrated HCl is 11.64 M. b. Calculate the volume of concentrated HCl needed to prepare 0.250 L of 2.00 M HCl

Step 1: Calculate the moles of HCl needed

Since we are given the desired molarity and volume, we can first find the moles of HCl needed. Moles of HCl = Molarity × volume = 2.00 M × 0.250 L = 0.500 mol

Step 2: Calculate the volume of concentrated HCl

Since we have the molarity of concentrated HCl (11.64 M) and the moles of HCl needed (0.500 mol), we can find the volume of concentrated HCl to be used. Volume = moles / molarity = 0.500 mol / 11.64 M = 0.0430 L or 43.0 mL b. The volume of concentrated HCl needed to prepare 0.250 L of 2.00 M HCl is 43.0 mL. c. Calculate the mass of sodium hydrogen carbonate needed to neutralize the spill

Step 1: Calculate the moles of HCl spilled

Given the volume of concentrated HCl spilled as 1.75 L, we can find the moles of HCl. Moles of HCl = molarity × volume = 11.64 M × 1.75 L = 20.37 mol

Step 2: Determine the stoichiometry of the reaction

The neutralization reaction between HCl and sodium hydrogen carbonate (NaHCO3) is as follows: HCl + NaHCO3 -> NaCl + H2O + CO2 From the balanced equation, 1 mol of NaHCO3 is required to neutralize 1 mol of HCl.

Step 3: Calculate the mass of NaHCO3 needed

Since 20.37 mol of HCl is spilled, 20.37 mol of NaHCO3 is required for neutralization. Calculate the mass of NaHCO3 using its molar mass (NaHCO3 = 23 + 1 + 12 + 3(16) = 84 g/mol). Mass of NaHCO3 = moles × molar mass = 20.37 mol × 84 g/mol = 1711 g c. 1711 g of sodium hydrogen carbonate is needed to neutralize the spill if a bottle containing 1.75 L of concentrated HCl dropped on a lab floor and broke open.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It helps us understand how much of each substance is involved and needed to achieve a balanced reaction.

When solving stoichiometric calculations, the following steps typically help:

• Write a balanced chemical equation to understand the mole ratio between reactants and products.
• Calculate the moles of the reactants or products using their given mass and molar mass.
• Use the mole ratio from the balanced equation to find the required moles of any other substance involved in the reaction.

In the exercise given, stoichiometry helps determine the mass of sodium hydrogen carbonate (NaHCO3) necessary to neutralize a spill of hydrochloric acid (HCl). The key is acknowledging that one mole of HCl reacts with one mole of NaHCO3, guiding us to calculate the exact amount required for neutralization.

Neutralization reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react to form water and a salt. It's a vital concept in chemistry that is often used to predict the outcome of mixing different substances.

In the context of the problem, when hydrochloric acid (HCl) spills, it needs to be neutralized safely with a base. Sodium hydrogen carbonate, commonly known as baking soda (NaHCO3), is a base that can effectively neutralize HCl. The reaction is:
\[\text{HCl} + \text{NaHCO}_3 \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\]
In this balanced equation, each mole of HCl neutralizes one mole of NaHCO3. Key points about neutralization include:

• Neutralization produces a salt (in this case, NaCl) and water.
• The reaction is often exothermic, meaning it releases heat.
• Appropriate stoichiometry calculations can ensure complete neutralization without excess unreacted acid or base.

By using stoichiometry and the balanced equation, the problem calculates the required mass of NaHCO3 to neutralize the acid spill, ensuring safety in the lab.

Hydrochloric acid density

Density is a measure of mass per unit volume and is critical when dealing with solutions. The density of a solution helps in understanding its concentration and is essential for calculating molarity.

In the exercise, the density of hydrochloric acid is given as 1.18 g/mL. This indicates that every milliliter of the solution has a mass of 1.18 grams. To calculate the molarity of the solution, the following steps are crucial:

• Use the density to find the total mass of a known volume, usually 1 liter, because molarity is expressed as moles per liter.
• Determine the mass of the solute (HCl in this case) using its mass percentage in the solution.
• Calculate molarity by dividing the moles of solute by the volume of the solution in liters.

This approach in the problem enables us to find that the molarity of concentrated HCl is 11.64 M, using both density and mass percent to reveal the concentration details crucial for further calculations, like dilutions and reactions.