Question

A flask containing \(450 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) HBr was accidentally knocked to the floor. How many grams of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) would you need to put on the spill to neutralize the acid according to the following equation? \(2 \mathrm{HBr}(a q)+\mathrm{K}_{2} \mathrm{CO}_{3}(a q) \longrightarrow 2 \mathrm{KBr}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\).

Short answer

15.5 grams of \(\mathrm{K}_2\mathrm{CO}_3\) are needed to neutralize the spill.

Step-by-step solution

Calculate Moles of HBr

First, calculate the moles of HBr using the volume and molarity of the solution. Moles of HBr = Molarity (\( M \)) \( \times \) Volume (in liters). The volume should be converted to liters from milliliters by dividing by 1000. Thus, \( \text{Moles of HBr} = 0.500 \text{ M} \times 0.450 \text{ L} = 0.225 \text{ moles}.\)

Determine Moles of \(\mathrm{K}_2\mathrm{CO}_3\) Required

Using the balanced chemical equation, relate moles of HBr to moles of \( \mathrm{K}_2\mathrm{CO}_3 \). From the equation, \(2\) moles of HBr react with \(1\) mole of \( \mathrm{K}_2\mathrm{CO}_3 \). Therefore, the moles of \( \mathrm{K}_2\mathrm{CO}_3 \) required is half the moles of HBr: \( \frac{0.225 \text{ moles of HBr}}{2} = 0.1125 \text{ moles}.\)

Calculate Mass of \(\mathrm{K}_2\mathrm{CO}_3\)

Calculate the mass of \( \mathrm{K}_2\mathrm{CO}_3 \) needed by multiplying the moles by its molar mass. The molar mass of \( \mathrm{K}_2\mathrm{CO}_3 \) is \(2(39.1) + 12.0 + 3(16.0) = 138.2\text{ g/mol}\). Therefore, the mass is \(0.1125 \text{ moles} \times 138.2 \text{ g/mol} = 15.5 \text{ grams}.\)

Conclusion

Summarize the calculations. To neutralize the spill, \(15.5\) grams of \(\mathrm{K}_2\mathrm{CO}_3\) are required.

Key concepts

Chemical Reactions

In chemistry, a chemical reaction occurs when substances called reactants transform into different substances known as products. This process involves the breaking and forming of chemical bonds, leading to new arrangements of atoms. The chemical reaction involved in this problem is an acid-base neutralization reaction where hydrobromic acid (HBr) reacts with potassium carbonate (\(\mathrm{K}_2\mathrm{CO}_3\)), producing potassium bromide (\(\mathrm{KBr}\)), carbon dioxide (\(\mathrm{CO}_2\)), and water (\(\mathrm{H}_2\mathrm{O}\)).
Understanding chemical reactions requires recognizing how reactants convert to products according to a balanced chemical equation. A balanced equation ensures that the number of each type of atom is conserved before and after the reaction.
In general, chemical reactions can be categorized into various types such as synthesis, decomposition, single replacement, and double replacement reactions. In the provided example, it is a **double replacement reaction** where exchange of components occurs between reactants.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles. The balanced chemical equation provides the stoichiometric relationship between reactants and products, which can be thought of as a 'recipe' for the reaction.
For the neutralization of HBr with \(\mathrm{K}_2\mathrm{CO}_3\), the balanced equation \(2 \mathrm{HBr} + \mathrm{K}_2\mathrm{CO}_3 \rightarrow 2 \mathrm{KBr} + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}\) gives us a stoichiometric ratio. This means for every 2 moles of HBr, 1 mole of \(\mathrm{K}_2\mathrm{CO}_3\) is required to fully react. By using such relationships, it becomes possible to accurately determine how much \(\mathrm{K}_2\mathrm{CO}_3\) is needed based on the amount of HBr present.

Acid-Base Neutralization

Acid-base neutralization is a type of chemical reaction where an acid and a base react to form water and a salt. This reaction often also produces carbon dioxide if a carbonate compound is involved, as seen in this scenario.
When HBr, a strong acid, reacts with \(\mathrm{K}_2\mathrm{CO}_3\), a carbonate base, the HBr donates protons (H\(^+\) ions) to the carbonate, releasing carbon dioxide gas and water as byproducts alongside the formation of potassium bromide (KBr), the salt.
This type of reaction is essential in real-world applications like titrations, where one solution is used to determine the concentration of another, and in environmental spill treatments, where acids can be neutralized before they cause harm.

Molarity Calculations

Molarity is a way to express the concentration of a solution - the amount of solute (HBr in this case) dissolved in a specific volume of solvent. Molarity (\(M\)) is defined as moles of solute per liter of solution, giving insight into how dilute or concentrated a solution is.
In the initial problem, we calculated moles of HBr using molarity and the volume of the solution. The formula is:\[ \text{Moles of solute} = \text{Molarity (M)} \times \text{Volume (L)} \]This calculation is straightforward yet crucial, as it allows the determination of exact quantities needed for further reactions \(\text{--such as neutralization.}\) Understanding and performing molarity calculations enables you to control the outcomes of reactions effectively, a skill vital in chemistry labs and industry settings.