Question

What volume of \(6.15 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) is required to exactly neutralize 1.25 L of \(0.265 \mathrm{M} \mathrm{NH}_{3}(\text { aq }) ?\) $$\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}$$

Short answer

53.9 mL of 6.15 M HCl solution is required to exactly neutralize 1.25 L of 0.265 M NH3 solution.

Step-by-step solution

Understanding the Reaction

First, we need to understand the reaction equation: \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\longrightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}\). Here, NH3 is acting as a base, and H3O+ (which can be released from the HCl in water) is functioning as an acid. In this case, the reaction involves a 1:1 stoichiometry, meaning one mole of NH3 reacts with one mole of H3O+.

Calculating the Moles of NH3

Now, we calculate the moles of NH3 in 1.25 L of 0.265 M NH3 solution. Based on the definition of Molarity, we find the moles by multiplying the volume by the molarity: \(1.25 \, \text{L} \times 0.265 \, \text{moles/L} = 0.33125 \, \text{moles}\).

Finding the Required Volume of HCl

Because of the 1:1 ratio in the reaction, 0.33125 moles of HCl is also required. To find the required volume of 6.15 M HCl, we divide the moles by the molarity: \(0.33125 \, \text{moles} ÷ 6.15 \, \text{moles/L} = 0.0539 \, \text{L} = 53.9 \, \text{mL}\).

Key concepts

Molarity Calculation

Molarity is a crucial concept in chemistry that represents the concentration of a solution. It's defined as the number of moles of solute per liter of solution, denoted as M. Understanding molarity is essential for determining how much of a solution is needed to achieve a particular reaction.

To find molarity, you use the formula:

• \[ ext{Molarity (M)} = \frac{ ext{moles of solute}}{ ext{liters of solution}} \]

In this exercise, we used molarity to find out how many moles of ammonia (\( \text{NH}_3 \)) we had in a given volume. By multiplying the volume (\( 1.25 \) L) by the molarity (\( 0.265 \) M), we found that there were \( 0.33125 \) moles of \(\text{NH}_3\).

Being able to calculate molarity allows you to accurately prepare solutions and predict the outcomes of chemical reactions.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In simple terms, it allows chemists to predict how much of each substance is involved in a reaction.

This is done using balanced chemical equations, which tell us the ratios in which substances react and are produced. For our problem, the balanced equation shows:

• \( \text{NH}_3(\text{aq}) + \text{H}_3\text{O}^+(\text{aq}) \rightarrow \text{NH}_4^+(\text{aq}) + \text{H}_2\text{O} \)

This equation demonstrates a 1:1 stoichiometric ratio. This means one mole of \( \text{NH}_3 \) reacts with one mole of \( \text{H}_3\text{O}^+ \).

In our exercise, we determined that we needed exactly \( 0.33125 \) moles of \( \text{HCl} \) because the stoichiometry dictates it's the same amount that reacts with \( \text{NH}_3 \). This understanding is vital for solving complex reaction-based problems.

Acid-Base Reaction

Acid-base reactions are a type of chemical reaction that involves the transfer of protons (H+) between reactants. They are characterized by an acid donating a proton to a base.

In the case of our exercise, ammonia (\( \text{NH}_3 \)) acts as a base, while hydrochloric acid (\( \text{HCl} \)) acts as the source of the acidic hydrogen ion (\( \text{H}_3\text{O}^+ \)). When dissolved in water, \( \text{HCl} \) dissociates completely into \( \text{H}^+ \) ions and \( \text{Cl}^- \) ions.

• The reaction here is: \( \text{NH}_3(\text{aq}) + \text{H}_3\text{O}^+(\text{aq}) \rightarrow \text{NH}_4^+(\text{aq}) + \text{H}_2\text{O} \)

In this reaction, the \( \text{NH}_3 \) accepts a proton from \( \text{H}_3\text{O}^+ \), forming \( \text{NH}_4^+ \) and water. This mechanism is essential in neutralization reactions where the mission is to neutralize the properties of an acid or a base.

Understanding acid-base reactions, such as this one, helps in predicting the outcome of similar reactions and is a fundamental part of studying chemistry.