Question

A student had \(1.00 \mathrm{~L}\) of a \(1.00-M\) acid solution. Much to the surprise of the student, it took \(2.00 \mathrm{~L}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) solution to react completely with the acid. Explain why it took twice as much \(\mathrm{NaOH}\) to react with all of the acid. In a different experiment, a student had \(10.0 \mathrm{~mL}\) of \(0.020 \mathrm{M}\) HCl. Again, much to the surprise of the student, it took only \(5.00 \mathrm{~mL}\) of \(0.020 M\) strong base to react completely with the HCl. Explain why it took only half as much strong base to react with all of the \(\mathrm{HCl}\).

Short answer

In the first case, it took twice as much NaOH to react with the acid because the stoichiometry of the reaction was 1:2, meaning two moles of NaOH reacted with one mole of the acid for complete neutralization. In the second case, it took only half as much strong base to react with all of the HCl because the strong base was twice as effective at neutralizing the acid due to its higher concentration of hydroxide ions in solution, resulting in a 1:1 stoichiometry.

Step-by-step solution

Case 1: 1.00 L of 1.00 M acid with 2.00 L of 1.00 M NaOH

The key to explaining this observation lies in the stoichiometry of the reaction between the acid and NaOH. Since 2.00 L of NaOH solution was required to react with the 1.00 L of acid solution, we likely have a 1:2 stoichiometry ratio. First, let's write down the general chemical reaction between an acid HA and NaOH: \( HA + NaOH \rightarrow NaA + H_2O \) In this case, two moles of NaOH would react with one mole of the acid for complete neutralization: \( HA + 2NaOH \rightarrow Na_2A + 2H_2O \) The required amount of NaOH is twice the amount of the acid because of the stoichiometry of the reaction.

Case 2: 10.0 mL of 0.020 M HCl with 5.00 mL of 0.020 M strong base

In this case, we will explain why it took only half the volume of the strong base to react with all of the HCl solution. Let's assume the strong base is a hydroxide, BOH. The general reaction between HCl and a base is: \( HCl + BOH \rightarrow BCl + H_2O \) In this case, the stoichiometry of the reaction between HCl and the strong base BOH is 1:1: \( HCl + BOH \rightarrow BCl + H_2O \) Here, the amount of strong base required for neutralization is half the volume but with the same concentration as the HCl solution. This means the strong base is twice as effective at neutralizing the acid because BOH has twice the concentration of hydroxide ions (OH⁻) as compared to a monovalent base like NaOH. The required amount of the strong base is half the volume of the HCl solution because it is twice as effective at neutralizing the acid due to its higher concentration of hydroxide ions in solution.

Key concepts

Neutralization Reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react to form a salt and water. This process is critical in acid-base chemistry and is often encountered in various scientific analyses and real-world applications.

In a typical scenario, you'll see a hydrochloric acid (HCl) reacting with sodium hydroxide (NaOH), a strong base. The balanced chemical equation looks like this:
\[ HCl + NaOH \rightarrow NaCl + H_2O \]

The 'neutral' part of neutralization is pivotal as it signifies that neither acidic nor basic properties are present in the resulting solution—it's chemically neutral. Now, the amount of acid and base required to reach this neutrality depends on their respective concentrations and the stoichiometry of the reaction. For example, stronger acids or bases might need less volume to neutralize a given amount of its counterpart, highlighting the importance of concentration and the stoichiometric coefficients in the balanced equation.

Stoichiometry

Stoichiometry is the aspect of chemistry that pertains to the calculation of reactants and products in chemical reactions. In acid-base reactions, it becomes especially useful in predicting the outcomes of neutralization.

For instance, in the given textbook exercise, the fact that it took twice as much sodium hydroxide (NaOH) to react with the acid suggests that the stoichiometry of that reaction was not 1:1. Instead, it implies a reaction where one acid molecule reacts with two NaOH molecules:
\[ HA + 2NaOH \rightarrow Na_2A + 2H_2O \]

Correctly interpreting the stoichiometry is essential because it informs us how to balance chemical equations, and in practical terms, it guides how much of each reactant is needed for the reaction to complete without leaving excess of one over the other.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute divided by the volume of solution in liters.
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liter of solution}} \]

In the context of our textbook problem, a solution with a molarity of 1.00 M contains one mole of solute—in this case, the acid or NaOH—per liter of solution. Understanding molarity is helpful in stoichiometry calculations, as it allows us to convert between the volume of a solution and the amount of substance it contains. This conversion is at the core of solving neutralization problems, as seen in the explanation why differing volumes of solutions were used to achieve neutralization in the student's experiments.

Concentration of Solutions

The concentration of a solution talks about how much of a given solute is dissolved in a certain amount of solvent. Concentration can be expressed in various ways, with molarity being just one of them. By understanding the concentration, students can control the reactiveness of solutions in chemical reactions.

In the textbook exercise, we are shown that different amounts of solutions were used to neutralize acids, which underlines the role of concentration in stoichiometry. The reason why a strong base could neutralize the HCl with half the volume is that the strong base had a higher concentration of hydroxide ions, making it more effective per unit of volume. When dealing with concentration in neutralization reactions, always consider the quantity of reactive species in the solution, not just the volume or the molarity by itself.

Important Note:

When you're performing laboratory work or stoichiometry calculations, always remember to account for the concentration of the solutions. It's not all about quantity but quality—the 'strength' of your reactants plays a key role in the reaction.