Question

A student had 1.00 L of a \(1.00-M\) acid solution. Much to the surprise of the student, it took \(2.00 \mathrm{L}\) of \(1.00 \mathrm{M}\) NaOH solution to react completely with the acid. Explain why it took twice as much NaOH to react with all of the acid. In a different experiment, a student had \(10.0 \mathrm{mL}\) of 0.020 M HC1. Again, much to the surprise of the student, it took only \(5.00 \mathrm{mL}\) of \(0.020 \mathrm{M}\) strong base to react completely with the HCI. Explain why it took only half as much strong base to react with all of the HCl.

Short answer

In the first experiment, it took twice as much NaOH to react with the acid because the acid is likely a diprotic acid, which donates two protons (H+) per molecule. In the second experiment, it took only half as much strong base to react with all the HCl because the strong base is likely a dibasic base, which accepts two protons (H+) per molecule.

Step-by-step solution

1. Identify the reactants and products

We have two different reactions in the two experiments. Let's call the acid in the first experiment Acid_A and the strong base in the second experiment Base_B. In both reactions, NaOH is used as the base. We write generic reactions as follows: Reaction 1: Acid_A + NaOH → Products Reaction 2: HCl + Base_B → Products

2. Determine the mole-to-mole ratios

In order to compare the amounts of reactants, we first need to determine the mole-to-mole ratios of the acid and the base in each reaction. For both reactions: Moles of Acid_A = Volume (L) * Concentration (M) Moles of NaOH = Volume (L) * Concentration (M) Moles of HCl = Volume (L) * Concentration (M) Moles of Base_B = Volume (L) * Concentration (M)

3. Calculate moles of reactants in both experiments

For Reaction 1: Moles of Acid_A = 1 L * 1 M = 1 mole Moles of NaOH = 2 L * 1 M = 2 moles For Reaction 2: Moles of HCl = 0.01 L * 0.02 M = 0.0002 mole Moles of Base_B = 0.005 L * 0.02 M = 0.0001 mole

4. Analyze the stoichiometry of both reactions

For Reaction 1, it took 2 moles of NaOH to react with 1 mole of Acid_A, which implies a 1:2 stoichiometry between Acid_A and NaOH. This suggests that Acid_A is a diprotic acid, meaning that it can donate two protons (H+) per molecule in the reaction. For example, Acid_A could be H2SO4. For Reaction 2, it took 0.0001 mole of Base_B to react with 0.0002 mole of HCl, indicating a 2:1 stoichiometry between HCl and Base_B. This suggests that Base_B is a dibasic base, meaning that it can accept two protons (H+) per molecule in the reaction. For example, Base_B could be Ba(OH)2.

5. Conclusion

In the first experiment, it took twice as much NaOH to react with the acid because the acid is likely a diprotic acid, which donates two protons per molecule. In the second experiment, it took only half as much strong base to react with all the HCl because the strong base is likely a dibasic base, which accepts two protons per molecule.

Key concepts

Diprotic Acid

A diprotic acid is a type of acid that has two hydrogen ions (protons) to donate in a reaction. This means it can undergo two phases of proton donation. Common examples of diprotic acids include sulfuric acid (\( ext{H}_2 ext{SO}_4 \)) and carbonic acid (\( ext{H}_2 ext{CO}_3 \)).
In a reaction with a base, a diprotic acid will often need two moles of a monobasic base, like sodium hydroxide (\( ext{NaOH} \)), to completely neutralize one mole of acid. The neutralization reaction proceeds in steps: the first hydrogen ion is donated and reacts, followed by the second.
This stoichiometry was observed in the original exercise, where the acid required twice the amount of \( ext{NaOH} \), suggesting it had two protons available for donation — hence, it was a diprotic acid.

Neutralization Reaction

In chemistry, a neutralization reaction happens when an acid and a base react to form water and a salt. During this reaction, the hydrogen ions from the acid combine with the hydroxide ions from the base, resulting in water.
The stoichiometry of a neutralization reaction depends on the number of reactive protons available in the acid and the number of hydroxides in the base. The first experiment demonstrated this concept, where a diprotic acid required twice as much \( ext{NaOH} \) to react completely compared to a monoprotic acid like \( ext{HCl} \).
Understanding the nature of both the acid and base involved in the reaction helps in predicting the stoichiometry and amount of base required to neutralize a given amount of acid.

Mole Concept

The mole concept is essential in chemistry for understanding reactions quantitatively. It refers to a specific measurement unit used to count particles, usually atoms or molecules, and is equivalent to Avogadro's number (∼6.022 × 10^{23} particles).
Using the mole concept allows chemists to equate volumes and concentrations (\( ext{M} = ext{mol/L} \)) to determine how substances will react in set proportions. For example, with 1 mole of a diprotic acid, it requires 2 moles of a monobasic base to achieve complete neutralization, demonstrating a 1:2 mole ratio.
By calculating moles from known concentrations and volumes, students can predict the exact amount of each reactant needed in chemical reactions, as illustrated in the original exercise with different acid and base scenarios.