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Chapter 6: Problem 22

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

Expert verified
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

01

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
02

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)
03

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
04

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.
05

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.

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Most popular questions from this chapter

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