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Chapter 4: Problem 7

Consider separate aqueous solutions of \(\mathrm{HCl}\) and $\mathrm{H}_{2} \mathrm{SO}_{4}$ with the same molar concentrations. You wish to neutralize an aqueous solution of NaOH. For which acid solution would you need to add more volume (in milliliters) to neutralize the base? a. the HCl solution b. the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution c. You need to know the acid concentrations to answer this question. d. You need to know the volume and concentration of the NaOH solution to answer this question. e. c and d Explain.

Short Answer

Expert verified
a. the HCl solution

Step by step solution

01

Write the balanced neutralization reactions for both acids

First, we need to write down the balanced chemical equations for the neutralization reactions of HCl and H₂SO₄ with NaOH. 1. HCl + NaOH -> NaCl + H₂O 2. H₂SO₄ + 2 NaOH -> Na₂SO₄ + 2 H₂O In equation 1, one mole of HCl reacts with one mole of NaOH. In equation 2, one mole of H₂SO₄ reacts with two moles of NaOH.
02

Determine the stoichiometry of the reactions

Comparing these reactions, we can infer that one mole of H₂SO₄ can neutralize twice the amount of NaOH as one mole of HCl. It's important to note that 1 mole of H₂SO₄ has twice the neutralizing power, given that it disassociates into two H⁺ ions.
03

Compare the volumes of acids required for neutralization

Since both the HCl and H₂SO₄ solutions have the same molar concentrations, the volume required to neutralize a given amount of NaOH would be inversely proportional to their neutralizing power. In this case, given that one mole of H₂SO₄ can neutralize twice the amount of NaOH as one mole of HCl, we can conclude that we would need a smaller volume of the H₂SO₄ solution to neutralize the same amount of NaOH.
04

Choose the correct answer from the available options

Based on our analysis, we can conclude that we would need to add more volume of the HCl solution to neutralize the NaOH solution compared to the H₂SO₄ solution. Hence, the correct answer is: a. the HCl solution

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Most popular questions from this chapter

The zinc in a 1.343 -g sample of a foot powder was precipitated as \(\mathrm{ZnNH}_{4} \mathrm{PO}_{4} .\) Strong heating of the precipitate yielded 0.4089 \(\mathrm{g} \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) . Calculate the mass percent of zinc in the sample of foot powder.

You are given a 1.50-g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 mL of water and then add an excess of 0.500 M silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.641 g. a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the mass percent of sodium chloride in the original unknown mixture.

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) A 0.205-g sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The BaSO\(_{4}\) that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0 -mL aliquot and adding 50.0 \(\mathrm{mL}\) of 0.213 \(\mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required 13.21 \(\mathrm{mL}\) of 0.103 \(\mathrm{M}\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) . Sulfuric acid has two acidic hydrogens.

What mass of barium sulfate can be produced when 100.0 mL of a 0.100-M solution of barium chloride is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate?
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