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Chapter 4: Problem 149

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.

Short Answer

Expert verified
The molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 0.00929 M.

Step by step solution

01

Calculate the moles of excess OH- ions

First, we find out the moles of excess \(\mathrm{OH}^{-}\) ions after the reaction. We have the volume and molarity of \(\mathrm{HCl}\) for neutralization of \(\mathrm{OH}^{-}\). n(HCl) = V(HCl) 脳 M(HCl) n(HCl) = Volume of HCl 脳 Molarity of HCl n(OH-) = n(HCl) (Since HCl and OH- ions react in 1:1 ratio) n(OH-) = 13.21 mL 脳 0.103 M Convert mL to L: n(OH-) = 0.01321 L 脳 0.103 M n(OH-) = 0.00136033 moles
02

Calculate the moles of NaOH initially added

Next, we need to find out the moles of \(\mathrm{NaOH}\) initially added. n(NaOH) = V(NaOH) 脳 M(NaOH) n(NaOH) = Volume of NaOH 脳 Molarity of NaOH n(NaOH) = 50 mL 脳 0.213 M Convert mL to L: n(NaOH) = 0.050 L 脳 0.213 M n(NaOH) = 0.01065 moles
03

Calculate the moles of OH- ions that reacted with H2SO4

Now, we will subtract the excess \(\mathrm{OH}^{-}\) ions to find out the moles of \(\mathrm{OH}^{-}\) ions that reacted with the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Since sulfuric acid has two acidic hydrogens, it reacts with two moles of OH- ions per mole. n(OH-)_{reacted} = n(NaOH) - n(OH-)_{excess} n(OH-)_{reacted} = 0.01065 moles - 0.00136033 moles n(OH-)_{reacted} = 0.00928967 moles
04

Calculate the moles of H2SO4

We have the moles of \(\mathrm{OH}^{-}\) ions that reacted with the \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Since each mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with two moles of \(\mathrm{OH}^{-} \) ions: n(H2SO4) = n(OH-)_{reacted} / 2 n(H2SO4) = 0.00928967 moles / 2 n(H2SO4) = 0.004644835 moles Finally, now we have the moles of the original \(\mathrm{H}_{2} \mathrm{SO}_{4}\) sample.
05

Calculate the molarity of H2SO4

We have the moles and volume of the original \(\mathrm{H}_{2} \mathrm{SO}_{4}\) sample. M(H2SO4) = n(H2SO4) / V(H2SO4) M(H2SO4) = 0.004644835 moles / 0.500 L M(H2SO4) = 0.00928967 M The molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 0.00929 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity Calculation
Understanding molarity is crucial when dealing with solution chemistry. Molarity (M) represents the concentration of a solute in a solution and is defined as the number of moles of solute per liter of solution. In practice, it allows us to know how much of a substance is present in a given volume of liquid.
To calculate molarity:
  • First, determine the number of moles of the solute. This involves using the formula: \( n = V \times M \), where \( n \) is the number of moles, \( V \) is the volume in liters, and \( M \) is the molarity.
  • Next, divide these moles by the volume of the solution in liters to find the molarity, \( M = \frac{n}{V} \).
This method is applied in our example, where sulfuric acid (\( \mathrm{H}_2\mathrm{SO}_4 \)) is the solute whose molarity is determined after reacting with sodium hydroxide (\( \mathrm{NaOH} \)) and hydrochloric acid (\( \mathrm{HCl} \)). Calculations like this assist in understanding the strength and concentration of acids and bases in solutions.
Stoichiometry
Stoichiometry involves calculating the quantities of reactants and products in chemical reactions. It is based on the balanced chemical equation and is foundational in understanding how different chemicals interact.
In the context of the acid-base titration with \( \mathrm{H}_2\mathrm{SO}_4 \), understanding stoichiometry becomes vital. Each molecule of sulfuric acid can donate two hydrogen ions (\( \mathrm{H}^+ \)), meaning every mole of \( \mathrm{H}_2\mathrm{SO}_4 \) can neutralize two moles of hydroxide ions (\( \mathrm{OH}^- \)). This 1:2 relation is critical in the calculations:
  • Determine the moles of \( \mathrm{NaOH} \) added initially through titration.
  • Calculate the excess \( \mathrm{OH}^- \) that remained after reacting with \( \mathrm{H}_2\mathrm{SO}_4 \).
  • Ascertain the moles of \( \mathrm{OH}^- \) that actually reacted with the \( \mathrm{H}_2\mathrm{SO}_4 \).
This stoichiometry-driven approach leads us to find the number of moles of \( \mathrm{H}_2\mathrm{SO}_4 \) that initially reacted, which is crucial for accurate molarity calculations.
Solution Chemistry
Solution chemistry focuses on how substances dissolve, react, and interact within a liquid medium. In acid-base titration, this is particularly relevant as it underlies the neutralization reactions and solubility principles you observe.
During the titration process:
  • A known concentration of \( \mathrm{NaOH} \) is slowly added to the \( \mathrm{H}_2\mathrm{SO}_4 \) solution, neutralizing the acid.
  • The reaction produces water and an excess of hydroxide ions if extra base is added.
  • To determine how much base was in excess, \( \mathrm{HCl} \) is carefully introduced until neutrality is again achieved.
This sequence involves a dynamic interplay of the components in solution. It's important to understand how ionic compounds like these behave in water, dissociating into their respective ions and participating in reactions that are influenced by the concentration and nature of each ion present. This knowledge is essential to predicting outcomes and ensuring accurate measurements in chemical reactions.

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Most popular questions from this chapter

Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the AuCl, \(^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid鈥揵ase reactions. a. \(\mathrm{HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow\) b. \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow\) c. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow\)

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.

What volume of 0.0200 M calcium hydroxide is required to neutralize 35.00 mL of 0.0500 M nitric acid?

Many plants are poisonous because their stems and leaves contain oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) or sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) . When ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-},\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride, \(\mathrm{CaCl}_{2},\) in aqueous solution.
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