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A \(10.00-\mathrm{mL}\) sample of sulfuric acid from an automobile battery requires \(35.08 \mathrm{~mL}\) of \(2.12 \mathrm{M}\) sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? Sulfuric acid contains two acidic hydrogens.

Short Answer

Expert verified
The molarity of the sulfuric acid is approximately \(3.72 M\).

Step by step solution

01

Calculate the moles of sodium hydroxide.

Using the volume and concentration of sodium hydroxide, we can find the moles of sodium hydroxide: Moles of NaOH = Volume of NaOH 脳 Concentration of NaOH Volume of NaOH = \(35.08 mL\) Concentration of NaOH = \(2.12 M\) Moles of NaOH = \(35.08 mL 脳 2.12 M\) Moles of NaOH = \(74.3696 mmol\)
02

Determine the moles of sulfuric acid.

Now that we have the moles of sodium hydroxide, we need to use the stoichiometry between sulfuric acid and sodium hydroxide to find the moles of sulfuric acid: Moles of H鈧係O鈧 = Moles of NaOH 脳 Stoichiometry Stoichiometry (ratio) of H鈧係O鈧 to NaOH is \(\frac{1}{2}\), As 1 mole of H鈧係O鈧 requires 2 moles of NaOH for complete neutralization. Moles of H鈧係O鈧 = \(74.3696mmol 脳 \frac{1}{2}\) Moles of H鈧係O鈧 = \(37.1848 mmol\)
03

Calculate the molarity of sulfuric acid.

Finally, we have the moles of sulfuric acid and its volume, and we can find its molarity: Molarity of H鈧係O鈧 = \(\frac{Moles \:of \: H鈧係O鈧剗{Volume \: of \: H鈧係O鈧剗\) Volume of H鈧係O鈧 = \(10.00 mL\) Molarity of H鈧係O鈧 = \(\frac{37.1848 mmol}{10.00 mL}\) = \(\frac{37.1848 mmol}{10.00 mL} 脳 \frac{1 L}{1000 mL} 脳 \frac{1 mol}{1000 mmol}\) Molarity of H鈧係O鈧 = \(3.71848 M\) The molarity of the sulfuric acid is approximately \(3.72 M\).

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Key Concepts

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

Stoichiometry
Stoichiometry is the fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It allows us to predict how much of a substance will be consumed or produced during a reaction. For instance, when we calculate the moles of sodium hydroxide needed to react with sulfuric acid, we use stoichiometry principles.

Stoichiometry relies on balanced chemical equations to determine these relationships. It is important to first balance the chemical equation to accurately understand the molar ratio between reactants and products.- In the case of sulfuric acid and sodium hydroxide, our balanced reaction is: \[ ext{H}_2 ext{SO}_4 + 2 ext{NaOH} ightarrow ext{Na}_2 ext{SO}_4 + 2 ext{H}_2 ext{O} \] This tells us that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.- By knowing the stoichiometric ratio, we can directly relate the moles of each substance involved.
This step is crucial in converting moles of one substance to moles of another, as seen in our exercise.
Neutralization Reaction
A neutralization reaction is a type of chemical reaction in which an acid and a base react to form water and a salt. This is a key concept in our exercise, where sulfuric acid (H鈧係O鈧) is neutralized by sodium hydroxide (NaOH).

In a neutralization reaction: - The hydrogen ions (H鈦) from the acid react with hydroxide ions (OH鈦) from the base to produce water (H鈧侽). - Additionally, the remaining ions form a salt, such as sodium sulfate (Na鈧係O鈧) in our scenario. The significance of neutralization in chemistry extends to various applications, such as titrations, where we measure the volume of a reactant needed to completely neutralize another. In our case, recognizing that sulfuric acid has two acidic hydrogens means each molecule can react with two molecules of NaOH, influencing the stoichiometry and calculations.
Sulfuric Acid Titration
Sulfuric acid titration involves the gradual addition of a base to a known volume of sulfuric acid until the acid is fully neutralized. This method helps us determine the concentration or molarity of the sulfuric acid solution in our exercise.

Here鈥檚 how a titration is generally performed: - A solution of known concentration, called the titrant (sodium hydroxide here), is slowly added to the analyte (sulfuric acid). - An indicator or a pH meter is used to detect the end point, which in this scenario is when the acid is completely neutralized. Titrations are vital in analytical chemistry for determining the concentration of an unknown. They rely on the precise measurement of volume, just as we measured the 35.08 mL of NaOH to titrate and neutralize the 10.00 mL sulfuric acid sample. Understanding the goal and endpoint of titration enhances our ability to perform stoichiometric calculations accurately.
Sodium Hydroxide
Sodium hydroxide (NaOH), commonly known as lye or caustic soda, is a strong, highly reactive base. It plays a crucial role in neutralization reactions, acting as the reagent in our sulfuric acid titration.

Sodium hydroxide is characterized by: - Its ability to dissociate completely in water, providing hydroxide ions (OH鈦) necessary for neutralizing acids. - Being highly soluble in water, which makes it a popular choice for preparing solutions of precise molarity for titrations. In our exercise, sodium hydroxide helps us determine the molarity of sulfuric acid by measuring how much NaOH is required to reach the endpoint. Moreover, understanding the properties of NaOH aids in recognizing its behavior during reactions and ensures safe handling. It鈥檚 essential to note these chemical properties to properly utilize NaOH in various applications, reinforcing its importance in laboratory and industrial settings.

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Most popular questions from this chapter

Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4}\), and \(\mathrm{NaOH}\). A precipitate forms only when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added. Which cations could be present in the unknown soluble ionic compound?

How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50 M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" \((18 M)\) sulfuric acid b. \(\mathrm{HCl}\) from "concentrated" \((12 \mathrm{M})\) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

In a 1-L beaker, \(203 \mathrm{~mL}\) of \(0.307 \mathrm{M}\) ammonium chromate was mixed with \(137 \mathrm{~mL}\) of \(0.269 \mathrm{M}\) chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate. Write the balanced chemical reaction occurring here. If the percent yield of the reaction was \(88.0 \%\), what mass of chromium(III) chromate was isolated?

A \(2.20-\mathrm{g}\) sample of an unknown acid (empirical formula \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\) ) is dissolved in \(1.0 \mathrm{~L}\) of water. A titration required \(25.0\) \(\mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)

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