Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 148

A 10.00-mL sample of sulfuric acid from an automobile battery requires 35.08 mL of 2.12 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 in the automobile battery is 3.718 M.

Step by step solution

01

Write the balanced equation for the reaction

The neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is given by: H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l) Notice that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid.
02

Calculate the moles of sodium hydroxide used

We are given the volume and concentration of the sodium hydroxide solution. We can calculate the moles of sodium hydroxide using the formula: Moles = Molarity × Volume Let's plug in the given values and calculate the moles of sodium hydroxide: Moles of NaOH = \( 2.12 M \times 0.03508 L = 0.07437 \,mol\)
03

Determine the moles of sulfuric acid using stoichiometry

From the balanced equation, we know that 2 moles of NaOH react with 1 mole of H2SO4. We can now use the moles of sodium hydroxide calculated in step 2 to determine the moles of sulfuric acid involved in the reaction using stoichiometry. Using the stoichiometric ratio of 2:1 (NaOH:H2SO4), we can calculate the moles of H2SO4 as follows: Moles of H2SO4 = \( \cfrac{Moles \, of\, NaOH}{2} = \cfrac{0.07437 \,mol}{2} = 0.03718 \,mol \)
04

Calculate the molarity of sulfuric acid

Now that we have the moles of sulfuric acid, we can determine its molarity. The molarity is given by the formula: Molarity = \( \cfrac{Moles}{Volume} \) We are given the volume of sulfuric acid as 10.00 mL. To use this value in the molarity calculation, we need to convert it to liters: Volume of H2SO4 in liters = \( 10.00 \,mL * \cfrac{1 \,L}{1000 \,mL} = 0.01000 \,L \) Now we can calculate the molarity of the sulfuric acid: Molarity of H2SO4 = \( \cfrac{0.03718 \,mol}{0.01000 \,L} = 3.718 \,M \) So, the molarity of the sulfuric acid in the automobile battery is 3.718 M.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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.

What volume of each of the following bases will react completely with 25.00 mL of 0.200 M HCl? a. 0.100\(M \mathrm{NaOH}\) b. 0.0500\(M \mathrm{Sr}(\mathrm{OH})_{2}\) c. 0.250 \(\mathrm{M} \mathrm{KOH}\)

Balance the following oxidation–reduction reactions that occur in basic solution. a. $\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)$ b. $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)$ c. $\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)$

A 1.00-g sample of an alkaline earth metal chloride is treated with excess silver nitrate. All of the chloride is recovered as 1.38 g of silver chloride. Identify the metal.

A 100.0-mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.200 M aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free