/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 148 A 10.00-mL sample of sulfuric ac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sulfuric Acid
Sulfuric acid, with the chemical formula \( H_2SO_4 \), is a strong mineral acid. It is widely used in various industrial processes due to its high reactivity and its ability to donate two protons (hydrogen ions). In aqueous solutions, sulfuric acid dissociates into two protons and a sulfate ion \( SO_4^{2-} \). This ability to donate more than one proton makes sulfuric acid a diprotic acid. Each mole of sulfuric acid can neutralize two moles of a strong base like sodium hydroxide (\( NaOH \)), which is important in stoichiometric calculations. It's crucial to understand these properties of sulfuric acid when performing chemical reactions, especially in neutralization processes.
Molarity Calculation
Molarity is a measure of the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula to calculate molarity is: \(\text{Molarity} = \cfrac{\text{Moles of solute}}{\text{Volume of solution in liters}} \).
This calculation is vital in chemistry, as it helps quantify the concentration of a solution, allowing for precise stoichiometric calculations in reactions. For instance, in the problem at hand, to find the molarity of sulfuric acid, we first need the number of moles present in the specified volume. The sample volume in liters must also be accurately considered to compute the correct molarity.
Stoichiometry
Stoichiometry is the section of chemistry that deals with calculating the relative quantities of reactants and products in chemical reactions. It is based on the principles of conservation of mass and the concept of moles. In the context of a neutralization reaction like \( H_2SO_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \), stoichiometry helps determine how much of each reactant is needed to completely react with the other. This process involves using the balanced chemical equation to find the mole ratio between reactants and products. Using stoichiometry, we can precisely deduce that two moles of sodium hydroxide are needed to neutralize one mole of sulfuric acid. This ratio is essential for calculating the exact amounts necessary in laboratory or industrial reactions.
Balanced Chemical Equation
A balanced chemical equation is a symbolic representation of a chemical reaction where the number of moles of each element is conserved before and after the reaction. Balancing involves ensuring that atoms are neither created nor destroyed, aligning with the law of conservation of mass. For the reaction between sulfuric acid and sodium hydroxide, the equation \( H_2SO_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \) reflects that each reactant produces a specific amount of product. The balanced equation provides the stoichiometric coefficients, which are critical for calculating the required reactant amounts. Mastery of balancing equations is fundamental in chemistry, enabling accurate predictions and calculations of chemical behavior.

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

Consider the reaction of 19.0 g of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and 29.0 g of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture

What mass of iron(III) hydroxide precipitate can be produced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide?

Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed. a. chromium(III) chloride and sodium hydroxide b. silver nitrate and ammonium carbonate c. copper(II) sulfate and mercury(I) nitrate d. strontium nitrate and potassium iodide

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: $$\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)$$ Triodide ion concentration is determined by titration with a sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)\) solution. The products are iodide ion and tetrathionate ion \(\left(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\right)\) a. Balance the equation for the reaction of \(\mathrm{IO}_{3}^{-}\) with \(\mathrm{I}^{-}\) ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HCl required to convert all of the 1 \(\mathrm{O}_{3}^{-}\) ions to \(\mathrm{I}_{3}^{-}\) ions? c. Write and balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{3}-\) in acidic solution. d. A 25.00 -mL sample of a 0.0100\(M\) solution of \(\mathrm{KIO}_{3}\) is reacted with an excess of \(\mathrm{KL}\) . It requires 32.04 \(\mathrm{mL}\) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution to titrate the \(\mathrm{I}_{3}^{-}\) ions present. What is the molarity of the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution? e. How would you prepare 500.0 \(\mathrm{mL}\) of the KIO\(_{3}\)solution in part d using solid \(\mathrm{KIO}_{3} ?\)

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$\mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }}$$ In the case of very dilute aqueous solutions, a concentration of 1.0 \(\mathrm{ppm}\) is equal to 1.0\(\mu \mathrm{g}\) of solute per 1.0 \(\mathrm{mL}\) , which equals 1.0 \(\mathrm{g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 \(\mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. 1.0 \(\mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. 10.0 \(\mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. 0.10 \(\mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{C} 1_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.