/*! 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 51 For each of the following acid-b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 51

For each of the following acid-base reactions, identify the acid and the base and then write the net ionic equation: a. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{CaSO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) b. \(\mathrm{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow\) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Short Answer

Expert verified
Question: Identify the acids and bases for each reaction, and write the net ionic equation. a. $\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{CaSO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$ b. $\mathrm{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$ c. $\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow \mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$ Answer: a. Acid: $\mathrm{H}_{2} \mathrm{SO}_{4}$, Base: $\mathrm{Ca}(\mathrm{OH})_{2}$, Net ionic equation: $2\mathrm{H^+}(a q)+ 2\mathrm{OH^-}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)$. b. Acid: $\mathrm{H}_{2} \mathrm{SO}_{4}$, Base: $\mathrm{PbCO}_{3}$, Net ionic equation: $2\mathrm{H^+}(a q)+\mathrm{CO}_{3}^{2-}(s)\rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$. c. Acid: $\mathrm{CH}_{3} \mathrm{COOH}$, Base: $\mathrm{Ca}(\mathrm{OH})_{2}$, Net ionic equation: $2\mathrm{OH^-}(a q)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow 2\mathrm{CH}_{3} \mathrm{COO^-}(a q)+ 2 \mathrm{H}_{2} \mathrm{O}(\ell)$.

Step by step solution

01

a. Identifying the Acid and the Base

For the reaction: $\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{CaSO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$, The acid is the substance that donates protons (H+ ions), while the base is the substance that accepts protons. In this case, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) donates two protons (\(2\mathrm{H}^+\)) and \(\mathrm{Ca(OH)_{2}}\) accepts two protons. Therefore, the acid is \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and the base is \(\mathrm{Ca}(\mathrm{OH})_{2}\).
02

a. Writing the Net Ionic Equation

To create the net ionic equation, we need to first write the full ionic equation, which includes all ions and molecular compounds. Then, we can eliminate any spectator ions (ions present on both sides of the equation) to get the net ionic equation. The full ionic equation is: $2\mathrm{H^+}(a q)+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{Ca^{2+}}(a q)+2\mathrm{OH^-}(a q) \rightarrow \mathrm{Ca^{2+}}(a q)+\mathrm{SO}_{4}^{2-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$. Now, let's eliminate the spectator ions. The \(\mathrm{Ca^{2+}}\) and \(\mathrm{SO}_{4}^{2-}\) ions appear on both sides of the equation, so they are spectator ions. The net ionic equation is: $2\mathrm{H^+}(a q)+ 2\mathrm{OH^-}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)$.
03

b. Identifying the Acid and the Base

For the reaction: $\mathrm{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$, The acid is \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and the base is \(\mathrm{PbCO}_{3}\), as they donate and accept protons respectively.
04

b. Writing the Net Ionic Equation

Following the same process as in part a, the full ionic equation is: $2\mathrm{H^+}(a q)+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{Pb^{2+}}(s)+\mathrm{CO}_{3}^{2-}(s) \rightarrow \mathrm{Pb^{2+}}(s)+\mathrm{SO}_{4}^{2-}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$. The net ionic equation is: \(2\mathrm{H^+}(a q)+\mathrm{CO}_{3}^{2-}(s)\rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\).
05

c. Identifying the Acid and the Base

For the reaction: $\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow \mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$, The acid is \(\mathrm{CH}_{3} \mathrm{COOH}\), and the base is \(\mathrm{Ca}(\mathrm{OH})_{2}\), as they donate and accept protons respectively.
06

c. Writing the Net Ionic Equation

Following the same process as in part a, the full ionic equation is: $\mathrm{Ca^{2+}}(a q)+2\mathrm{OH^-}(a q)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow \mathrm{Ca^{2+}}(aq)+2\mathrm{CH}_{3} \mathrm{COO^-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$. The net ionic equation is: $2\mathrm{OH^-}(a q)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow 2\mathrm{CH}_{3} \mathrm{COO^-}(a q)+ 2 \mathrm{H}_{2} \mathrm{O}(\ell)$.

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 91影视!

Key Concepts

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

Net Ionic Equation
In chemistry, the net ionic equation is a simplified version of a chemical equation that only shows the chemical species that are involved in a reaction, excluding spectator ions. These are ions that do not participate in the actual chemical change. By focusing on the active elements, a net ionic equation clearly highlights the essence of the chemical reaction.

To write a net ionic equation, follow these steps:
  • Begin with the full ionic equation, which shows all ions and compounds as they exist in the solution.
  • Identify and eliminate the spectator ions 鈥� these are ions that remain unchanged on both sides of the reaction.
  • What remains is the net ionic equation, consisting only of ions and compounds that undergo a chemical change.
For instance, in the chemical reaction between sulfuric acid (\( \mathrm{H}_2 \mathrm{SO}_4 \)) and calcium hydroxide (\( \mathrm{Ca(OH)_2} \)), the net ionic equation becomes:\[2\mathrm{H^+}(a q) + 2\mathrm{OH^-}(a q) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\ell)\]
This highlights the fundamental acid-base reaction resulting in water.
Acid Identification
Acid identification is about recognizing substances that release protons, or \( \mathrm{H^+} \) ions, in a chemical reaction. Acids are key players in many reactions and understanding them is crucial for mastering acid-base chemistry.

Characteristics of acids include:
  • They are proton donors, often releasing \( \mathrm{H^+} \) ions in solution.
  • They have a pH of less than 7.
  • Common examples include \( \mathrm{HCl} \), \( \mathrm{H_2SO_4} \), and \( \mathrm{CH_3COOH} \).
In the reactions given, \( \mathrm{H}_2 \mathrm{SO}_4 \) acts as an acid in two situations, donating protons to the respective bases.
Base Identification
Base identification involves recognizing substances that accept protons, or produce hydroxide ions (\( \mathrm{OH^-} \)) in solution. Bases are essential for neutralizing acids and play a vital role in various chemical processes.

Characteristics of bases include:
  • They are typically proton acceptors.
  • They often have a pH greater than 7.
  • Common bases include \( \mathrm{NaOH} \), \( \mathrm{KOH} \), and \( \mathrm{Ca(OH)_2} \).
In the provided exercises, \( \mathrm{Ca(OH)_2} \) is identified as a base in reactions involving both \( \mathrm{H}_2 \mathrm{SO}_4} \) and \( \mathrm{CH}_3 \mathrm{COOH} \), accepting protons to form water and yield the resulting products.

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

Tap water in North America from groundwater sources contains an average of \(48 \mathrm{mg} / \mathrm{L} \mathrm{Ca}^{2+}\) ion. What is the molarity of calcium ion in this water?

Some people who prefer natural foods make their own apple cider vinegar. They start with freshly squeezed apple juice that contains about \(6 \%\) natural sugars. These sugars, which all have nearly the same empirical formula, \(\mathrm{CH}_{2} \mathrm{O},\) are fermented with yeast in a chemical reaction that produces equal numbers of moles of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\) and carbon dioxide. The product of fermentation, called hard cider, undergoes an acid fermentation step in which ethanol and dissolved oxygen gas react together to form acetic acid (CH \(_{3} \mathrm{COOH}\) ) and water. This acetic acid is the principal solute in vinegar. a. Write a balanced chemical equation for the fermentation of natural sugars to ethanol and carbon dioxide. You may use in the equation the empirical formula given in the preceding paragraph. b. Write a balanced chemical equation for the acid fermentation of ethanol to acetic acid. c. What are the oxidation states of carbon in the reactants and products of the two fermentation reactions? d. If a sample of apple juice contains \(1.00 \times 10^{2} \mathrm{g}\) of natural sugar, what is the maximum quantity of acetic acid that could be produced by the two fermentation reactions?

Gold does not dissolve in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) but readily dissolves in \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) (selenic acid). Which acid is the stronger oxidizing agent?

A puddle of coastal seawater, caught in a depression formed by some coastal rocks at high tide, begins to evaporate on a hot summer day as the tide goes out. If the volume of the puddle decreases to \(23 \%\) of its initial volume, what is the concentration of \(\mathrm{Na}^{+}\) after evaporation if it was \(0.449 \mathrm{M}\) initially?

How many milliliters of \(0.100 M\) NaOH are required to neutralize the following solutions? a. \(10.0 \mathrm{mL}\) of \(0.0500 M \mathrm{HCl}\) b. \(25.0 \mathrm{mL}\) of \(0.126 M \mathrm{HNO}_{3}\) c. \(50.0 \mathrm{mL}\) of \(0.215 M \mathrm{H}_{2} \mathrm{SO}_{4}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

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.