/*! 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 58 Complete and balance each of the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 58

Complete and balance each of the following neutralization reactions, name the products, and write the overall ionic and net ionic equations. a. \(\mathrm{HBr}(a q)+\mathrm{KOH}(a q) \rightarrow\) b. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) c. \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \rightarrow\) d. \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{Sr}(\mathrm{OH})_{2}(a q) \rightarrow\)

Short Answer

Expert verified
a. HBr(aq) + KOH(aq) b. H3PO4(aq) + Ba(OH)2(aq) c. Al(OH)3(s) + HCl(aq) d. CH3COOH(aq) + Sr(OH)2(aq) Answer: a. H^(+)(aq) + OH^(-)(aq) -> H2O(l) b. 6H^(+)(aq) + 6OH^(-)(aq) -> 6H2O(l) c. 3H^(+)(aq) + 3OH^(-)(aq) -> 3H2O(l) d. 2H^(+)(aq) + 2OH^(-)(aq) -> 2H2O(l)

Step by step solution

01

Balancing the chemical equation

The balanced equation will be: \(\mathrm{HBr}(aq) + \mathrm{KOH}(aq) \rightarrow \mathrm{KBr}(aq) + \mathrm{H_2O}(l)\)
02

Naming the products

The products are: Potassium bromide (KBr) and Water (H2O).
03

Writing the ionic and net ionic equations

Ionic equation: \(\mathrm{H}^+(aq) + \mathrm{Br}^-(aq) + \mathrm{K}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{K}^+(aq) + \mathrm{Br}^-(aq) + \mathrm{H_2O}(l)\) Net ionic equation: \(\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{H_2O}(l)\) b. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\)
04

Balancing the chemical equation

The balanced equation will be: \(2 \mathrm{H}_{3} \mathrm{PO}_{4}(aq) + 3 \mathrm{Ba}(\mathrm{OH})_{2}(aq) \rightarrow \mathrm{Ba}_3(\mathrm{PO_4})_2(s) + 6 \mathrm{H_2O}(l)\)
05

Naming the products

The products are: Barium phosphate (Ba3(PO4)2) and Water (H2O).
06

Writing the ionic and net ionic equations

Ionic equation: \(6 \mathrm{H}^+(aq) + 6 \mathrm{OH}^-(aq) + 2 \mathrm{PO}_4^{3-}(aq) + 3 \mathrm{Ba}^{2+}(aq) \rightarrow \mathrm{Ba}_3(\mathrm{PO_4})_2(s) + 6 \mathrm{H_2O}(l)\) Net ionic equation: \(6 \mathrm{H}^+(aq) + 6 \mathrm{OH}^-(aq) \rightarrow 6\mathrm{H_2O}(l)\) c. \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \rightarrow\)
07

Balancing the chemical equation

The balanced equation will be: \(\mathrm{Al}(\mathrm{OH})_{3}(s) + 3\mathrm{HCl}(aq) \rightarrow \mathrm{AlCl_3}(aq) + 3\mathrm{H_2O}(l)\)
08

Naming the products

The products are: Aluminum chloride (AlCl3) and Water (H2O).
09

Writing the ionic and net ionic equations

Ionic equation: \(\mathrm{Al}^{3+}(aq) + 3 \mathrm{OH}^-(aq) + 3 \mathrm{H}^+(aq) + 3\mathrm{Cl}^-(aq) \rightarrow \mathrm{Al}^{3+}(aq) + 3\mathrm{Cl}^-(aq) + 3 \mathrm{H_2O}(l)\) Net ionic equation: \(3 \mathrm{H}^+(aq) + 3 \mathrm{OH}^-(aq) \rightarrow 3 \mathrm{H_2O}(l)\) d. \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{Sr}(\mathrm{OH})_{2}(a q) \rightarrow\)
10

Balancing the chemical equation

The balanced equation will be: \(2 \mathrm{CH}_{3} \mathrm{COOH}(aq) + \mathrm{Sr}(\mathrm{OH})_{2}(aq) \rightarrow \mathrm{Sr}(\mathrm{CH}_3 \mathrm{COO})_{2}(aq) + 2 \mathrm{H_2O}(l)\)
11

Naming the products

The products are: Strontium acetate (Sr(CH3COO)2) and Water (H2O).
12

Writing the ionic and net ionic equations

Ionic equation: \(2 \mathrm{CH}_{3} \mathrm{COO}^-(aq) + 2 \mathrm{H}^+(aq) + \mathrm{Sr}^{2+}(aq) + 2 \mathrm{OH}^-(aq) \rightarrow \mathrm{Sr}^{2+}(aq) + 2 \mathrm{CH}_{3} \mathrm{COO}^-(aq) + 2 \mathrm{H_2O}(l)\) Net ionic equation: \(2 \mathrm{H}^+(aq) + 2 \mathrm{OH}^-(aq) \rightarrow 2 \mathrm{H_2O}(l)\)

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.

Balancing Chemical Equations
To understand neutralization reactions, one must start with the basics of balancing chemical equations. It's a fundamental practice in chemistry that ensures the law of conservation of mass is upheld鈥攖hat is, the same number of atoms of each element is present on both sides of the equation.

When balancing equations, first write down the unbalanced equation. Then count the number of atoms of each element on both the reactant and product sides. Adjust coefficients鈥攏ever subscripts, as this changes the substance鈥攖o balance the atoms. For example, in the reaction between hydrobromic acid (HBr) and potassium hydroxide (KOH), balancing leads to one molecule of HBr reacting with one molecule of KOH to form one molecule of potassium bromide (KBr) and one molecule of water (H2O).
Ionic Equations
In neutralization reactions involving strong acids and bases, it's useful to represent what's actually happening in solution. This is where ionic equations come in, which show the ions in an aqueous solution.

Consider the reaction between HBr and KOH. In an ionic equation, each strong electrolyte (substances that dissociate completely into ions in solution) is written as its constituent ions. So HBr and KOH dissociate into H+, Br-, K+, and OH-. The ionic equation is a realistic representation of the species that are in the solution before and after the reaction.
Net Ionic Equations
The net ionic equation simplifies the ionic equation by removing spectator ions鈥攖hose that do not participate in the actual reaction. They appear unchanged on both sides of the ionic equation.

For the same reaction between HBr and KOH, K+ and Br- are spectator ions. Removing them leaves us with the essence of the reaction, where H+ ions react with OH- ions to form water, H2O. Thus, the net ionic equation is H+(aq) + OH-(aq) 鈫� H2O(l), providing a clear picture of the chemical change.
Acid-Base Reactions
Neutralization reactions are a type of acid-base reaction in which an acid reacts with a base to produce a salt and usually water. In essence, these reactions are about the transfer of protons (H+) from the acid to the base.

This proton transfer neutralizes the acid and base, often resulting in a pH that is closer to neutral (7), depending on the strength of the reactants. In our examples, we have strong acids and strong bases, so the products will tend to be neutral. However, if a weak acid reacts with a strong base or vice versa, the resulting solution will have a pH that is not neutral.

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

Nitrogen-Fixing Bacteria Bacteria found among the roots of legumes perform an important biological function, converting nitrogen to ammonia in a process known as nitrogen fixation. The electrons required for this redox reaction are supplied by transition metal-containing enzymes called nitrogenases. The unbalanced chemical equation for this process is \(\mathrm{N}_{2}(g)+\mathrm{H}^{+}(a q)+\mathrm{M}^{2+}(a q) \rightarrow \mathrm{NH}_{3}(a q)+\mathrm{H}_{2}(g)+\mathrm{M}^{3+}(a q)\) where M represents a transition metal such as iron. Nitrification is a multistep process in which the nitrogen in organic and inorganic compounds is biochemically oxidized. Bacteria and fungi are responsible for a part of the nitrification process described by the reaction: $$ \mathrm{NH}_{4}^{+}(a q)+\mathrm{M}^{3+}(a q) \rightarrow \mathrm{NO}_{2}^{-}(a q)+\mathrm{M}^{2+}(a q) $$ a. What are the oxidation numbers of nitrogen in the reactants and products of each reaction? b. Which compounds or ions are being reduced in each reaction? c. Balance the equations in acidic solution.

3\. Many nonmetal oxides react with water to form acidic solutions. Give the formula and name for the acids produced from the following reactions: a. \(P_{4} O_{10}+6 H_{2} O \rightarrow\) b. \(\mathrm{SeO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow\) c. \(\mathrm{B}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\)

Balance the following half-reactions by adding the appropriate number of electrons. Which are oxidation half-reactions and which are reduction half- reactions? a. \(\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)\) b. \(\operatorname{Ag}\left[(s) \rightarrow \operatorname{Ag}(s)+\bar{I}^{-}(a q)\right.\) c. \(\mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\) d. \(\mathrm{I}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{IO}_{3}^{-}(a q)+12 \mathrm{H}^{+}(a q)\)

Write a balanced molecular equation and a net ionic equation for the following reactions: a. Solid magnesium hydroxide reacts with a solution of sulfuric acid. b. Solid magnesium carbonate reacts with a solution of hydrochloric acid. c. Ammonia gas reacts with hydrogen chloride gas. d. Gaseous sulfur trioxide is dissolved in water and reacts with a solution of sodium hydroxide.

What name is given to a proton donor?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

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.