EBK INTRODUCTORY CHEMISTRY
EBK INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 9780134553153
Author: CORWIN
Publisher: PEARSON CO
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Chapter 14, Problem 73E
Interpretation Introduction

(a)

Interpretation:

A balanced net ionic equation for the acid-base reaction, HCl(aq)+ KOH(aq) KCl(aq)+H2O(l) is to be stated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation, the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. The chemical equation representing the reaction in which only the species actually participating are written is termed as an ionic reaction.

Expert Solution
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Answer to Problem 73E

A balanced net ionic equation for the acid-base reaction, HCl(aq)+ KOH(aq) KCl(aq)+H2O(l) is shown below.

H+(aq)+OH(aq)H2O(l)

Explanation of Solution

The chemical reaction is given below.

HCl(aq)+ KOH(aq) KCl(aq)+H2O(l)

It is balanced as the number of atoms on each side is the same. HCl, KOH, and KCl are strong electrolytes. H2O is a weak electrolyte. The balanced nonionic equation given above is converted into a total ionic equation by writing the strong electrolytes in ionized form and the weak electrolytes in nonionized form.

H+(aq)+Cl(aq)+ K+(aq)+OH(aq) K+(aq)+Cl(aq)+H2O(l)

In the above total ionic equation, K+ and Cl act as spectator ions and are removed to obtain the net ionic equation shown below.

H+(aq)+OH(aq)H2O(l)

Conclusion

A balanced net ionic equation for the given acid-base reaction has been rightfully stated.

Interpretation Introduction

(b)

Interpretation:

A balanced net ionic equation for the acid-base reaction, HC2H3O2(aq) + Ca(OH)2(aq) Ca(C2H3O2)2(aq) +H2O(l) is to be stated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation, the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. The chemical equation representing the reaction in which only the species actually participating are written is termed as an ionic reaction.

Expert Solution
Check Mark

Answer to Problem 73E

A balanced net ionic equation for the acid-base reaction, HC2H3O2(aq) + Ca(OH)2(aq) Ca(C2H3O2)2(aq) +H2O(l) is shown below.

2HC2H3O2(aq)+2OH(aq)2C2H3O2(aq) +H2O(l)

Explanation of Solution

The chemical reaction is given below.

HC2H3O2(aq) + Ca(OH)2(aq) Ca(C2H3O2)2(aq) +H2O(l)

The balanced chemical reaction is given below.

2HC2H3O2(aq) + Ca(OH)2(aq) Ca(C2H3O2)2(aq) +H2O(l)

It is balanced as the number of atoms on each side is the same. Ca(OH)2, and Ca(C2H3O2)2 are strong electrolytes. HC2H3O2 and H2O are weak electrolytes. The balanced nonionic equation given above is converted into a total ionic equation by writing the strong electrolytes in ionized form and the weak electrolytes in nonionized form.

2HC2H3O2(aq) + Ca2+(aq)+2OH(aq) Ca2+(aq)+2C2H3O2(aq) +H2O(l)

In the above total ionic equation, Ca2+ acts as a spectator ion and is removed to obtain the net ionic equation shown below.

2HC2H3O2(aq)+2OH(aq)2C2H3O2(aq) +H2O(l)

Conclusion

A balanced net ionic equation for the given acid-base reaction has been rightfully stated.

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Chapter 14 Solutions

EBK INTRODUCTORY CHEMISTRY

Ch. 14 - Prob. 11CECh. 14 - Prob. 12CECh. 14 - Prob. 13CECh. 14 - Prob. 14CECh. 14 - Prob. 15CECh. 14 - Prob. 16CECh. 14 - Prob. 17CECh. 14 - Prob. 1KTCh. 14 - Prob. 2KTCh. 14 - Prob. 3KTCh. 14 - Prob. 4KTCh. 14 - Prob. 5KTCh. 14 - Prob. 6KTCh. 14 - Prob. 7KTCh. 14 - Prob. 8KTCh. 14 - Prob. 9KTCh. 14 - Prob. 10KTCh. 14 - Prob. 11KTCh. 14 - Prob. 12KTCh. 14 - Prob. 13KTCh. 14 - Prob. 14KTCh. 14 - Prob. 15KTCh. 14 - Prob. 16KTCh. 14 - Prob. 17KTCh. 14 - Prob. 18KTCh. 14 - Prob. 19KTCh. 14 - Prob. 20KTCh. 14 - Prob. 21KTCh. 14 - Prob. 22KTCh. 14 - Prob. 23KTCh. 14 - Prob. 1ECh. 14 - Prob. 2ECh. 14 - Prob. 3ECh. 14 - Prob. 4ECh. 14 - Prob. 5ECh. 14 - Prob. 7ECh. 14 - Prob. 8ECh. 14 - Prob. 9ECh. 14 - Prob. 10ECh. 14 - Prob. 11ECh. 14 - Prob. 12ECh. 14 - Prob. 13ECh. 14 - Prob. 14ECh. 14 - Prob. 15ECh. 14 - Prob. 16ECh. 14 - Prob. 17ECh. 14 - Prob. 18ECh. 14 - Prob. 19ECh. 14 - Prob. 20ECh. 14 - Prob. 21ECh. 14 - Prob. 22ECh. 14 - Prob. 23ECh. 14 - Prob. 24ECh. 14 - Prob. 25ECh. 14 - Prob. 26ECh. 14 - Prob. 27ECh. 14 - Prob. 28ECh. 14 - Prob. 29ECh. 14 - Prob. 30ECh. 14 - Prob. 31ECh. 14 - Prob. 32ECh. 14 - Prob. 33ECh. 14 - Prob. 34ECh. 14 - Prob. 35ECh. 14 - Prob. 36ECh. 14 - Prob. 37ECh. 14 - Prob. 38ECh. 14 - Prob. 39ECh. 14 - Prob. 40ECh. 14 - Prob. 41ECh. 14 - Prob. 42ECh. 14 - Prob. 43ECh. 14 - Prob. 44ECh. 14 - Prob. 45ECh. 14 - Prob. 46ECh. 14 - Prob. 47ECh. 14 - Prob. 48ECh. 14 - Prob. 49ECh. 14 - Prob. 50ECh. 14 - Prob. 51ECh. 14 - Prob. 52ECh. 14 - Prob. 53ECh. 14 - Prob. 54ECh. 14 - Prob. 55ECh. 14 - Prob. 56ECh. 14 - Prob. 57ECh. 14 - Prob. 58ECh. 14 - Prob. 59ECh. 14 - Prob. 60ECh. 14 - Prob. 61ECh. 14 - Prob. 62ECh. 14 - Prob. 63ECh. 14 - Prob. 64ECh. 14 - Prob. 65ECh. 14 - Prob. 66ECh. 14 - Prob. 67ECh. 14 - Prob. 68ECh. 14 - Prob. 69ECh. 14 - Prob. 70ECh. 14 - Prob. 71ECh. 14 - Prob. 72ECh. 14 - Prob. 73ECh. 14 - Prob. 74ECh. 14 - Prob. 75ECh. 14 - Prob. 76ECh. 14 - Prob. 77ECh. 14 - Prob. 78ECh. 14 - Prob. 79ECh. 14 - Prob. 80ECh. 14 - Prob. 81ECh. 14 - Prob. 82ECh. 14 - Prob. 83ECh. 14 - Prob. 84ECh. 14 - Prob. 85ECh. 14 - Prob. 86ECh. 14 - Prob. 87ECh. 14 - Prob. 88ECh. 14 - Prob. 89ECh. 14 - Prob. 90ECh. 14 - Prob. 1STCh. 14 - Prob. 2STCh. 14 - Prob. 3STCh. 14 - Prob. 4STCh. 14 - Prob. 5STCh. 14 - Prob. 6STCh. 14 - Prob. 7STCh. 14 - Prob. 8STCh. 14 - Prob. 9STCh. 14 - Prob. 10STCh. 14 - Prob. 11STCh. 14 - Prob. 12STCh. 14 - Prob. 13STCh. 14 - Prob. 14STCh. 14 - Prob. 15STCh. 14 - Prob. 16ST
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