INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN
INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN
8th Edition
ISBN: 9780134580289
Author: CORWIN
Publisher: PEARSON
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Chapter 14, Problem 76E
Interpretation Introduction

(a)

Interpretation:

A balanced net ionic equation for the reaction, Zn(NO3)2(aq)+ NaOH(aq)Zn(OH)2(s)+NaNO3(aq), 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 76E

A balanced net ionic equation for the solution reaction, Zn(NO3)2(aq)+ NaOH(aq)Zn(OH)2(s)+NaNO3(aq) is shown below.

Zn2+(aq)+2OH(aq)Zn(OH)2(s)

Explanation of Solution

The chemical reaction is given below.

Zn(NO3)2(aq)+ NaOH(aq)Zn(OH)2(s)+NaNO3(aq)

The balanced reaction is given below.

Zn(NO3)2(aq)+ 2NaOH(aq)Zn(OH)2(s)+2NaNO3(aq)

It is balanced as the number of atoms on each side is the same. Zn(NO3)2, NaOH, and NaNO3 are strong electrolytes. Zn(OH)2 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.

Zn2+(aq)+2NO3(aq)+ 2Na+(aq)+2OH(aq)Zn(OH)2(s)+2Na+(aq)+2NO3(aq) In the above total ionic equation, Na+ and NO3acts as spectator ions and are removed to obtain the net ionic equation shown below.

Zn2+(aq)+2OH(aq)Zn(OH)2(s)

Conclusion

A balanced net ionic equation for the given solution reaction has been rightfully stated.

Interpretation Introduction

(b)

Interpretation:

A balanced net ionic equation for the solution reaction, MgSO4(aq)+NH4OH(aq)Mg(OH)2(s)+(NH4)2SO4(aq) 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 76E

A balanced net ionic equation for the solution reaction, MgSO4(aq)+NH4OH(aq)Mg(OH)2(s)+(NH4)2SO4(aq) is shown below.

Mg2+(aq)++2OH(aq)Mg(OH)2(s)

Explanation of Solution

The chemical reaction is given below.

MgSO4(aq)+NH4OH(aq)Mg(OH)2(s)+(NH4)2SO4(aq)

The balanced chemical equation is given below.

MgSO4(aq)+2NH4OH(aq)Mg(OH)2(s)+(NH4)2SO4(aq)

It is balanced as the number of atoms on each side is the same. MgSO4, NH4OH and (NH4)2SO4 are strong electrolytes. Mg(OH)2 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.

Mg2+(aq)+SO42(aq)+2NH4+(aq)+2OH(aq)Mg(OH)2(s)+2NH4+(aq)+SO42(aq) In the above total ionic equation, SO42, and NH4+ act as spectator ions and are removed from the total ionic equation. Therefore, the net ionic equation is given below.

Mg2+(aq)++2OH(aq)Mg(OH)2(s)

Conclusion

A balanced net ionic equation for the given solution reaction has been rightfully stated.

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Chapter 14 Solutions

INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN

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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