Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
Question
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Chapter 19, Problem 67QRT

(a)

Interpretation Introduction

Interpretation:

Oxidation state of nitrogen in the compounds present in given reaction has to be determined.

    N2H4(l)+HNO2(aq)HN3(aq)+2H2O(l)

(a)

Expert Solution
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Explanation of Solution

Oxidation state of nitrogen in N2H4:

Hydrogen has an oxidation state of +1.  Sum of oxidation state of all the atoms in a compound will be zero.  Therefore, oxidation state of nitrogen can be calculated as shown below,

    2x+4(+1) =0 2x =4 x =2

Oxidation state of nitrogen in HNO2:

Hydrogen has an oxidation state of +1.  Oxygen has an oxidation state of 2.  Sum of oxidation state of all the atoms in a compound will be zero.  Therefore, oxidation state of nitrogen can be calculated as shown below,

    1+x+2(2) =0 x3 =0 x =+3

Oxidation state of nitrogen in HN3:

Hydrogen has an oxidation state of +1.  Sum of oxidation state of all the atoms in a compound will be zero.  Therefore, oxidation state of nitrogen can be calculated as shown below,

    1+3x =03x+1 =0x =1/3

(b)

Interpretation Introduction

Interpretation:

Oxidizing agent in the below given reaction has to be identified.

    N2H4(l)+HNO2(aq)HN3(aq)+2H2O(l)

(b)

Expert Solution
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Explanation of Solution

Oxidation state of nitrogen in N2H4 is 2, oxidation state of nitrogen in HNO2 is +3 and that of HN3 is 1/3.  Oxidizing agent is the one that gets reduced in the reaction.  This means the oxidizing agent gains electrons.  This in turn means that the oxidation state becomes reduced.  Therefore, the oxidizing agent in the given reaction is found to be HNO2 because, the oxidation state of nitrogen is reduced from +3 to 1/3 by gaining electrons.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.010M solution of HN3 has to be calculated.

(c)

Expert Solution
Check Mark

Explanation of Solution

Ionization constant for hydrazoic acid is given as 1.0×105.

Equation for the reaction of hydrazoic acid with water at equilibrium can be given as,

    HN3(aq)+H2O(l)H3O+(aq)+N3-(aq)

Equilibrium expression for this can be given as,

    Ka = [H3O+][N3-][HN3](1)

Concentration of the products increases till the equilibrium is reached.

                HN3(aq) H3O+(aq) N3(aq)
Initial Concentration (M)        0.010        5.00            0.00
Change as Reaction occurs (M)    x        +x            +x
Equilibrium Concentration (M)    0.010x    +x            +x

Substituting these values in equation (1),

     Ka = (x)(x)(0.010-x)1.0×105 = x2(0.010) x2 = (1.0×105)(0.010) = 1.0×105×0.010 = 0.00316×0.1 = 3.16×104M

Therefore, concentration of hydronium ion is 3.16×10-4M.

It is known that pH can be calculated from the concentration of hydronium ion using the equation shown below,

    pH = -log[H3O+] = -log(3.2×10-4) = 3.494

Therefore, pH of solution of hydrazoic acid is calculated to be 3.494.

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Chapter 19 Solutions

Chemistry: The Molecular Science

Ch. 19.5 - Prob. 19.9ECh. 19.5 - Prob. 19.3PSPCh. 19.5 - Use the terms oxidation, reduction, oxidizing...Ch. 19.5 - Prob. 19.11ECh. 19.6 - Prob. 19.13ECh. 19.6 - Prob. 19.14ECh. 19.6 - Prob. 19.15CECh. 19.6 - Prob. 19.16CECh. 19.6 - Prob. 19.4PSPCh. 19.6 - Prob. 19.5PSPCh. 19.6 - Prob. 19.17ECh. 19.6 - Prob. 19.6PSPCh. 19.6 - Prob. 19.7PSPCh. 19.6 - Prob. 19.8PSPCh. 19 - Prob. 1QRTCh. 19 - Prob. 2QRTCh. 19 - Prob. 3QRTCh. 19 - Prob. 4QRTCh. 19 - Prob. 5QRTCh. 19 - Prob. 6QRTCh. 19 - Prob. 7QRTCh. 19 - Prob. 8QRTCh. 19 - Prob. 9QRTCh. 19 - Prob. 10QRTCh. 19 - Prob. 11QRTCh. 19 - Prob. 12QRTCh. 19 - Prob. 13QRTCh. 19 - Prob. 14QRTCh. 19 - Prob. 15QRTCh. 19 - Prob. 16QRTCh. 19 - Prob. 17QRTCh. 19 - Prob. 18QRTCh. 19 - Prob. 19QRTCh. 19 - Prob. 20QRTCh. 19 - Prob. 21QRTCh. 19 - Prob. 22QRTCh. 19 - Prob. 23QRTCh. 19 - Prob. 24QRTCh. 19 - Prob. 25QRTCh. 19 - Prob. 26QRTCh. 19 - Identify the substance or substances produced by...Ch. 19 - Prob. 28QRTCh. 19 - Prob. 29QRTCh. 19 - Prob. 30QRTCh. 19 - Prob. 31QRTCh. 19 - Prob. 32QRTCh. 19 - Prob. 33QRTCh. 19 - Prob. 34QRTCh. 19 - Prob. 35QRTCh. 19 - Prob. 36QRTCh. 19 - Prob. 37QRTCh. 19 - Prob. 38QRTCh. 19 - Prob. 39QRTCh. 19 - Prob. 40QRTCh. 19 - Prob. 41QRTCh. 19 - Prob. 42QRTCh. 19 - A human body contains approximately 5 L of blood....Ch. 19 - Prob. 44QRTCh. 19 - Prob. 45QRTCh. 19 - Prob. 46QRTCh. 19 - Prob. 47QRTCh. 19 - Prob. 48QRTCh. 19 - Prob. 49QRTCh. 19 - Prob. 50QRTCh. 19 - Prob. 51QRTCh. 19 - Prob. 52QRTCh. 19 - Prob. 53QRTCh. 19 - Prob. 54QRTCh. 19 - Prob. 55QRTCh. 19 - Prob. 56QRTCh. 19 - Prob. 57QRTCh. 19 - Prob. 58QRTCh. 19 - Prob. 59QRTCh. 19 - Prob. 60QRTCh. 19 - Prob. 61QRTCh. 19 - Prob. 62QRTCh. 19 - Prob. 63QRTCh. 19 - Prob. 64QRTCh. 19 - Prob. 65QRTCh. 19 - Prob. 66QRTCh. 19 - Prob. 67QRTCh. 19 - Prob. 68QRTCh. 19 - Prob. 69QRTCh. 19 - Prob. 70QRTCh. 19 - Prob. 71QRTCh. 19 - Prob. 72QRTCh. 19 - Prob. 73QRTCh. 19 - Prob. 74QRTCh. 19 - Use the phase diagram for sulfur for Question 75....Ch. 19 - Prob. 76QRTCh. 19 - Prob. 77QRTCh. 19 - Prob. 78QRTCh. 19 - Prob. 79QRTCh. 19 - Prob. 80QRTCh. 19 - A natural brine found in Arkansas has a bromide...Ch. 19 - Prob. 82QRTCh. 19 - Prob. 83QRTCh. 19 - Prob. 84QRTCh. 19 - At 20. C the vapor pressure of white phosphorus is...Ch. 19 - Prob. 86QRTCh. 19 - Assume that the radius of Earth is 6400 km, the...Ch. 19 - Prob. 88QRTCh. 19 - Prob. 89QRTCh. 19 - Prob. 90QRTCh. 19 - Prob. 91QRTCh. 19 - Prob. 92QRTCh. 19 - Prob. 93QRTCh. 19 - Prob. 94QRTCh. 19 - Prob. 95QRTCh. 19 - Use a Born-Haber cycle (Sec. 5-13) to calculate...Ch. 19 - Prob. 97QRTCh. 19 - Elemental analysis of a borane indicates this...Ch. 19 - Prob. 99QRTCh. 19 - Prob. 100QRT
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