Chapter 16, Problem 75PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Hydrazine, N2H4, can interact with water in two steps.N2H4(aq) + H2O(ℓ) ⇄ N2H5+(aq) + OH−(aq) Kb1 = 8.5 × 10−7N2H5+(aq) + H2O(ℓ) ⇄ N2H62+(aq) + OH−(aq) Kb2 = 8.9 × 10−16 (a) What is the concentration of OH−, N2H5+ and N2H62+ in a 0.010M aqueous solution of hydrazine? (b) What is the pH of the 0.010M solution hydrazine?

(a)

Interpretation Introduction

Interpretation:

The concentration of OH-, N2H5+ and N2H62+ in a 0.010M aqueous solution of hydrazine has to be determined.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb: It is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Explanation

Hydrazine, N2H4 can interact with water in two steps.

First ionization:

N2H4(aq)Â +Â H2O(l)Â â‡ŒN2H5+(aq)Â +Â OH-(aq)Â Â Â Kb1=8.5Ã—10-7

EquilibriumÂ expression:Kb1=Â [N2H5+][OH-][N2H4]

Second ionization:

N2H5+(aq)Â +Â H2O(l)â‡ŒN2H62+(aq)Â +Â OH-(aq)Â Â Kb2=8.9Ã—10-16

EquilibriumÂ expression:Kb2=Â [N2H62+][OH-][N2H5+]

From the Kb1Â andÂ Kb2 values, Â Kb2 is smaller than the Â Kb1.

Therefore, OH- is almost produced entirely from.

Letâ€™s calculate the OH- from Â Kb1.

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

Â Â Â Â Â Â Â Â N2H4(aq)Â +Â H2OÂ (aq)â‡ŒÂ N2H5+(aq)Â +Â OHâˆ’(aq)IÂ Â Â Â Â Â Â 0.010Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --CÂ Â Â Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â +xEÂ Â Â Â Â Â Â (0.010-x)Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

EquilibriumÂ expression:Kb1=Â [N2H5+][OH-][N2H4]

8.5Ã—10-7Â =Â (x)(x)0.010-Â x8.5Ã—10-7Â =Â (x)20.010-Â x(0.010-x)Â approximatelyÂ equalsÂ toÂ 0.0108.5Â Ã—10âˆ’7Â =Â (x)20.010Â Â Â Â Â Â Â Â Â Â x2=Â Â (0

(b)

Interpretation Introduction

Interpretation:

pH of the 0.010M solution hydrazine has to be calculated

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb: It is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

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