   # Hydrazine, N 2 H 4 , can interact with water in two steps. N 2 H 4 (aq) + H 2 O( ℓ ) ⇄ N 2 H 5 + (aq) + OH − (aq) K b1 = 8.5 × 10 −7 N 2 H 5 + (aq) + H 2 O( ℓ ) ⇄ N 2 H 6 2+ (aq) + OH − (aq) K b2 = 8.9 × 10 −16 (a) What is the concentration of OH − , N 2 H 5 + and N 2 H 6 2+ in a 0.010M aqueous solution of hydrazine? (b) What is the pH of the 0.010M solution hydrazine? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 73PS
Textbook Problem
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## 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 of Solution

Hydrazine, N2H4 can interact with water in two steps.

First ionization:

N2H4(aq) + H2O(l) N2H5+(aq) + OH-(aq)   Kb1=8.5×10-7

Equilibrium expression:Kb1[N2H5+][OH-][N2H4]

Second ionization:

N2H5+(aq) + H2O(l)N2H62+(aq) + OH-(aq)  Kb2=8.9×10-16

Equilibrium expression:Kb2[N2H62+][OH-][N2H5+]

From the Kb1 and Kb2 values,  Kb2 is smaller than the  Kb1.

Therefore, OH- is almost produced entirely from.

Let’s calculate the OH- from  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.

N2H4(aq) + H2O (aq) N2H5+(aq) + OH(aq)I       0.010               --                       --                --C           -x                --                     +x              +xE       (0.010-x)         --                      x                x

Equilibrium expression:Kb1[N2H5+][OH-][N2H4]

8.5×10-7 = (x)(x)0.010- x8.5×10-7 = (x)20.010- x(0.010-x) approximately equals to 0.0108.5 ×107 = (x)20.010          x2=  (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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