   Chapter 14, Problem 96E

Chapter
Section
Textbook Problem

# For the reaction of hydrazine (N2H4) in water, H 2 NNH 2 ( a q ) + H 2 O ( l ) ⇌ H 2 NNH 3 + ( a q ) + OH − ( a q ) Kb is 3.0 × 10−6. Calculate the concentrations of all species and the pH of a 2.0-M solution of hydrazine in water.

Interpretation Introduction

Interpretation: The concentration of all the species present in the given 2.0M solution of hydrazine in water and the pH value of the solution is to be calculated.

Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The pOH of a solution is calculated by the formula, pOH=log[OH]

The sum, pH+pOH=14

At equilibrium, the equilibrium constant expression is expressed by the formula,

Kb=ConcentrationofproductsConcentrationofreactants

To determine: The concentration of all the species present in the given 2.0M solution of hydrazine in water and the pH value of the solution.

Explanation

Explanation

The major species present in the given solution of hydrazine (H2NNH2) in water are H2NNH3+ , OH and H2O .

H2NNH2 is a weak base.

In the presence of water, the reaction that takes place is,’

H2NNH2(aq)+H2O(l)H2NNH3+(aq)+OH(aq)

The major species present in the given solution of H2NNH3+ , OH and H2O .

The equilibrium constant expression for the given reaction is, Kb=[H2NNH3+][OH][H2NNH2]

At equilibrium, the equilibrium constant expression is expressed by the formula,

Kb=ConcentrationofproductsConcentrationofreactants

Where,

• Kb is the base ionization constant.

The equilibrium constant expression for the given reaction is,

Kb=[H2NNH3+][OH][H2NNH2] (1)

The [OH] is 2.45×10-3M_ .

The change in concentration of H2NNH2 is assumed to be x .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

H2NNH2(aq)H2NNH3+(aq)+OH(aq)Inititialconcentration2.000Changex+x+xEquilibriumconcentration2.0xxx

The equilibrium concentration of [H2NNH2] is (2.0x)M .

The equilibrium concentration of [H2NNH3+] is xM .

The equilibrium concentration of [OH] is xM .

The Kb value is given to be 3.0×106 .

Substitute the value of Kb , [H2NNH2] , [H2NNH3+] and [OH] in equation (1).

3.0×106=[x][x][2

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