   Chapter 16.10, Problem 2.3ACP

Chapter
Section
Textbook Problem

Write an equation for the reaction of the amide ion (a stronger base than OH−) and water. Does the equilibrium favour products or reactants?

Interpretation Introduction

Interpretation:

The equation for the reaction of an amide ion with water has to be writtern and the reaction is reactant or product favored at equilibrium has to be exlained.

Concept introduction:

An equilibrium constant (K) is the ratio of the concentration of products and reactants raised to appropriate stoichiometric coefficient at equilibrium.

The reaction of an acid HA with water is written as,

HA(aq)+H2O(l)H3O+(aq)+A(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

Ka=[H3O+][A][HA]

An equilibrium constant (K) with subscript a indicates that it is an equilibrium constant of an acid in water.

The reaction of any base B with water is written as,

B(aq)+H2O(l)BH(aq)+OH(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

Kb=[BH][OH][B]

An equilibrium constant (K) with subscript b indicates that it is an equilibrium constant of the base in water.

Explanation

The value of dissociation constant is given below.

The dissociation constant (Kb) for NH2 is very large.

The dissociation constant (Kb) for OH is 1.

The dissociation constant (Ka) for H2O is 1×1014.

The dissociation constant (Ka) for NH3 is very small.

The reaction between NH2 and H2O is written as,

NH2(aq)+H2O(aq)NH3(aq)+OH(aq)

Since larger the value of Ka stronger the acid and larger the value of Kb stronger the base

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