Living by Chemistry
Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
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Chapter U6, Problem C23.1RE
Interpretation Introduction

(a)

Interpretation:

Whether hydrazine is a strong acid, strong base, weak acid or weak base needs to be determined.

Concept introduction:

An equilibrium reaction is represented as follows:

A + BAB

For the above reaction, the change in concentration of A and B is equal to that of AB.

For the above reaction, equilibrium constant can be represented as follows:

K=[AB][A][B]

Here, K is equilibrium constant.

Expert Solution
Check Mark

Explanation of Solution

The ionization equation for the hydrazine is represented as follows:

N2H4(aq) + H2O(l)N2H5 + (aq) + OH(aq)

The pH of 0.1 M hydrazine solution is 10. For solutions with pH value more than 7 are basic in nature thus, hydrazine is basic in nature.

Also, the given reaction is reversible thus, hydrazine does not completely react with water to release hydroxide ion. Thus, it is a weak base.

Interpretation Introduction

(b)

Interpretation:

All the compounds and ions present in 0.1 M solution of hydrazine solution needs to be listed.

Concept introduction:

An equilibrium reaction is represented as follows:

A + BAB

For the above reaction, the change in concentration of A and B is equal to that of AB.

For the above reaction, equilibrium constant can be represented as follows:

K=[AB][A][B]

Here, K is equilibrium constant.

Expert Solution
Check Mark

Explanation of Solution

The ionization equation for the hydrazine is represented as follows:

N2H4(aq) + H2O(l)N2H5 + (aq) + OH(aq)

The pH of solution is 10, thus, hydroxide ion concentration can be calculated as follows:

From pH, pOH of solution will be:

pOH=14pH=1410=4

Thus, hydroxide ion concentration will be:

[OH]=10pOH=104 M

The ICE table for the equation can be represented as follows:

         N2H4(aq) + H2O(l)N2H5 + (aq) + OH(aq)I          0.1              -               -                   -C           x           -              + x                   + xE       0.1x           -                x                  x

Here,

x=[OH]=104 M

Also,

x=[OH]=[N2H5 + ]=104 M

The equilibrium concentration of  N2H4 will be:

[ N2H4]=0.1104=0.099 M

There is no change in the concentration of liquid water thus, the species present in 0.1 M solution are N2H4(aq), N2H5 + (aq) and OH(aq).

(c)

Interpretation Introduction

Interpretation:

The equilibrium equation for hydrazine needs to be written.

Concept introduction:

An equilibrium reaction is represented as follows:

A + BAB

For the above reaction, the change in concentration of A and B is equal to that of AB.

For the above reaction, equilibrium constant can be represented as follows:

K=[AB][A][B]

Here, K is equilibrium constant.

(c)

Expert Solution
Check Mark

Explanation of Solution

The ionization equation for the hydrazine is represented as follows:

N2H4(aq) + H2O(l)N2H5 + (aq) + OH(aq)

The equilibrium constant equation for the above reaction can be represented as follows:

K=[N2H5 + (aq)][OH(aq)][N2H4(aq)]

Here, water is not included in the equilibrium expression because it is in liquid form and the species in solid and liquid form are not represented in the equilibrium expression.

The given value of equilibrium constant K for the reaction of hydrazine is 8.7×107

Thus,

8.7×107=[N2H5 + (aq)][OH(aq)][N2H4(aq)]

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Chapter U6 Solutions

Living by Chemistry

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