Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
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Chapter 13, Problem 67QAP

Using the equilibrium constants listed in Table 13.2, arrange the following 0.1 M aqueous solutions in order of increasing pH (from lowest to highest).

(a) NaNO2

(b) HCl

(c) NaF

(d) Zn(H2O)3(OH)(NO3)

Expert Solution & Answer
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Interpretation Introduction

Interpretation: The order of increasing pH value needs to be determined.

Concept Introduction: The dissociation reaction of a weak base is represented as follows:

  BOHB++OH

The expression for the base dissociation constant will be as follows:

  Kb=[B+][OH][BOH]

Here, [B+] is equilibrium concentration of conjugate acid, [OH] is equilibrium concentration of hydroxide ion and [BOH] is the equilibrium concentration of weak base.

From hydroxide ion concentration, pOH of the solution can be calculated as follows:

  pOH=log[OH]

From pOH, pH of the solution can be calculated as follows:

  pH=14pOH

Answer to Problem 67QAP

0.1 M HCl<0.1 M Zn(H2O)3(OH)(NO3)<0.1 M NaF<0.1 M NaNO2

Explanation of Solution

The given aqueous solutions are as follows:

M NaNO2 , 0.1 M HCl , 0.1 M NaF , 0.1 M Zn(H2O)3(OH)(NO3)

Here, HCl is a strong acid thus, it completely dissociates in the solution.

From the table 13.2, the equilibrium constants for NO2 is 1.7×1011 , F is 1.4×1011 and Zn(H2O)3(OH)+ is 3.0×105 .

(a)

The dissociation reaction of NaNO2 is represented as follows:

  NaNO2+H2OHNO2+NaOH0.1             -            -              -0.1-x          -           x              x

The equilibrium constant can be calculated as follows:

  Kb=[HNO2][NaOH][NaNO2]

Putting the values,

  1.7×1011=(x)(x)(0.1x)

Thus,

  x=1.304×106,+1.304×106

Therefore, the value of [NaOH] is 1.304×106 .

Thus,

  [OH]=1.304×106

The pOH of the solution can be calculated as follows:

  pOH=log[OH]

Putting the values,

  pOH=log(1.304×106)=5.88

The pH of the solution can be calculated as follows:

  pH=14pOH

Putting the values,

  pH=145.88=8.12

(b)

HCl is a strong acid, the dissociation can be represented as follows:

  HClH++Cl

Since, molarity of HCl is 0.1 M thus, concentration of hydrogen ion is 0.1 M.

The pH of the solution can be calculated as follows:

  pH=log[H+]

Putting the value,

  pH=log(0.1)=1

(c)

0.1 M NaF solution.

The dissociation reaction can be represented as follows:

  NaF+H2OHF+NaOH0.1      -            -        -0.1-x    -           x         x

The equilibrium constant for the reaction will be:

  kb=[HF][NaOH][NaF]

Putting the values,

  1.4×1011=(x)(x)(0.1x)

Thus,

  x=1.18×106,+1.18×106

The value of x cannot be negative thus, it is equal to 1.18×106 .

Thus,

  [NaOH]=[OH]=1.18×106

The pOH of the solution can be calculated as follows:

  pOH=log[OH]

Putting the value,

  pOH=log(1.18×106)=5.93

The pH of the solution can be calculated as follows:

  pH=14pOH

Putting the value,

  pH=145.93=8.07

(d)

0.1 M Zn(H2O)3(OH)(NO3)

The decomposition reaction can be represented as follows:

  Zn( H 2O)3(OH)(NO3)+H2OZn( H 2O)3(OH)2+HNO30.1                                     -                       -                    -0.1-x                                -                         x                    x

The equilibrium constant for the reaction will be:

  kb=[Zn( H 2 O)3( OH)2][HNO3][Zn( H 2 O)3(OH)(N O 3)]

Putting the values,

  3.0×105=(x)(x)(0.1x)

Thus,

  x=0.00174,+0.00712

Since, the value of x cannot be negative thus, it is equal to 0.00712 .

  [HNO3]=0.00712

Since, HNO3 is a strong acid thus, [H+]=0.00712

The pH of the solution can be calculated as follows:

  pH=log[H+]

Putting the values,

  pH=log(0.00712)=2.15

The order of increasing pH value can be represented as follows:

0.1 M HCl<0.1 M Zn(H2O)3(OH)(NO3)<0.1 M NaF<0.1 M NaNO2

Conclusion

Thus, the order of increasing pH value can be represented as follows:

0.1 M HCl<0.1 M Zn(H2O)3(OH)(NO3)<0.1 M NaF<0.1 M NaNO2

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Chapter 13 Solutions

Chemistry: Principles and Reactions

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