EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 8220101443908
Author: Harris
Publisher: MAC HIGHER
Question
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Chapter 11, Problem 11.9P

(a)

Interpretation Introduction

Interpretation:

Value of pH and concentrations of H2M, HM and M2 in 0.100 M H2M solution has to be calculated.

Concept Introduction:

Ka2 for reaction is defined as second dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. Kb2 for reaction is defined as second association constant for base. It helps to determine the amount of hydroxide ions in base.

(a)

Expert Solution
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Explanation of Solution

First equilibrium reaction for malonic acid that is designated as Ka1 is as follows:

  H2MKa1HM+H+

The concentration of malonic acid is 0.100 M. Consider concentration of H+ to be x, thus value of [HM] is also x and that of [H2M] is 0.100x. The expression for equilibrium constant Ka1 for malonic acid is given as follows:

  Ka1=[HM][H+][H2M]        (1)

Rearrange equation (1) for [H+].

  [H+]=Ka1[H2M][HM]        (2)

Here,

[HM] is concentration of HM.

[H+] is concentration of H+.

[H2M] is concentration of H2M.

Substitute 1.42×103 for Ka1, x for [H+], x for [HM] and 0.100x for [H2M] in equation (2).

  x=(1.42×103)(0.100x)xx2=(1.42×103)(0.100x)x2+(1.42×103)x(1.42×104)=0x=1.12×102

Therefore, value of [H+] is 1.12×102 M, [HM] is 1.12×102 M and [H2M] is 0.089 M.

Second equilibrium reaction for malonic acid that is designated as Ka2 is as follows:

  HMKa2M2+H+

The expression for equilibrium constant Ka2 for malonic acid is given as follows:

  Ka2=[M2][H+][HM]        (3)

Rearrange equation (3) for [M2].

  [M2]=Ka2[HM][H+]        (4)

Here,

[HM] is concentration of HM.

[H+] is concentration of H+.

[M2] is concentration of M2.

Substitute 2.01×106 for Ka2, 1.12×102 for [H+] and 1.12×102 for [HM] in equation (6).

  [M2]=(2.01×106)(1.12×102)(1.12×102)=2.01×106 M

Therefore, value of [M2] is 2.01×106 M.

(b)

Interpretation Introduction

Interpretation:

Value of pH and concentrations of H2M, HM and M2 in 0.100 M NaHM solution has to be calculated.

Concept Introduction:

Ka2 for reaction is defined as second dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. Kb2 for reaction is defined as second association constant for base. It helps to determine the amount of hydroxide ions in base.

(b)

Expert Solution
Check Mark

Explanation of Solution

The equilibrium reaction for malonic acid is as follows:

  NaHMNa++HM

The expression used for calculation of concentration of H+ is as follows:

  [H+]=Ka1Ka2[HM]+Ka1KwKa1+[HM]        (5)

Here,

[HM] is concentration of HM.

[H+] is concentration of H+.

Ka1 is first acid equilibrium constant.

Ka2 is second equilibrium constant.

Kw is dissociation constant of water.

Substitute 1.42×103 for Ka1, 2.01×106 for Ka2, 0.100 for [HM] and 1014 for Kw in equation (6).

  [H+]=(1.42×103)(2.01×106)(0.100)+(1.42×103)(1014)1.42×103+(0.100)=2.85×1010+1.42×10170.10142=5.30×105 M

Therefore, value of [H+] is 5.30×105 M.

The expression used to calculate pH of [H+] is as follows:

  pH=log[H+]        (7)

Here,

[H+] is concentration of H+.

Substitute 5.30×105 for [H+] in equation (7).

  pH=log[5.30×105]=log5.30+5log10=4.28

Hence, pH of [H+] is 4.28.

First equilibrium reaction for malonic acid that is designated as Ka1 is as follows:

  H2MKa1HM+H+

The expression for equilibrium constant Ka1 for malonic acid is given as follows:

  Ka1=[HM][H+][H2M]        (8)

Rearrange equation (8) for [H2M].

  [H2M]=[HM][H+]Ka1        (9)

Here,

[HM] is concentration of HM.

[H+] is concentration of H+.

[H2M] is concentration of H2M.

Substitute 1.42×103 for Ka1, 5.30×105 for [H+] and 0.100 for [HM]  in equation (9).

  [H2M]=(0.100)(5.30×105)1.42×103=3.73×103 M

Therefore, value of [H2M] is 3.73×103 M.

Second equilibrium reaction for malonic acid that is designated as Ka2 is as follows:

  HMKa2M2+H+

The expression for equilibrium constant Ka2 for malonic acid is given as follows:

  Ka2=[M2][H+][HM]        (10)

Rearrange equation (10) for [M2].

  [M2]=Ka2[HM][H+]        (11)

Here,

[HM] is concentration of HM.

[H+] is concentration of H+.

[M2] is concentration of M2.

Substitute 2.01×106 for Ka2, 5.30×105 for [H+] and 0.100 for [HM] in equation (11).

  [M2]=(2.01×106)(0.100)(5.30×105)=3.9×103 M

Therefore, value of [M2] is 3.9×103 M.

(c)

Interpretation Introduction

Interpretation:

Value of pH and concentrations of H2M, HM and M2 in 0.100 M Na2M solution has to be calculated.

Concept Introduction:

Ka2 for reaction is defined as second dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. Kb2 for reaction is defined as second association constant for base. It helps to determine the amount of hydroxide ions in base.

(c)

Expert Solution
Check Mark

Explanation of Solution

The equilibrium reaction for Na2M is as follows:

  Na2MNa++M2        (12)

Equilibrium reaction for M2 is as follows:

  M2+H2OKb1HM+OHHM+H2OKb2H2M+OH

The relation between Ka2 and Kb1 is as follows:

  (Ka2)(Kb1)=Kw        (13)

Rearrange equation (13) for Kb1.

  Kb1=KwKa2        (14)

Here,

Ka2 is second acid equilibrium constant.

Kb1 is first base equilibrium constant.

Kw is equilibrium constant for water.

Substitute 2.06×106 for Ka2 and 1014 for Kw in equation (14).

  Kb1=10142.06×106=4.85×109

Thus, value of Kb1 is 4.85×109.

Consider concentration of OH to be x, thus value of [HM] is also x and that of [M2] is 0.100x. The expression for equilibrium constant Kb1 for malonic acid is given as follows:

  Kb1=[HM][OH][M2]        (15)

Rearrange equation (15) for [OH].

  [OH]=Kb1[M2][HM]        (16)

Here,

[HM] is concentration of HM.

[OH] is concentration of OH.

[M2] is concentration of M2.

Kb1 is first base equilibrium constant.

Substitute 4.85×109 for Kb1, x for [OH], x for [HM] and 0.100x for [M2] in equation (16).

  x=(4.85×109)(0.100x)xx2=(4.85×109)(0.100x)x2+(4.85×109)x(4.85×1010)=0x=2.2×105

Therefore, value of [OH] is 2.2×105 M and [HM] is 2.2×105 M.

The expression used for calculation of pH is as follows:

  pH=log[H+]        (17)

The expression used to calculate Kw is as follows:

  Kw=[H+][OH]        (18)

Rearrange equation (18) for [H+].

  [H+]=Kw[OH]        (19)

Substitute value of [H+] is equation (17).

  pH=logKw[OH]        (20).

Substitute 1014 for Kw and 2.2×105 for [OH] in equation (20).

  pH=log10142.2×105=9.34

The relation between Ka1 and Kb2 is as follows:

  (Ka1)(Kb2)=Kw        (21)

Rearrange equation (21) for Kb2.

  Kb2=KwKa1        (22)

Here,

Ka1 is first acid equilibrium constant.

Kb2 is second equilibrium constant.

Kw is equilibrium constant for water.

Substitute 1.42×103 for Ka1 and 1014 for Kw in equation (22).

  Kb2=10141.42×103=7.04×1012

Thus, value of Kb2 is 7.04×1012.

The expression for equilibrium constant Kb2 is given as follows:

  Kb2=[OH][H2M][HM]        (23)

Rearrange equation (23) for [H2M].

  [H2M]=Kb2[HM][OH]        (24)

Here,

[HM] is concentration of HM.

[OH] is concentration of OH.

[H2M] is concentration of H2M.

Kb2 is second base equilibrium constant.

Substitute 7.04×1012 for Kb2, 2.2×105 for [OH] and 2.2×105 for [HM] in equation (24).

  [H2M]=(7.04×1012)(2.2×105)(2.2×105)=7.04×1012

Therefore, value of [H2M] is 7.04×1012 M.

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