EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
Question
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Chapter 11, Problem 11.10P

(a)

Interpretation Introduction

Interpretation:

Value of pH and concentrations of B, BH+ and BH22+ in 0.100 M B 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

The compound B is a dibasic compound thus, it requires two protons. Acceptance of one proton produces BH+ and other produces BH22+. The equilibrium reaction for addition of one proton is as follows:

  B+H2OKb1BH++OH

The ICE table for concentration of molecules in reaction is as follows:

  B+H2OKb1BH++OH-Initial,M0.10000Change,MxxxFinal,M0.100xxx

Concentration of B is 0.100x and of BH+ and OH is be x.

The expression for equilibrium constant Kb1 for dibasic acid is given as follows:

  Kb1=[BH+][OH][B]        (1)

Rearrange equation (1) for [BH+].

  [BH+]=Kb1[B][OH]        (2)

Here,

[BH+] is concentration of BH+.

[B] is concentration of B.

[OH] is concentration of OH.

Kb1 is basic equilibrium constant for dibasic compound.

Substitute 1.00×105 for Kb1, x for [BH+], x for [OH] and 0.100x for [B] in equation (2).

  x=(1.00×105)(0.100x)xx2=(1.00×105)(0.100x)x2+(1.00×105)x(1.00×106)=0x=9.95×104

Therefore, value of [BH+] is 9.95×104, [OH] is 9.95×104 and [B] is 0.09 M.

Second equilibrium reaction for dibasic compound B that is designated as Kb2 is as follows:

  BH+Kb2BH22++OH

The expression for equilibrium constant Kb2 is given as follows:

  Kb2=[BH22+][OH][BH+]        (5)

Rearrange equation (5) for [BH22+].

  [BH22+]=Kb2[BH+][OH]        (6)

Here,

[BH+] is concentration of BH+.

[BH22+] is concentration of BH22+.

[OH] is concentration of OH.

Kb2 is basic equilibrium constant for dibasic compound.

Substitute 1.00×109 for Kb2, 9.95×104 for [BH+], 9.95×104 for [OH] and in equation (6).

  [BH22+]=(1.00×109)(9.95×104)(9.95×104)=1.00×109

Therefore, value of [BH22+] is 1.00×109 M.

The expression used for calculation of pH is as follows:

  pH=log[H+]        (7)

The expression for Kw is as follows:

  Kw=[H+][OH]        (8)

Rearrange equation (8) for [H+].

  [H+]=Kw[OH]        (9)

Substitute value of [H+] is equation (7).

  pH=logKw[OH]        (10).

Substitute 1014 for Kw and 9.95×104 for [OH] in equation (10).

  pH=log10149.95×104=11

Thus, pH is 11, concentration of BH+ is 9.95×104, of B is 0.09 M and of BH22+ is 1.00×109 M

(b)

Interpretation Introduction

Interpretation:

Value of pH and concentrations of B, BH+ and BH22+ in 0.100 M BH+Br solution has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
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Explanation of Solution

BH+Br is an intermediate form from that removal of one proton produces B and addition of one proton produces BH22+. The equilibrium reaction is as follows:

  BH+B+H+BH++H+BH22+

The formula to calculate value of pH for given reaction is as follows:

  pH=12(pKb1+pKb2)        (11)

The expression for pKa is as follows:

  pKb=logKb        (12)

Substitute equation (12) in equation (11).

  pH=12(logKb1+(logKb2))        (13)

Substitute 1.00×105 for Kb1 and 1.00×109 M for Kb2 in equation (13).

  pH=12(log(1.00×105)+(log(1.00×109 M)))=12(5+9)=7

The formula to calculate pH is as follows:

  pH=log[H+]        (14)

Rearrange equation (14) for [H+].

  [H+]=10pH        (15)

Substitute 7 for pH in equation (15).

  [H+]=107 M

The reaction for BH22+ is given as follows:

  BH22+BH++H+

The expression for equilibrium constant Kb1 is as follows:

  Kb1=[BH+][H+][BH22+]        (16)

Rearrange equation (16) for [BH22+].

  [BH22+]=[BH+][H+]Kb1        (17)

Substitute 1.00×105 for Kb1, 107 for [H+] and 0.100 for [BH+] in equation (17).

  [BH22+]=(0.100)(107)1.00×105=1.0×103 M

The reaction for BH+ is given as follows:

  BH+B+H+

The expression for equilibrium constant Kb2 is as follows:

  Kb2=[B][H+][BH+]        (18)

Rearrange equation (18) for [B].

  [B]=Kb2[BH+][H+]        (19)

Substitute 1.00×109 for Kb2, 107 for [H+] and 0.100 for [BH+] in equation (19).

  [B]=(1.00×109)(0.100)107=1.0×103

Thus, pH is 7, concentration of BH22+ is 1.0×103 M, of BH+ is 0.100 M and B is 1.0×103 M.

(c)

Interpretation Introduction

Interpretation:

Value of pH and concentrations of B, BH+ and BH22+ in 0.100 M BH22+(Br)2 solution has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

The compound B is a dibasic compound thus, it requires two protons. Acceptance of one proton produces BH+ and other produces BH22+. The equilibrium reaction for addition of one proton is as follows:

  BH22+K1BH++H+

The ICE table for concentration of molecules in reaction is as follows:

BH22+K1BH++H+Initial0.10000ChangexxxFinal0.100xxx

Concentration of BH22+ is 0.100x and of BH+ and H+ is be x.

The expression for equilibrium constant Kb1 for dibasic acid is given as follows:

  Kb1=[BH+][H+][BH22+]        (20)

Here,

[BH+] is concentration of BH+.

[BH22+] is concentration of BH22+.

[H+] is concentration of H+.

Kb1 is basic equilibrium constant for dibasic compound.

Substitute 1.00×105 for Kb1, x for [BH+], x for [H+] and 0.100x for [BH22+] in equation (20).

  1.00×105=(x)(x)0.100xx2+(1.00×105)x(1.00×106)=0

Sove above equation for x.

  x=9.95×104

Therefore, value of [BH+] is 9.95×104, [H+] is 9.95×104 and [BH22+] is 0.09 M.

Second equilibrium reaction for hydrolysis of dibasic compound B that is designated as Kb2 is as follows:

  BH+Kb2B+H+

The expression for equilibrium constant Kb2 is given as follows:

  Kb2=[B][H+][BH+]        (21)

Rearrange equation (21) for [B].

  [B]=Kb2[BH+][H+]        (22)

Substitute 1.00×109 for Kb2, 9.95×104 for [BH+], 9.95×104 for [H+] in equation (22).

  [B]=(1.00×109)(9.95×104)(9.95×104)=1.00×109

Therefore, value of [B] is 1.00×109 M.

The expression used for calculation of pH is as follows:

  pH=log[H+]        (23)

Substitute 9.95×104 for [H+] in equation (23).

  pH=log(9.95×104)=log9.95+4=0.998+4=3

Thus, concentration of BH+ is 9.95×104, of BH22+ is 0.09 M and of B is 1.00×109 M and pH is 3.

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