Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 18, Problem 18.184P

(a)

Interpretation Introduction

Interpretation:

pH of the saturated solution of quinine in water has to be calculated.

Concept Introduction:

The self-dissociation constant for water can be written as follows,

Kw = Ka × Kb 

Where,

  Kw = water dissociation constantKa  = acid dissociation constant Kb = base dissociation constant

  Kw=1014

Also, Kw=[H3O+][OH]

pH can be calculated from the following equation if [H3O+] is known.

pH=log10[H3O+]

(a)

Expert Solution
Check Mark

Explanation of Solution

Given,

    pKb=5.1 for more basic nitrogen. So corresponding Kb value is Kb=105.1=7.94328x10-6.

The base dissociation reaction for quinine can be written as follows,

C20H24N2O2(aq) + H2O(l)OH-(aq) + HC20H24N2O2+(aq)

The reaction table for this reaction is shown below.

Initial:1.6×103M00Change:x+x+xEquilibrium:1.6×103xxx

The base dissociation constant Kb can be written as follows,

Kb=[HC20H24N2O2+][OH-][H2C20H24N2O2]

Where,

                [HC20H24N2O2+] = 1.6×10-3 - x[OH-] = x[C20H24N2O2] = x

On substituting these in the above equation x is obtained as follows,

Kb=[HC20H24N2O2+][OH-][H2C20H24N2O2]=[x][x][1.6×103x]=7.94328×106

x is obtained by solving the following quadratic equation.

  x2=(7.94328×106)(1.6×103x)=1.27092×1087.94328×106xx2=7.94328×106x1.270925×108=0x=b±b24ac2awhere,a=1b=7.94328×106c=1.270925×108x=7.94328×106±(7.94328×106)24(1)(1.270925×108)2(1)=1.08834×104M

1.08834×104M is the concentration of [OH-]. Concentration of H3O+ can be calculated from the following relation.

  Kw=[H3O+][OH][H3O+]=Kw[OH]=10×141.08834×104=9.18830513×1011M

  pH can be calculated using the following equation.

  pH=log10[H3O+]=log10(9.18830513×1011)=10.03676=10.0

(b)

Interpretation Introduction

Interpretation:

Less contribution of aromatic N to the pH has to be shown.

Concept Introduction:

The self-dissociation constant for water can be written as follows,

Kw = Ka × Kb 

Where,

  Kw = water dissociation constantKa  = acid dissociation constant Kb = base dissociation constant

  Kw=1014

pH can be calculated from the following equation if [H3O+] is known.

pH=log10[H3O+]

Also, Kw=[H3O+][OH]

(b)

Expert Solution
Check Mark

Explanation of Solution

Given,

    pKb=9.7 for aromatic nitrogen. So corresponding Kb value is Kb=109.7=1.995262x1010.

The base dissociation reaction for quinine can be written as follows,

C20H24N2O2(aq) + H2O(l)OH-(aq) + HC20H24N2O2+(aq)

The reaction table for this reaction is shown below.

Initial:1.6×103M00Change:x+x+xEquilibrium:1.6×103xxx

The base dissociation constant Kb can be written as follows,

Kb=[HC20H24N2O2+][OH-][C20H24N2O2]

Where,

    [HC20H24N2O2+] = 1.6×10-3 - x[OH-] = x[C20H24N2O2] = x

On substituting these in the above equation x is obtained as follows,

Kb=[HC20H24N2O2+][OH-][C20H24N2O2]=[x][x][1.6×103x]=1.995262×1010

Assuming x is small compared to 1.6×103 the x in the denominator can be neglected.

[x][x][1.6×103]=1.995262×1010x2=(1.995262×1010)(1.6×103)x=5.65015×107MOH

Concentration of H3O+ can be calculated from the following relation.

Kw=[H3O+][OH][H3O+]=Kw[OH]=10×145.65015×107=1.769×10×8M

pH can be calculated using the following equation.

  pH=log10[H3O+]=log10(1.769×10×8)=7.75

(c)

Interpretation Introduction

Interpretation:

pH of quinine hydrochloride at given concentration has to be calculated.

Concept Introduction:

The self-dissociation constant for water can be written as follows,

Kw = Ka × Kb 

Where,

  Kw = water dissociation constantKa  = acid dissociation constant Kb = base dissociation constant

  Kw=1014

pH can be calculated from the following equation if [H3O+] is known.

pH=log10[H3O+]

Also, Kw=[H3O+][OH]

(c)

Expert Solution
Check Mark

Explanation of Solution

The acid dissociation reaction for qunine can be written as follows,

HC20H24N2O2+(aq) +H2O(l) H3O+(aq) + C20H24N2O2(aq)

The reaction table for this reaction is shown below.

Initial:0.33M00Change:x+x+xEquilibrium:0.33xxx

Ka is obtained from following relation.

Kw = Ka × Kb Ka=KwKb=10147.94328×106=1.25893×109

From the reaction Ka is obtained as follows,

Ka=1.25893×109=[H3O+][C20H24N2O2][HC20H24N2O2+]=(x)(x)(0.33x)

Assume x is very much smaller than 0.33, so x can be neglected from the denominator. And x is obtained as follows,

  (x)(x)(0.33)=1.25893×109x2=(1.25893×109)(0.33)x =[H3O+] = 2.038252× 10-5M

pH can be calculated using the following equation.

pH=log10[H3O+]=log10(2.038252× 10-5)=4.69074=4.7

(d)

Interpretation Introduction

Interpretation:

pH of 1.5% of quinine hydrochloride has to be calculated.

Concept Introduction:

Molarity of a solution can be calculated using the following equation.

Molarity=numberofmolesofsoluteVolumeofsolutioninlitre

And number of moles of solute can be calculated using the following equation.

Number of moles = Given mass of soluteMolecular mass of the solute

Therefore molarity can be calculated as follows,

Molarity=numberofmolesofsoluteVolumeofsolutioninlitre=W2×1000M2×V(ml)

Where,

  W2 = given mass of soluteM2= molecular mass of soluteV(ml) = volume of solution in ml

(d)

Expert Solution
Check Mark

Explanation of Solution

Use QHCl to represent quinine hydrochloride.

Given that 1.5% of QHCl is present in the solution. This means 1.5 g of QHCl in 100 ml of solution. O molarity can be calculated as follows,

Molarity of a solution can be calculated using the following equation.

Molarity=numberofmolesofQHClVolumeofsolutioninlitre

And number of moles of solute can be calculated using the following equation.

Molarity=numberofmolesofQHClVolumeofsolutioninlitre=W2×1000M2×V(ml)=1.5×1000360.87×100=0.041566M

The acid dissociation reaction for quinine can be written as follows,

HC20H24N2O2+(aq) +H2O(l) H3O+(aq) + C20H24N2O2(aq)

The reaction table for this reaction is shown below.

Initial:0.041566M00Change:x+x+xEquilibrium:0.041566xxx

Ka is obtained from following relation.

Kw = Ka × Kb Ka=KwKb=10147.94328×106=1.25893×109

From the reaction Ka is obtained as follows,

Ka=1.25893×109=[H3O+][C20H24N2O2][HC20H24N2O2+]=(x)(x)(0.041566x)

Assume x is very much smaller than 0.041566, so x can be neglected from the denominator. And x is obtained as follows,

(x)(x)(0.041566)=1.25893×109x2=(1.25893×109)(0.041566)x =[H3O+] = 7.233857× 10-6M

pH can be calculated using the following equation.

pH=log10[H3O+]=log10(7.233857× 10-6)=5.1406=5.1

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Chapter 18 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

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