Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Textbook Question
Chapter 17, Problem 82GQ

For each of the following cases, decide whether the pH is less than 7, equal to 7, or greater than 7.

(a) Equal volumes of 0.10 M acetic acid, CH3CO2H, and 0.10 M KOH are mixed.

(b) 25 mL of 0.015 M NH3 is mixed with 12 mL of 0.015 M HCl.

(c) 150 mL of 0.20 M HNO3, is mixed with 75 mL of 0.40 M NaOH.

(d) 25 mL of 0.45 M H2SO4 is mixed with 25 mL of 0.90 M NaOH.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH has to be calculated when equal volume of CH3COOH(0.10molL1) is titrated with NaOH(0.10molL1).

Concept introduction:

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of CH3COOH with NaOH. The equilibrium can be represented as,

CH3COOH(aq)+NaOH(aq)H2O(l)+CH3COONa(aq)

Explanation of Solution

The pH calculation at equivalence point is given below.

Given:

Refer to table 16.2 in the textbook for the value of Ka.

The value of Ka for acetic acid is 1.8×105.

The value of Kw for water is 1.0×1014.

The pKa value is calculated as follows;

pKa=log(Ka)

Substitute, 1.8×105 for Ka.

pKa=log(1.8×105)=4.74

Therefore, pKa values is 4.74.

The initial concentration of CH3COOH is 0.100molL1.

The initial concentration of NaOH is 0.100molL1.

The volume of NaOH added is equal to the volume of CH3COOH.

Therefore, let the volume of CH3COOH is 100mL.

Thus volume of NaOH added  will be 100mL.

totalvolume=volumeofCH3COOH(L) + volume of NaOH(L)

totalvolume = 0.1(L)+0.1(L)=0.2L

Therefore, total volume after reaction is 0.2L.

The calculation of moles is done by using the expression,

Numberof moles=concentration(molL1)volume(L)

The ICE table (1) for the reaction between NaOH and CH3COOH is given below,

EquationCH3COOH(aq)+NaOH(aq)H2O(l)+CH3COONa(aq)Initial(mol)0.01000.01000Change(mol)0.01000.01000.0100Afterreaction(mol)000.0100

From ICE table (1),

Calculate the concentration of acetate ion after reaction.

Substitute, 0.0100mol for Numberof moles and 0.20L for volume in equation (4).

concentration = 0.01000.20(molL1)=0.05molL1

The concentration of acetate ion after reaction is 0.05molL1.

The acetate ion produced will undergo hydrolysis in presence of water and the reaction equilibrium is written as,

CH3COO(aq)+H2O(l)OH(aq)+CH3COOH(aq)

The hydrolysis equilibrium is represented in ICE table (2).

EquationCH3COO(aq)+H2O(l)OH(aq)+CH3COOH(aq)Initial(molL1)0.050000Change(molL1)x+x+xAfterreaction(molL1)0.0500x+x+x

From ICE table (2),

There is an approximation, that the value of x is very small as comparison to 0.0500 thus it can be neglected with respect to it.

Therefore, Concentration of acetate ion left after reaction is 0.0500molL1.

Calculate the concentration of OH by using the equation (3).

Kw=(Ka)(Kb)

Rearrange it for Kb

Kb=KwKa (1)

The expression of Kb for acetate ion from the ICE table (3) will be written as,

Kb=[CH3COOH](eq)[OH](eq)[CH3COO](eq) (2)

Equate equation (5) and (6),

KwKa=[CH3COOH](eq)[OH](eq)[CH3COO](eq)

Substitute, 1.8×105 for Ka, 1.0×1014 for Kw, x for [OH](eq), x for [CH3COOH](eq), 0.0500 for [CH3COO](eq).

1.0×10141.8×105=(x)(x)(0.0500)

(0.55×109)=x2(0.0500)

Rearrange for x2,

x2=(0.555×109)(0.0500)=0.027×109

Rearrange for x,

x=0.27×1010=0.519×10-5

The value of OH concentration is 0.519×10-5molL1.

Calculate the value of pOH by using the expression, pOH = log[OH].

Substitute, 0.519×105 for [OH]

pOH = log(0.519×105)=(5.28)=5.28

Therefore, the value of pOH is 5.28.

Thus, the value of pH is calculated by using expression,

pH + pOH =14

Rearrange for pH,

pH  =14 pOH

Substitute, 5.28 for pOH

pH  =145.288.72

Therefore, the value of pH at equivalence point is 8.72.

The value of pH is 8.72. This value is above the neutral pH value 7, thus at this point the solution will be basic in nature.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH when 25mL,NH3(0.015molL1) is titrated with 12 mL,HCl(0.015molL1) has to be calculated.

Concept introduction:

For weak base-strong acid titration the pH value can be calculated at various points before and after equivalence point.

The equilibrium established during the titration of NH3 with HCl. The equilibrium can be represented as,

NH3(aq)+ H3O+(aq)H2O(l)+NH4+(aq)

Explanation of Solution

The pH calculation just before the equivalence point is done by using Henderson-Hasselbalch equation.

As the addition of HCl is done there will be formation of buffer solution NH3/NH4+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugateacid][base] (3)

The pH value will be calculated by using expression, pH + pOH = 14.

Given:

Refer to table 16.2 in the textbook for the value of Kb.

The value of Kb for ammonia is 1.8×105.

The pKb value is calculated as follows;

pKb=log(Kb)

Substitute, 1.8×105 for Kb.

pKb=log(1.8×105)=4.74

Therefore, pKb value is 4.74.

The initial concentration of NH3 is 0.015molL1.

The initial concentration of HCl is 0.015molL1.

The volume of NH3 is 25 mL.

The volume of HCl is 12 mL.

The total volume after the reaction is calculated as,

totalvolume=volumeofNH3(L) + volume of HCl(L)

totalvolume = 0.025(L)+0.012(L)=0.037L

Therefore, total volume after reaction is 0.037 L.

The calculation of moles is done by using the expression,

Numberof moles=concentration(molL1)volume(L)

The ICE table (2) for the reaction between NH3 and HCl is given below,

EquationNH3(aq)+H3O+(aq)H2O(l)+NH4+(aq)Initial(mol)0.0003750.000180Change(mol)0.000180.00018+0.00018Afterreaction(mol)0.00019500.00018

From ICE table (2),

Number of moles of ammonia left after reaction are 0.000195.

Number of moles of H3O+ left after reaction are 0mol.

Number of moles of ammonium ion (NH4+)  produced after the reaction are 0.00018.

Concentration calculations is done by using the expression,

concentration = Numberof molestotal volume(molL1)

Therefore, concentration of ammonium ion (NH4+) after reaction is 0.0048molL1.

Concentration of ammonia after reaction is 0.0052molL1.

Calculate the pOH value using equation (3).

pOH=pKb+log[conjugateacid][base]

Substitute, 0.0048 for [conjugateacid], 0.0052 for [base] and 4.74 for pKb.

pOH=4.74+log(0.0048)(0.0052)=4.74+log(0.923)=4.740.0347=4.70

Therefore, the value of pOH is 4.70.

Thus pH value is calculated by using expression, pH + pOH = 14

Substitute, 4.70 for pOH.

pH +4.70 = 14

Rearrange for pH,

pH  = 144.709.30

Therefore, the value of pOH when 25 mL,NH3(0.015molL1) is titrated with 12 mL,HCl(0.015molL1) is 9.30.

The value of pH is 9.30. This value is above the neutral pH value 7, thus at this point the solution will be basic in nature.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH when 150mL,HNO3(0.20molL1) is titrated with 75 mL,NaOH(0.40molL1) has to be calculated.

Concept introduction:

Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, HNO3 is titrated against NaOH. The reaction between HNO3 and NaOH can be written as follows.

HNO3(aq)+NaOH(aq)NaNO3(aq)+H2O(l)

Explanation of Solution

The pH calculation for the strong acid – strong base titration is given below.

Given:

The initial concentration of HNO3 is 0.20molL1.

The volume of HNO3 is 150 mL or 0.150 L.

The initial concentration of NaOH is 0.40molL1.

The volume of NaOH is 75 mL or 0.075 L.

The table gives the moles of reactant and product for the reaction between HNO3 and NaOH.

EquationHNO3(aq)+NaOH(aq)H2O(l)+NaNO3(aq)Initial(mol)0.030.030Change(mol)0.030.03+0.03Afterreaction(mol)000.03

After the reaction there is no moles of acid and base are present in the reaction mixture. That means all acid and base has been neutralized and titration has reached the equivalence point.

For the strong acid and strong base titration the pH value is 7.0 at the equivalence point as it is a neutral solution having water and NaNO3.

 The value of pH is 7.0. Thus the solution will be neutral in nature.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pH when 25mL,H2SO4(0.45molL1) is titrated with 25 mL,NaOH(0.90molL1) has to be calculated.

Concept introduction:

Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, H2SO4 is titrated against NaOH. The reaction between H2SO4 and NaOH can be written as follows.

H2SO4(aq)+2NaOH(aq)Na2SO4(aq)+2H2O(l)

The sulphuric acid is a dibasic acid as it has two hydrogen atoms which can be donated.

Therefore two equivalents of NaOH are required to balance the reaction.

Explanation of Solution

The pH calculation for the strong acid – strong base titration is given below.

Given:

The initial concentration of H2SO4 is 0.45molL1.

The volume of H2SO4 is 25 mL or 0.025 L.

The initial concentration of NaOH is 0.90molL1.

The volume of NaOH is 25 mL or 0.025 L.

The table gives the moles of reactant and product for the reaction between H2SO4 and NaOH.

EquationH2SO4(aq)+2NaOH(aq)2H2O(l)+Na2SO4(aq)Initial(mol)22.522.50Change(mol)22.522.5+22.5Afterreaction(mol)0022.5

After the reaction there is no moles of acid and base are present in the reaction mixture. That means all acid and base has been neutralized and titration has reached the equivalence point.

For the strong acid and strong base titration the pH value is 7.0 at the equivalence point as it is a neutral solution having water and Na2SO4.

The value of pH is 7.0. Thus the solution will be neutral in nature.

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