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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

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)

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

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

(b)

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)

(c)

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)

(d)

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.

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Chapter 17 Solutions

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Sect-17.4 P-17.11CYUSect-17.5 P-17.12CYUSect-17.5 P-17.13CYUSect-17.5 P-17.14CYUSect-17.6 P-17.15CYUSect-17.6 P-17.16CYUSect-17.6 P-1.1ACPSect-17.6 P-1.2ACPSect-17.6 P-1.3ACPSect-17.6 P-1.4ACPSect-17.6 P-1.5ACPSect-17.6 P-2.1ACPSect-17.6 P-2.2ACPCh-17 P-1PSCh-17 P-2PSCh-17 P-3PSCh-17 P-4PSCh-17 P-5PSCh-17 P-6PSCh-17 P-7PSCh-17 P-8PSCh-17 P-9PSCh-17 P-10PSCh-17 P-11PSCh-17 P-12PSCh-17 P-13PSCh-17 P-14PSCh-17 P-15PSCh-17 P-16PSCh-17 P-17PSCh-17 P-18PSCh-17 P-19PSCh-17 P-20PSCh-17 P-21PSCh-17 P-22PSCh-17 P-23PSCh-17 P-24PSCh-17 P-25PSCh-17 P-26PSCh-17 P-27PSCh-17 P-28PSCh-17 P-29PSCh-17 P-30PSCh-17 P-31PSCh-17 P-32PSCh-17 P-33PSCh-17 P-35PSCh-17 P-36PSCh-17 P-37PSCh-17 P-38PSCh-17 P-39PSCh-17 P-40PSCh-17 P-41PSCh-17 P-42PSCh-17 P-43PSCh-17 P-44PSCh-17 P-45PSCh-17 P-46PSCh-17 P-47PSCh-17 P-48PSCh-17 P-49PSCh-17 P-50PSCh-17 P-51PSCh-17 P-52PSCh-17 P-53PSCh-17 P-54PSCh-17 P-55PSCh-17 P-56PSCh-17 P-57PSCh-17 P-58PSCh-17 P-59PSCh-17 P-60PSCh-17 P-61PSCh-17 P-62PSCh-17 P-63PSCh-17 P-64PSCh-17 P-65PSCh-17 P-66PSCh-17 P-67PSCh-17 P-68PSCh-17 P-69PSCh-17 P-70PSCh-17 P-71PSCh-17 P-72PSCh-17 P-73PSCh-17 P-74PSCh-17 P-75PSCh-17 P-76PSCh-17 P-77GQCh-17 P-78GQCh-17 P-79GQCh-17 P-80GQCh-17 P-81GQCh-17 P-82GQCh-17 P-83GQCh-17 P-84GQCh-17 P-85GQCh-17 P-86GQCh-17 P-87GQCh-17 P-88GQCh-17 P-89GQCh-17 P-90GQCh-17 P-91GQCh-17 P-92GQCh-17 P-93GQCh-17 P-94GQCh-17 P-95GQCh-17 P-96GQCh-17 P-97GQCh-17 P-98GQCh-17 P-99GQCh-17 P-100GQCh-17 P-101ILCh-17 P-102ILCh-17 P-103ILCh-17 P-104ILCh-17 P-105ILCh-17 P-106ILCh-17 P-107ILCh-17 P-108ILCh-17 P-109ILCh-17 P-110ILCh-17 P-111ILCh-17 P-112ILCh-17 P-113SCQCh-17 P-114SCQCh-17 P-115SCQCh-17 P-116SCQCh-17 P-117SCQCh-17 P-118SCQCh-17 P-119SCQCh-17 P-120SCQ

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