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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

The base ethylamine (CH 3 CH 2 NH 2 ) has a Kb of. A closely related base, ethanolamine (HOCH 2 CH 2 NH 2 ) , has a Kb of 3 .2×10 5 .

(a) Which of the two bases is stronger?

(b) Calculate the pH of a 0.10M solution of the strong base?

(a)

Interpretation Introduction

Interpretation:

The stronger has to be identified from the given bases.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka: it is a acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb: It is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

The stronger acids have high ka and weak acids have low ka values.

As well as stronger bases have high Kb values and weak bases have low Kb vales.

Equilibrium shifted to the more ka and kb values side of the reactants or products.

Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation pH and pOH

 pH + pOH =14

Explanation

If the Kb value is high, basicity will be high, the given Kb values are given below,

Kb for Ethylamine  = 4

(b)

Interpretation Introduction

Interpretation:

pH of a 0.10M solution of the strong base has to be calculated.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka: it is a acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb: It is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

The stronger acids have high ka and weak acids have low ka values.

As well as stronger bases have high Kb values and weak bases have low Kb vales.

Equilibrium shifted to the more ka and kb values side of the reactants or products.

Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation pH and pOH

 pH + pOH =14

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Chapter 16 Solutions

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Sect-16.10 P-1.2ACPSect-16.10 P-1.3ACPSect-16.10 P-2.1ACPSect-16.10 P-2.2ACPSect-16.10 P-2.3ACPSect-16.10 P-2.4ACPSect-16.10 P-2.5ACPCh-16 P-1PSCh-16 P-2PSCh-16 P-3PSCh-16 P-4PSCh-16 P-5PSCh-16 P-6PSCh-16 P-7PSCh-16 P-8PSCh-16 P-9PSCh-16 P-10PSCh-16 P-11PSCh-16 P-12PSCh-16 P-13PSCh-16 P-14PSCh-16 P-15PSCh-16 P-16PSCh-16 P-17PSCh-16 P-18PSCh-16 P-19PSCh-16 P-20PSCh-16 P-21PSCh-16 P-22PSCh-16 P-23PSCh-16 P-24PSCh-16 P-25PSCh-16 P-26PSCh-16 P-27PSCh-16 P-28PSCh-16 P-29PSCh-16 P-30PSCh-16 P-31PSCh-16 P-32PSCh-16 P-33PSCh-16 P-34PSCh-16 P-35PSCh-16 P-36PSCh-16 P-37PSCh-16 P-38PSCh-16 P-39PSCh-16 P-40PSCh-16 P-41PSCh-16 P-42PSCh-16 P-43PSCh-16 P-44PSCh-16 P-45PSCh-16 P-46PSCh-16 P-47PSCh-16 P-48PSCh-16 P-49PSCh-16 P-50PSCh-16 P-51PSCh-16 P-52PSCh-16 P-53PSCh-16 P-54PSCh-16 P-55PSCh-16 P-56PSCh-16 P-57PSCh-16 P-58PSCh-16 P-59PSCh-16 P-60PSCh-16 P-61PSCh-16 P-62PSCh-16 P-63PSCh-16 P-64PSCh-16 P-65PSCh-16 P-66PSCh-16 P-67PSCh-16 P-68PSCh-16 P-69PSCh-16 P-70PSCh-16 P-71PSCh-16 P-72PSCh-16 P-73PSCh-16 P-74PSCh-16 P-75PSCh-16 P-76PSCh-16 P-77PSCh-16 P-78PSCh-16 P-79PSCh-16 P-80PSCh-16 P-81PSCh-16 P-82PSCh-16 P-83PSCh-16 P-84PSCh-16 P-85GQCh-16 P-86GQCh-16 P-87GQCh-16 P-88GQCh-16 P-89GQCh-16 P-90GQCh-16 P-91GQCh-16 P-92GQCh-16 P-93GQCh-16 P-94GQCh-16 P-95GQCh-16 P-96GQCh-16 P-97GQCh-16 P-98GQCh-16 P-99GQCh-16 P-100GQCh-16 P-101GQCh-16 P-102GQCh-16 P-103GQCh-16 P-104GQCh-16 P-105GQCh-16 P-106GQCh-16 P-107GQCh-16 P-108GQCh-16 P-109GQCh-16 P-110GQCh-16 P-111ILCh-16 P-112ILCh-16 P-113ILCh-16 P-114ILCh-16 P-115ILCh-16 P-116ILCh-16 P-117ILCh-16 P-118ILCh-16 P-119SCQCh-16 P-120SCQCh-16 P-121SCQCh-16 P-122SCQCh-16 P-123SCQCh-16 P-124SCQCh-16 P-125SCQCh-16 P-126SCQCh-16 P-127SCQCh-16 P-128SCQCh-16 P-129SCQ

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