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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

Ethylene diamine, H2NCH2CH2NH2, can interact with water in two steps, forming OH in each step (Appendix I). If you have a 0.15 M aqueous solution of the amine, calculate the concentration of [H3NCH2CH2NH3]2+ and OH.

Interpretation Introduction

Interpretation:

Concentration of [H3NCH2CH2NH3]2+ and OH- has to be calculated form 0.15 M aqueous solution of the amine.

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]

Explanation

The ionization reactions of Ethylenediamine are as follows.

First ionization:

H2NCH2CH2NH2(aq) + H2O(l)NH2CH2CH2NH3+(aq) + OH-(aq)Equilibrium expression:Kb1[NH2CH2CH2NH3+][OH-]H2NCH2CH2NH2]Kb1= 8.5×10-5Second ionization:NH2CH2CH2NH3+(aq) + H2O(l)[NH3CH2CH2NH3]2+(aq) + OH-(aq)Equilibrium expression:Kb2[NH3CH2CH2NH32+][OH-]H2NCH2CH2NH3+]Kb2= 3.57×10-3

Compare the Kb1 and Kb2,Kb1  is greater than Kb2 . Therefore most of the OH- is produced from the Kb1 

Let’s calculate the concentration of OH- from Kb1 .

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

        NH2CH2CH2NH2(aq) + H2O (l) NH2CH2CH2NH3+(aq) + OH(aq)I                      0.15                    --                       --                                 --C                       -x                      --                     +x                                 +xE                 (0.15-x)                  --                       x                                   x

Kb1[NH2CH2CH2NH3+][OH-]H2NCH2CH2NH2]Kb1= 8.5×10-5

8.5×10-5 = (x)(x)0.15- x8.5×10-5 = (x)20.15- x(0.15-x) approximately equals to 0.158.5 ×10-5 = (x)20.15          x2=  (0.15)(8.5×10-5)          x  =  (0

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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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