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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

Chloroacetic acid (ClCH2CO2H) has Ka = 1.41 × 10−3. What is the value of Kb for the chloroacetate ion (ClCH2CO2)?

Interpretation Introduction

Interpretation:

The value of ionization constant, Kb for conjugate base Chloroacetate ion has to be calculated by using the acid dissociation constant Ka of Chloroacetic acid.

Concept introduction:

An acid undergoes dissociation in an aqueous medium. The acid dissocition gives  ions H3O+(aq.) and A(aq.). The conjugate base A(aq.) may interact with H2O and forms HA(aq.) and it can be represented as following equilibrium.

HA(aq.)+ H2O(l)H3O+(aq.)+ A1(aq.) (1)

The dissociation constant for the acid is Ka,

  Ka=[H3O+][A][HA]

Interaction of anion A(aq.) is represented as,

A1(aq.)+ H2O(l)HA(aq.)+ OH1(aq.) (2)

The dissociation constant for the conjugate anion is Kb,

  Kb=[HA][OH][A]

From equation (1) and (2) the net equilibrium will be

2H2O(l)H3O+(aq.)+ OH1(aq.) (3)

The equation (3) is known as the auto-ionization of water molecule. The dissociation constant for water is written as follows,

  Kw=[H3O+][OH]

Thus from equation (1) and (2), there is an important relation established between ionization constant Ka and ionization constant Kb of conjugate base, i.e.

  Kw=Ka×Kb                                                                                                 (4)

The ionisation constant of water has a constant value i.e. Kw=1×10-14. On taking negative logarithm of equation (4), a new relation established in terms of pKa,pKb and pKw as follows,

-log(Kw)=-log(Ka×Kb)

 pKw= pKa+pKb (5)

Equation (5) is used to determine

Explanation

In an aqueous solution,Chloroacetic acid undergoes ionisation and its conjugate base Chloroacetate ion interact with water and again forms Chloroacetic acid. There are two equilibrium established which can be written as,

ClCH2COOH(aq.)+ H2O(l)H3O+(aq.)+ ClCH2COO1(aq.)

ClCH2COO1(aq.)+ H2O(l)ClCH2COOH(aq.)+OH1(aq.)

The net equilibrium will be

2H2O(l)H3O+(aq.)+ OH1(aq.)

Therefore using the equation (4), Kb for Chloroacetate ion is calculated below

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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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