BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

Solutions

Chapter
Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

Calculate the hydronium ion concentration and pH of the solution that results when 22.0mL of 0.15M acetic acid CH 3 CO 2 H is mixed with 22.0 mL of 0.15M NaOH .

Interpretation Introduction

Interpretation:

The hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15M acetic acid, CH3CO2H is mixed with 22.0 mL of 0.15MNaOH 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 an 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]

Ionic product constant for water  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]Reltion between pH and pOH pH + pOH =14

Explanation

Given volume of 0.1M CH3CO2H = 22.0 mL

And it is mixed with 22.0 mL of 0.1M OH-. Therefore, the base converts into acetic acid to its conjugate base, the acetate ion.

In this reaction the entire amount of base added converts CH3CO2H completely into conjugate base CH3COO-.

Let’s calculate the amount of consumed acetic acid:

(22.0mL×1L1000mL) × (0.15molL)=  0.0033 mol

Let’s calculate the consumed amount of NaOH

(22.0mL×1L1000mL) × (0.15molL)=  0.0033 mol

Let’s calculate the produced amount of CH3COO- after completion of reaction.

=0.0033 mol CH3COOH × 1mol CH3COO-1mol CH3COOH=0.0033 mol

The total volume of solution= 22.0mL + 22.0mL= 44.0mL × 1L1000mL= 0.044L

Concentration ofCH3COO-=0.0033mol0.044L= 0.075M

Therefore, 0.75M is the final volume of CH3COO- after completion of the reaction.

The CH3COO- ions react with water. The resulting solution is basic in nature.

The equilibrium reaction is as follows.

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

The equilibrium expression:

Kb[CH3COOH][OH-][CH3COO-]

Kb of CH3COO- is 5.6×10-10

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.

         CH3COO-(aq) + H2O(l) CH3CO2H(aq) + OH-(aq)   I                 0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Chapter 16 Solutions

Show all chapter solutions add
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

Additional Science Solutions

Find more solutions based on key concepts

Show solutions add

For each atomic symbol, give the name of the element. a Be b Ag c Si d C

General Chemistry - Standalone book (MindTap Course List)

Fasting hypoglycemia maybe caused by all except a. pancreatic tumors. b. poorly controlled diabetes. c. overuse...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

________ is lifes primary source of energy. a. Food b. Water c. Sunlight d. ATP

Biology: The Unity and Diversity of Life (MindTap Course List)

Identify the 3 and 5 ends of the DNA segment AGTCAT.

Chemistry for Today: General, Organic, and Biochemistry