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Science
ChemistryChemistry & Chemical Reactivity10th Edition

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Chapter 17, Problem 11PS
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Chemistry & Chemical Reactivity

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

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Problem 2PS
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Problem 4PS
Problem 5PS
Problem 6PS
Problem 7PS
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Problem 18PS
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Problem 21PS
Problem 22PS
Problem 23PS
Problem 24PS
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Problem 40PS
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Problem 43PS
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Problem 45PS
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Problem 47PS
Problem 48PS
Problem 49PS
Problem 50PS
Problem 51PS
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Problem 54PS
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Problem 100GQ
Problem 101IL
Problem 102IL
Problem 103IL
Problem 104IL
Problem 105IL
Problem 106IL
Problem 107IL
Problem 108IL
Problem 109IL
Problem 110IL
Problem 111IL
Problem 112IL
Problem 113SCQ
Problem 114SCQ
Problem 115SCQ
Problem 116SCQ
Problem 117SCQ
Problem 118SCQ
Problem 119SCQ
Problem 120SCQ
BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

Calculate the pH of a solution that has an acetic acid concentration of 0.050 M and a sodium acetate concentration of 0.075 M.

Interpretation Introduction

Interpretation:

The value of pH is to be calculated for the buffer solution containing 0.05 M CH3COOH and 0.075 M CH3COONa by using the Henderson-Hasselbalch equation.

Concept introduction:

The Henderson-Hasselbalch equation relates pH of a buffer with pKa of acid, concentration of conjugate base and concentration of acid. The expression is written as,

pH=pKa+log[conjugate base][acid] (1)

This equation shows that pH of buffer solution is controlled by two major factors. First, is strength of the acid which can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium.

Explanation

The equilibrium between CH3COOH and its conjugate base CH3COONa is written as follows;

  CH3COOH(aq)+H2O(l)⇌H3O+(aq)+ CH3COO−(aq)  (acid)                                                 (conjugate base)

The calculation of pH is done by using Henderson-Hasselbalch equation is given below.

Given:

Refer to table 16.2 in the textbook for the value of Ka.

The value of Ka for CH3COOH is 1.8×10−5.

Negative logarithm of the Ka value gives the pKa value of the acid.

pKa=−log(Ka)=−log(1.8×10−5)=4

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Sect-17.1 P-17.1CYUSect-17.2 P-17.2CYUSect-17.2 P-17.3CYUSect-17.2 P-17.4CYUSect-17.2 P-17.5CYUSect-17.3 P-17.6CYUSect-17.3 P-17.7CYUSect-17.4 P-17.8CYUSect-17.4 P-17.9CYUSect-17.4 P-17.10CYU
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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