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Science
ChemistryChemistry & Chemical Reactivity10th Edition

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Chapter 17, Problem 20PS
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Chemistry & Chemical Reactivity

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

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Problem 2PS
Problem 3PS
Problem 4PS
Problem 5PS
Problem 6PS
Problem 7PS
Problem 8PS
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Problem 10PS
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Problem 13PS
Problem 14PS
Problem 15PS
Problem 16PS
Problem 17PS
Problem 18PS
Problem 19PS
Problem 20PS
Problem 21PS
Problem 22PS
Problem 23PS
Problem 24PS
Problem 25PS
Problem 26PS
Problem 27PS
Problem 28PS
Problem 29PS
Problem 30PS
Problem 31PS
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Problem 35PS
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Problem 39PS
Problem 40PS
Problem 41PS
Problem 42PS
Problem 43PS
Problem 44PS
Problem 45PS
Problem 46PS
Problem 47PS
Problem 48PS
Problem 49PS
Problem 50PS
Problem 51PS
Problem 52PS
Problem 53PS
Problem 54PS
Problem 55PS
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Problem 65PS
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Problem 72PS
Problem 73PS
Problem 74PS
Problem 75PS
Problem 76PS
Problem 77GQ
Problem 78GQ
Problem 79GQ
Problem 80GQ
Problem 81GQ
Problem 82GQ
Problem 83GQ
Problem 84GQ
Problem 85GQ
Problem 86GQ
Problem 87GQ
Problem 88GQ
Problem 89GQ
Problem 90GQ
Problem 91GQ
Problem 92GQ
Problem 93GQ
Problem 94GQ
Problem 95GQ
Problem 96GQ
Problem 97GQ
Problem 98GQ
Problem 99GQ
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
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Chemistry & Chemical Reactivity

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

Describe how to prepare a buffer solution from NH3 and NH4Cl to have a pH of 9.5.

Interpretation Introduction

Interpretation:

To make the buffer solution contains NH3 and NH4+ and its pH value is 9.5 has to be explained.

Concept introduction:

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

pOH=pKb+log[conjugate acid][base] (1)

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

Explanation

The equilibrium between  NH3 and its conjugate base  NH4+ is written as follows;

NH3(aq)+H2O(l)⇌NH4+(aq)+ OH−(aq)  (base)                   (conjugate acid)

The value of pOH is calculated by using expression,pH + pOH = 14.

Given:

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

The value of Kb for  NH3 is 1.8×10−5.

Negative logarithm of the Kb value gives the pKb value of the acid.

pKb=−log(Kb)=−log(1.8×10−5)=4.74

Therefore, pKb value for the  NH3 is 4.74.

The pH for the given buffer solution is 9.5.

The value of pOH is calculated by using expression,

pH + pOH = 14

Rearrange for pOH,

pOH = 14−pH  

Substitute, 9.5 for pH,

pOH = 14−9.5= 4.5

Therefore, pOH value for the given buffer is 4.5.

The value of concentration ratio of base and its conjugate acid is calculated by using equation (1).

pOH=pKb+log[conjugate acid][base]

Substitute, 4.74 for pKb, 4.5 for pOH, [NH4+]  for [conjugate acid], [NH3] for

[base]

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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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