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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

A buffer solution with it pH of 12.00 consists of Na3PO4 and Na2HPO4. The volume of solution is 200.0 mL.

  1. (a) Which component of the buffer is present in a larger amount?
  2. (b) If the concentration of Na3PO4 is 0.400 M, what mass of Na2HPO4 is present?
  3. (c) Which component of the buffer must be added to change the pH to 12.25? What mass of that component is required?

(a)

Interpretation Introduction

Interpretation:

For the given buffer solution of Na3PO4 and Na2HPO4, Component which is present in larger amount has to be determined.

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,

This equation shows that pH of buffer solution is controlled by two major factors. First, Strength of the acid can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium. It can be seen from equation (1) that pH of the buffer solution is comparable to pKa values. So this equation can be used to establish a relation between pH and pKa value of acid.

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

pOH=pKb+log[conjugateacid][base] (1)

Explanation

The calculation for amount of Na3PO4 is done by using Henderson-Hasselbalch equation.

Given:

Refer to table no. 16.2 in the textbook for the value of Kb.

The value of Kb for Na3PO4 is 2.8×102.

The value of Kb for Na2HPO4 is 1.6×107.

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

Therefore, pKb value for the Na3PO4 is 1.55 and for Na2HPO4 pKb value is 6.80.

The value of pH for the solution is 12.

The pOH value of the solution is calculated.

pOH = 1412=2.0

Therefore, pOH value of the solution is 2.0.

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

pOH=pKb+log[conjugateacid][base]

Substitute 1.55 for pKb, [Na2HPO4] for [conjugate acid], [Na3PO4] for [base], 2

(b)

Interpretation Introduction

Interpretation:

For the given buffer solution of Na3PO4 and Na2HPO4,

Mass of the Na3PO4 has to be calculated when concentration of Na2HPO4

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,

This equation shows that pH of buffer solution is controlled by two major factors. First, Strength of the acid can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium. It can be seen from equation (1) that pH of the buffer solution is comparable to pKa values. So this equation can be used to establish a relation between pH and pKa value of acid.

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

pOH=pKb+log[conjugateacid][base] (1)

(c)

Interpretation Introduction

Interpretation:

The Component of the buffer must be added to change pH to 12.25 has to be predicted; mass of that component has to be calculated.

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,

This equation shows that pH of buffer solution is controlled by two major factors. First, Strength of the acid can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium. It can be seen from equation (1) that pH of the buffer solution is comparable to pKa values. So this equation can be used to establish a relation between pH and pKa value of acid.

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

pOH=pKb+log[conjugateacid][base] (1)

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Chapter 17 Solutions

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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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