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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 111CP
Textbook Problem
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What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00?

Interpretation Introduction

Interpretation: The formation of a buffer solution by the addition of NaOH to a solution of HOCl is given. The volume of NaOH that must be added to the solution to achieve a pH equal to 8 is to be calculated.

Concept introduction: The solution of weak acid and their conjugate base which upon addition of an acid or base resists the change in pH is called buffer solution. The pH of the buffer solution remains constant.

Explanation of Solution

Explanation

Given

The concentration of NaOH is 0.0100M .

The concentration of HOCl is 0.0500M .

The volume of HOCl is 1.00L .

The pH of the solution is 8.0 .

The reaction between NaOH and HOCl is represented as,

NaOH+HOClNaOCl+H2O

The Henderson-Hasselbalch equation is represented as shown below.

pH=pKa+log[A][HA]

Where,

  • pH is the negative logarithm of H+ ions concentration in the solution.
  • pKa is the negative logarithm of dissociation constant of the acid.
  • [A] is the concentration of conjugate base of the given weak acid.
  • [HA] is the concentration of given weak acid.

Substitute the value of pH and pKa in the above equation as,

pH=pKa+log[OCl][HOCl]8.0=7.45+log[OCl][HOCl]0.55=log[OCl][HOCl][OCl][HOCl]=3.548

Let the volume of NaOH solution is xl

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres

Rearrange the above equation to obtain the value of number of moles.

Numberofmoles=Concentration×Volumeofsolutioninlitres

Substitute the value of Concentration and volume of solution of NaOH in the above equation as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.0100M×xL=0.0100xmoles

Similarly substitute the value of Concentration and volume of solution of HOCl in the above equation as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.0500M×1.00L=0.0500moles

Addition of NaOH to HOCl leads to the formation of OCl

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Chapter 14 Solutions

Chemistry: An Atoms First Approach
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Ch. 14 - What are the major species in solution after...Ch. 14 - A friend asks the following: Consider a buffered...Ch. 14 - Mixing together solutions of acetic acid and...Ch. 14 - Could a buffered solution be made by mixing...Ch. 14 - Sketch two pH curves, one for the titration of a...Ch. 14 - Sketch a pH curve for the titration of a weak acid...Ch. 14 - You have a solution of the weak acid HA and add...Ch. 14 - You have a solution of the weak acid HA and add...Ch. 14 - The common ion effect for weak acids is to...Ch. 14 - Consider a buffer solution where [weak acid] ...Ch. 14 - A best buffer has about equal quantities of weak...Ch. 14 - Consider the following pH curves for 100.0 mL of...Ch. 14 - An acid is titrated with NaOH. 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