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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with 25.0 mL of 0.100 M NaOH. The student then adds 13.0 mL of 0.100 M HCl. The pH of the resulting solution is 4.70. How is the value of pKa for the unknown acid related to 4.70?

Interpretation Introduction

Interpretation: The pH of solution of a weak acid HA with NaOH to which HCl is given. The relationship of pKa to the value of 4.70 is to be stated.

Concept introduction: A strong acid has a weak conjugate base while a weak acid has a strong conjugate base.

Explanation

Explanation

Given

The concentration of NaOH is 0.100M.

The volume of NaOH is 25.0mL.

The concentration of HCl is 0.100M

The volume of HCl is 13.0mL.

The pH of the solution is 4.70

The pH of the solution is shown below.

pH=log[H+]

Where,

  • [H+] is the concentration of Hydrogen ions.

Substitute the value of pH in the above equation to find the concentration of [H+].

pH=log[H+]4.7=log[H+][H+]=104.7=2×105M

Therefore, the number of moles of [H+] present per litre in final solution =2×105.

The total volume of final solution

=(25.0+13.0)mL=38.0mL

The conversion of mL into L is done as,

1mL=0.001L

Hence, the conversion of 38.0mL, 13.0mL and 25.0mL into L is done as,

38.0mL=38.0×0.001L=0.038L25.0mL=25.0×0.001L=0.025L13.0mL=13.0×0.001L=0.013L

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (1)

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

Numberofmoles=Concentration×Volumeofsolutioninlitres

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

Numberofmoles=Concentration×Volumeofsolutioninlitres=2.0×105moleL×0.038L=7.6×107moles

Similarly, substitute the value of concentration and volume of NaOH in above equation as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100moleL×0.025L=0.0025moles

The reaction between acid HA and base NaOH is represented as,

HA+NaOHNaA+H3O+

As it is seen from the equation that 1mole of HA reacts with 1mole of NaOH to produce 1mole of NaA

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