9th Edition
Steven S. Zumdahl
ISBN: 9781133611097




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.




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.



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

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


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

The total volume of final solution


The conversion of mL into L is done as,


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


The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (1)

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


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


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


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


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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