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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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BuyFindarrow_forward

Chemistry

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

Potassium hydrogen phthalate, known as KHP (molar mass = 204.22 g/mol), can be obtained in high purity and is used to determine the concentration of solutions of strong bases by the reaction

HP ( a q ) + OH ( a q ) H 2 O ( l ) + P 2 ( a q )

If a typical titration experiment begins with approximately 0.5 g KHP and has a final volume of about 100 mL, what is an appropriate indicator to use? The pKa for HP is 5.51.

Interpretation Introduction

Interpretation:

The final volume of the titration experiment and the initial amount of potassium hydrogen phthalate (KHP) is given. An appropriate indicator that can be used in the stated titration is to be determined.

Concept introduction:

A substance that is expected to change its color in response to a change in the chemical properties of a solution is termed as an indicator. The end point in a titration corresponds to a color change of the solution.

To determine: An appropriate indicator that can be used in the stated titration.

Explanation

Explanation

To find the value of Kb

The mass of KHP is 0.5g .

The molar mass of KHP is 2044.22g/mol .

Volume of the solution is 0.1L .

The number of moles is calculated by the formula,

Moles=MassMolarmass

Substitute the value of mass and molar mass of KHP in the above expression.

Moles=0.5g204.22g/mol=0.00245mol

The given value of pKa of HP is 5.51 .

The Ka is calculated by the formula,

Ka=10pKa

Substitute the value of pKa in the above expression.

Ka=105.51=3.1×106

The value of Kb is calculated by the formula,

Kb=KwKa

Substitute the value of Ka in the above expression.

Kb=1.0×10143.1×106=3.22×10-9_

To find the [OH]

The given salt KHP will give 0.0244M K+ and 0.0244M HP ions.

The volume of the solution is 0.1L .

The concentration is calculated by the formula,

Concentration=NumberofmolesVolume(L)

Substitute the value of number of moles of KHP and the volume of the solution in the above expression.

Concentration=0.00245mol0.1L=0.0245M

The change in the concentration of P2 is assumed to be x .

The ICE table is formed for the given reaction.

P2(aq)HP(aq)+OH(aq)Initialconcentration0.024500Changex+x+xEquilibriumconcentration0.0245xxx

The equilibrium concentration of [P2] is (0

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