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lution 01 0.0500 M NaOH Volumes: Initial 45.16 mL 42.93 mL Final 21.38 mL 18.78 mL Used Average Used mL

lution 01 0.0500 M NaOH Volumes: Initial 45.16 mL 42.93 mL Final 21.38 mL 18.78 mL Used Average Used mL

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 88QAP
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Label as KHTar in .01M KCl. Add approximately 100 mL of .01 M KCl and 2 grams of KHTar

Solve for [HTar] and amount used.

2. Solution of KHTar in 0.10 M NaCl: 0.0500 M NaOH Volumes: Initial 45.16 mL 42.93 mL Final 21.38 mL 18.78 mL Used Average Used mL
Transcribed Image Text:2. Solution of KHTar in 0.10 M NaCl: 0.0500 M NaOH Volumes: Initial 45.16 mL 42.93 mL Final 21.38 mL 18.78 mL Used Average Used mL
Expert Solution
Step 1

The values as calculated

Initial 45.16ml 42.93ml
Final 21.38ml 18.78ml
Used 23.78ml 24.15ml

Average NaOH used

=23.78+24.152=23.96ml

 

Step 2

Dissociation of HTar

KHTar(s)⇔K+(aq)+HTar-(aq)

from KCl solution K+=0.01M               (I)

now this Htar reacts with NaOH solution.

HTar-(aq) + OH-(aq)⇔ Tar2-(aq) +H2O(l)        (II)

now volume of NaOH used=23.96 ml

and  molarity of NaOH solution used =0.05 M

therefore moles of NaOH ==23.96×0.05 m mol=1.196mmol

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