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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

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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.

### 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** This section details the usage of NaOH in a titration involving KHTar in a 0.10 M NaCl solution. The solution's concentration of NaOH is 0.0500 M. The initial and final volumes for two trials are given, where the volume of NaOH used needs to be calculated along with the average volume used.
Transcribed Image Text:### 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** This section details the usage of NaOH in a titration involving KHTar in a 0.10 M NaCl solution. The solution's concentration of NaOH is 0.0500 M. The initial and final volumes for two trials are given, where the volume of NaOH used needs to be calculated along with the average volume used.
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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