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EXPERIMENT : CONDUCTOMETRIC TITRATION OF A MIXTURE OF ACIDS WITH A STRONG BASE I want the analysis for this experiment

EXPERIMENT : CONDUCTOMETRIC TITRATION OF A MIXTURE OF ACIDS WITH A STRONG BASE I want the analysis for this experiment

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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EXPERIMENT : CONDUCTOMETRIC TITRATION OF A MIXTURE OF ACIDS WITH A STRONG BASE

I want the analysis for this experiment 

Calculations: For HCl: Given the volume of the strong acid (HCI) = 15 M1= 0.01 M From the experimental data: Volume of NaOH (V1) = 14 ml M1XV1 0.01 x14 So, the molarity of HCI = 15 = 9.333 x 10-3 M 15 molarity of HCI = 9.333 x 10-3 M For weak acid ( CH3COOH ): Given the volume of the weak acid (CH3COOH) = 15 M1= 0.01 M From the experimental data: Volume of NaOH (V2) = 18 ml So, the molarity of CH3COOH = М1x(V2-V1) 0.01 x(18-14) = 2.666 x 10-3 M 15 15 molarity of CH3COOH = 2.666 × 10-3 M
Transcribed Image Text:Calculations: For HCl: Given the volume of the strong acid (HCI) = 15 M1= 0.01 M From the experimental data: Volume of NaOH (V1) = 14 ml M1XV1 0.01 x14 So, the molarity of HCI = 15 = 9.333 x 10-3 M 15 molarity of HCI = 9.333 x 10-3 M For weak acid ( CH3COOH ): Given the volume of the weak acid (CH3COOH) = 15 M1= 0.01 M From the experimental data: Volume of NaOH (V2) = 18 ml So, the molarity of CH3COOH = М1x(V2-V1) 0.01 x(18-14) = 2.666 x 10-3 M 15 15 molarity of CH3COOH = 2.666 × 10-3 M
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