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Use the References to access important values if needed for this question. In the laboratory, a general chemistry student measured the pH of a 0.432 M aqueous solution of hydrocyanic acid to be 4.868. Use the information she obtained to determine the K, for this acid. K(experiment) =

Use the References to access important values if needed for this question. In the laboratory, a general chemistry student measured the pH of a 0.432 M aqueous solution of hydrocyanic acid to be 4.868. Use the information she obtained to determine the K, for this acid. K(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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Use the References to access important vallues if needed for this question. In the laboratory, a general chemistry student measured the pH of a 0.432 M aqueous solution of hydrocyanic acid to be 4.868. Use the information she obtained to determine the K, for this acid. K,(experiment) =|
Transcribed Image Text:Use the References to access important vallues if needed for this question. In the laboratory, a general chemistry student measured the pH of a 0.432 M aqueous solution of hydrocyanic acid to be 4.868. Use the information she obtained to determine the K, for this acid. K,(experiment) =|
In the laboratory, a general chemistry student measured the pH of a 0.469 M aqueous solution of acetic acid to be 2.523. Use the information she obtained to determime the K, for this acid. K,(experiment) =
Transcribed Image Text:In the laboratory, a general chemistry student measured the pH of a 0.469 M aqueous solution of acetic acid to be 2.523. Use the information she obtained to determime the K, for this acid. K,(experiment) =
Expert Solution
Step 1

(1.) 

Given datas –

[HCN] = 0.432 M 

pH = 4.868  

-log [H+] = 4.868 

[H+] = 10-4.868 

[H+] = 1.355 × 10-5 M 

 HCN <---> H+ +  CN–  -------(I) 

Therefore, by eqn 

[CN–] = 1.355 × 10-5 M = [H+] 

Hence, 

Ka = [H+] [CN–] / [HCN] 

Ka = (1.355 × 10-5)2 / 0.432 

Ka = 4.25 × 10-10 

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