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A 24.4 mL sample of 0.2124 M hydrotelluric acid, H₂Te, is titrated with 0.2124 M potassium hydroxide. What is the pH after 12.2 mL, 24.4 mL, 36.6 mL and 48.8 mL of KOH have been added? step mL KOH added pH 1 2 3 4 Not submitted 12.2 24.4 36.6 48.8 second ionization constants for H₂Te are 2.3x10-3 and 1.6x10-11

A 24.4 mL sample of 0.2124 M hydrotelluric acid, H₂Te, is titrated with 0.2124 M potassium hydroxide. What is the pH after 12.2 mL, 24.4 mL, 36.6 mL and 48.8 mL of KOH have been added? step mL KOH added pH 1 2 3 4 Not submitted 12.2 24.4 36.6 48.8 second ionization constants for H₂Te are 2.3x10-3 and 1.6x10-11

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
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A 24.4 mL sample of 0.2124 M hydrotelluric acid, H₂Te, is titrated with 0.2124 M potassium hydroxide. What is the pH after 12.2 mL, 24.4 mL, 36.6 mL and 48.8 mL of KOH have been added? step mL KOH added pH 1 12.2 24.4 36.6 48.8 Not submitted 2 3 4 second ionization constants for H₂Te are 2.3x10-3 and 1.6×10-11
Transcribed Image Text:A 24.4 mL sample of 0.2124 M hydrotelluric acid, H₂Te, is titrated with 0.2124 M potassium hydroxide. What is the pH after 12.2 mL, 24.4 mL, 36.6 mL and 48.8 mL of KOH have been added? step mL KOH added pH 1 12.2 24.4 36.6 48.8 Not submitted 2 3 4 second ionization constants for H₂Te are 2.3x10-3 and 1.6×10-11
Expert Solution
Step 1

H2Te is a diprotic acid with Ka1 = 2.3*10-3 and Ka2 = 1.6*10-11 

The dissociation equations are:

H2Te(aq) ⇌ HTe-(aq) + H+(aq) ; pKa1 = -log(Ka1) = - log(2.3*10-3) = 2.64

HTe-(aq) ⇌ Te2-(aq) + H+(aq) ; pKa2 = -log(Ka2) = - log(1.6*10-11) = 10.8 

potassium hydroxide, KOH is a monoacidic base. 

 

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