Explanation a. As given that the initial concentration of HF = 0.250 M. As per the below reaction: - HF (aq) + H2O (l) → H3O + (aq) + F − (aq) I (M) 0 .250 0 0 C (M) -X +X +X E (M) 0 .250-X X X K a = [ H3O + ] [ F - ] [ HF ] 3 .5 × 10 − 4 = ( X ) ( X ) ( 0.250 − X ) ( 0.250 − X ) ≈ 0 .250 X = 3.5 × 10 − 4 × 0 .250 X = 9 .4 × 10 − 3 M = [ H3O + ] Now pH will be calculated as below: - pH = - log [ H3O + ] = - log(9 .4 × 10 − 3 ) = 2 .02 The below formula will be used to calculate the % Ionization % Ionization = [ H3O + ] [ HCHO2 ] × 100 % = 9.4 × 10 − 3 0.250 × 100 % = 3 .7 % Therefore, the pH and the percent ionization of 0.250 M HF is 2.02 and 3.7 % respectively. b. As given that the initial concentration of HF = 0.100 M. As per the below reaction: - HF (aq) + H2O (l) → H3O + (aq) + F − (aq) I (M) 0 .100 0 0 C (M) -X +X +X E (M) 0 .100-X X X K a = [ H3O + ] [ F - ] [ HF ] 3 .5 × 10 − 4 = ( X ) ( X ) ( 0...

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ISBN: 9780321974617

Author: Nivaldo J. Tro

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Chapter 17, Problem 79E

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To determine: the pH and percent ionization of each HF solution.(Ka of HF is 6.8 × 10⁻¹)

a. 0.250 MHF

b. 0.100 MHF

c. 0.050 MHF

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Chapter 17, Problem 78E

Chapter 17, Problem 80E

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the pH and percent ionization of each HF solution.(Ka of HF is 6.8 × 10 −1 ) a. 0.250 MHF b. 0.100 MHF c. 0.050 MHF

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To determine: the pH and percent ionization of each HF solution.(Ka of HF is 6.8 × 10−1) a. 0.250 MHF b. 0.100 MHF c. 0.050 MHF

Want to see the full answer?

Chapter 17 Solutions

To determine: the pH and percent ionization of each HF solution.(Ka of HF is 6.8 × 10⁻¹)

a. 0.250 MHF

b. 0.100 MHF

c. 0.050 MHF