Explanation Here, we are given a 0.140 M F − solution. We can say that F − reacts with water to give HF according to the reaction F − (aq) + H 2 O(l) ↔ HF (aq) + OH − (aq) Now, from the above reaction, we can see that HF is the weak acid obtained from fluoride anion on hydrolysis. The dissociation constant for HF theoritically is given as K a = 7.2 × 10 −4 The dissociation constant for F − can be calculated using the relation K a × K b = K w We can calculate the [ OH − ] by constructing the ICE table. After calculating the [ OH − ] we can calculate the pOH using the logarithimic table. Finally we can calculate the pH using the pOH value. Given: Initial concentration of F − is 0.140 M Formula Used: The relation between dissociation constant K b and equilibrium concentration is K b = [ P r o d u c t ] [ R e a c t a n t ] The pOH can be calculated as pOH = -log[ OH − ] The pH can be calculated as pH + pOH = 14 The dissociation constant for F − can be calculated using the relation K a × K b = K w Where K w is the ionic product of water K a is the dissociation constant for HF K b is the dissociation constant for F − Calculation: The given reaction is F − (aq) + H 2 O(l) ↔ HF (aq) + OH − (aq) Here HF is the weak acid, theoretically the dissociation constant K a = 7.2 × 10 −4 Now, K a × K b = K w K a = 7.2 × 10 −4 × K b = 1 × 10 −14 K b = [ 1 × 10 − 14 ] [ 7.2 × 10 − 4 ] K b = 1.3888 × 10 −11 Therefore, The dissociation constant for F − is K b = 1...

Chemistry: Structure and Properties Custom Edition for Rutgers University General Chemistry

15th Edition

ISBN: 9781269935678

Author: Nivaldo J. Tro

Publisher: Pearson Education

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Chapter 17, Problem 99E

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To determine:

The [OH⁻] and pH of a solution that is 0.140 M in F⁻

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Chapter 17, Problem 98E

Chapter 17, Problem 100E

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Chemistry: Structure and Properties Custom Edition for Rutgers University General Chemistry

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