(a) Explanation Acid dissociation constant is 6.8 × 10 − 4 . The ionic product for water is 1.0 × 10 − 14 . Dissociation reaction of NaF is, NaF ( a q ) → Na + ( a q ) + F − ( a q ) F − is a conjugate base of a weak acid HF . When F − reacts with water it gives HF and OH − . The reaction that takes place is, F − ( a q ) + H 2 O ( a q ) → HF ( a q ) + OH − ( a q ) The ICE table for the equilibrium concentration of the F − , H 2 O , HF and OH − is, F − ( a q ) + H 2 O ( a q ) → HF ( a q ) + OH − ( a q ) Initial 0 .105 M 0 0 0 Change − x M 0 x M x M Equilibrium ( 0.105 − x ) M 0 x M x M The equilibrium concentration of water is neglected because it is not involved in the equilibrium constant expression. The equilibrium constant ( K b ) is calculated by the formula, K b = Concentration of product concentration of reactant Substitute the value of concentration of product and concentration of reactant in the above equation, K b = [ HF ] [ OH − ] [ F − ] …… (b) Interpretation Introduction To determine: The concentration of [ OH − ] and the pH of the solution. (c) Interpretation Introduction To determine: The concentration of [ OH − ] and the pH of the solution.

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Chapter 16, Problem 80E

(a)

Interpretation Introduction

To determine: The concentration of [OH−] and the pH of the solution.

(b)

Interpretation Introduction

To determine: The concentration of [OH−] and the pH of the solution.

(c)

Interpretation Introduction

To determine: The concentration of [OH−] and the pH of the solution.

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Chapter 16, Problem 79E

Chapter 16, Problem 81E

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Ch. 16.2 - Practice Exercise 1 Consider the following...Ch. 16.2 - Prob. 16.1.2PE Ch. 16.2 - Prob. 16.2.1PE Ch. 16.2 - Practice Exercise 2 When lithium oxide (Li2O) is...Ch. 16.2 - Based on information in Figure 16.4, place the...Ch. 16.2 - Practice Exercise 2 For each reaction, use Figure...Ch. 16.3 - Practice Exercise 1 In a certain acidic solution...Ch. 16.3 - Practice Exercise 2 Indicate whether solutions...Ch. 16.3 - Prob. 16.5.1PE Ch. 16.3 - Prob. 16.5.2PE

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