Interpretation Introduction To determine: The value of molar concentration of OH − and pH of the solution. Explanation Given The value of K b is 4 .0 × 10 − 6 . Concentration of the solution is 0.724 M . The dominant equilibrium reaction that takes place in the given case is, BrO − ( a q ) + H 2 O ( l ) ⇌ HBro ( a q ) + OH − The change in the concentration of BrO − is assumed to be x . The ICE table is formed for the given reaction. BrO − ( a q ) + H 2 O ( l ) ⇌ HBrO ( a q ) + OH − ( a q ) Initial : 0.724 0 0 Change : − x + x + x Equilibrium: 0.724 − x x x The liquid component does not effect the value of equilibrium constant. The equilibrium concentration of BrO − is ( 0.724 − x ) M . The equilibrium concentration of HBrO is x M . The equilibrium concentration of OH − is ( x ) M . At equilibrium, the equilibrium constant expression is expressed by the formula, K b = Concentration of products Concentration of reactants Where, K a is the base ionization constant

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Chapter 16, Problem 72E

Interpretation Introduction

To determine: The value of molar concentration of OH− and pH of the solution.

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Chapter 16, Problem 71E

Chapter 16, Problem 73E

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Chapter 16 Solutions

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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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