(a) Interpretation Introduction To determine: The pH of the pure water at 50 ∘ C . Explanation Given The ionic product constant for water is 5.48 × 10 − 14 at 50 ∘ C . The ionic product constant ( K w ) is 5.48 × 10 − 14 . The ionization reaction of water is, 2H 2 O → H 3 O + ( a q ) + OH − ( a q ) Formula used: The equilibrium constant is calculated by the formula, K w = [ Concentration of products ] [ Concentration of reactants ] pH= − log [ H + ] Calculation: The equilibrium constant expression for the given reaction is, K w = [ OH − ] [ H 3 O + ] Substitute the value of concentration of product and concentration of reactant in the above equation (b) Interpretation Introduction To determine: The sign of the Δ H for the given reaction.

19th Edition

ISBN: 9781323849996

Author: Brown

Publisher: Pearson Custom Publishing

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Chapter 16, Problem 120IE

(a)

Interpretation Introduction

To determine: The pH of the pure water at 50∘ C .

(b)

Interpretation Introduction

To determine: The sign of the ΔH for the given reaction.

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Chapter 16, Problem 119IE

Chapter 16, Problem 121IE

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Chapter 16 Solutions

CHEMISTRY VOL. 2 W/MASTERING CHEM. >IC<

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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