Explanation Given The number of moles of the solution is 0 .5 mol . The volume of the solution is 0 .05 L . The dissociation reaction of NaF is, NaF → Na + + F − The F − ion is a strong conjugate base of weak acid HF . The dissociation reaction of HF is, HF ⇌ H + + F − The K a value of HF is 6.6 × 10 − 4 . Formula used: The value of K b is calculated by the formula, K b = K w K a The molarity is calculated by the formula, Molarity = Number of moles Volume Calculation: Substitute the value of K w and K a in the above equation, K b = 1.0 × 10 − 14 6.6 × 10 − 4 = 1.5 × 10 − 11 The molarity is calculated by the formula, Molarity = Number of moles Volume Substitute the value of number of moles and volume of the solution in the above equation, Molarity = 0 .05 0 .5 = 0.1 M The reaction of F − with water is, F − + H 2 O → HF + OH − The ICE table for the equilibrium concentration of F − , HF , and OH − is, F − + H 2 O ⇌ HF + OH − Initial 0.1 M 0 0 0 Change − x 0 + x + x Equilibrium ( 0.1 − x ) M 0 x x The solid and liquid species do not affect the value of equilibrium constant. The equilibrium constant is calculated by the formula, K b = Product concentration Reactant concentration Where, K b is the base ionization constant. Substitute the value of K b , concentration of the product species and the concentration of the reactant species in the above equation, 1.5 × 10 − 11 = ( x ) ( x ) ( 0.1 − x ) The x is neglected as the value of x is small as compare to the value of equilibrium constant. ( x ) 2 = 1.5 × 10 − 12 ( x ) = 1.22 × 10 − 6 Therefore, the concentration of OH − is 1.22 × 10 − 6 . The value of pH is calculated by the formula, pH = 14 − ( − log [ OH ] ) Substitute the value of OH − in the above equation, pH = 14 − ( − log [ 1.22 × 10 − 6 ] ) = 14 − 5.91 = 8.09 Therefore, the pH of the NaF is 8

Chemistry: The Central Science (14th Edition)

14th Edition

ISBN: 9780134414232

Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus

Publisher: PEARSON

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Chapter 16, Problem 85E

Interpretation Introduction

Interpretation: The salt for the given pH and given concentration is to be determined.

Concept introduction: The salt of weak acid and strong base forms a basic solution, since on dissociation it forms the strong conjugate base of weak acid that reacts with water to produce hydroxide ions.

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Chapter 16, Problem 84E

Chapter 16, Problem 86E

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Chapter 16 Solutions

Chemistry: The Central Science (14th Edition)

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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The salt for the given pH and given concentration is to be determined. Concept introduction: The salt of weak acid and strong base forms a basic solution, since on dissociation it forms the strong conjugate base of weak acid that reacts with water to produce hydroxide ions.

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Interpretation: The salt for the given pH and given concentration is to be determined. Concept introduction: The salt of weak acid and strong base forms a basic solution, since on dissociation it forms the strong conjugate base of weak acid that reacts with water to produce hydroxide ions.

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Chapter 16 Solutions

Interpretation: The salt for the given pH and given concentration is to be determined.

Concept introduction: The salt of weak acid and strong base forms a basic solution, since on dissociation it forms the strong conjugate base of weak acid that reacts with water to produce hydroxide ions.