Explanation Given The value of K a for hydrofluoric acid is 6.8 × 10 − 4 . The pH of the solution of HF is 3.65 . The chemical reaction for the ionization of hydrofluoric acid in aqueous solution with H + ion as a product is, HF ( a q ) ⇌ H + ( a q ) + F − ( a q ) pH is the negative logarithm in the base 10 of [ H + ] . The concentration of H + ion is calculated by the formula, pH = − log [ H + ] [ H + ] = antilog ( − pH ) Where, [ H + ] is the concentration of H + ions. Substitute the value of pH in the above formula. [ H + ] = antilog [ − 3.65 ] = 2.23 × 10 − 4 M K a is the dissociation constant of acids. Weak acids ionize partially in aqueous solution. So, the dissociation constant of acids is used for determining the extent to which a weak acid ionizes. The concentration of H + ion after dissociation is 2.23 × 10 − 4 M and the initial concentration of hydrofluoric acid is taken as x M . The ICE table for the dissociation reaction of hydrofluoric acid is, HF ( a q ) ⇌ H + ( a q ) + F − ( a q ) Initial concentration ( M ) : x 0 0 Change in concentration ( M ) : − 2.23 × 10 − 4 + 2.23 × 10 − 4 + 2.23 × 10 − 4 Equilibrium concentration ( M ) : ( x − 2

Laboratory Experiments For Chemistry: The Central Science, Si Edition

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ISBN: 9781292221335

Author: Theodore E. Brown

Publisher: PEARSON

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Chapter 16, Problem 56E

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To determine: The concentration of hydrofluoric acid, if the pH of the solution of HF is given to be 3.65 .

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Chapter 16 Solutions

Laboratory Experiments For Chemistry: The Central Science, Si 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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