Explanation Given The value of K a for acetic acid is 1.8 × 10 − 5 . The pH of a sample of vinegar is 2.90 . Vinegar contains the acetic acid. The chemical reaction for the ionization of acetic acid in aqueous solution with H + ion as a product is, CH 3 COOH ( a q ) ⇌ H + ( a q ) + CH 3 COO − ( 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 [ − 2.90 ] = 1.25 × 10 − 3 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 1.25 × 10 − 3 M and the initial concentration of acetic acid in vinegar is taken as x M . The ICE table for the dissociation reaction of acetic acid is, CH 3 COOH ( a q ) ⇌ H + ( a q ) + CH 3 COO − ( a q ) Initial concentration ( M ) : x 0 0 Change in concentration ( M ) : − 1.25 × 10 − 3 + 1.25 × 10 − 3 + 1.25 × 10 − 3 Equilibrium concentration ( M ) : ( x − 1

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Chapter 16, Problem 55E

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To determine: The concentration of acetic acid in the vinegar if the pH of a given sample of vinegar is 2.90 .

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Chapter 16, Problem 54E

Chapter 16, Problem 56E

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Chapter 16 Solutions

CHM 101 VOL 1 2014 >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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