Explanation Given The concentration of phenylacetic acid solution is 0.085 M . The pH of the solution is 2.68 . Phenylacetic acid ( C 6 H 5 CH 2 COOH ) is a weak acid so it will ionize partially and release H + ions in the aqueous solution. The chemical reaction of phenylacetic acid in aqueous solution with H + ion as a product is, C 6 H 5 CH 2 COOH ( a q ) ⇌ H + ( a q ) + C 6 H 5 CH 2 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 + ] = 10 − pH Where, [ H + ] is the concentration of H + ions. Substitute the value of pH in the above formula. [ H + ] = 10 − 2 .68 = 2.08 × 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 initial concentration of phenylacetic acid is 0.085 M and the concentration of H + ion after dissociation is 2.08 × 10 − 3 M . The ICE table for the dissociation reaction of lactic acid is, C 6 H 5 CH 2 COOH ( a q ) ⇌ H + ( a q ) + C 6 H 5 CH 2 COO − ( a q ) Initial: 0.085 M 0 0 Change: − 2.08 × 10 − 3 M 2.08 × 10 − 3 M 2

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Chapter 16, Problem 52E

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To determine: The dissociation constant (Ka) for phenylacetic acid.

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Chapter 16, Problem 51E

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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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The dissociation constant ( K a ) for phenylacetic acid.

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To determine: The dissociation constant (Ka) for phenylacetic acid.

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