Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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**Topic: Calculating the pH of a Weak Acid Solution**

Phenylacetic acid (\(C_6H_5CH_2CO_2H\)) is a weak monoprotic acid with an acid dissociation constant (\(K_a\)) of \(4.90 \times 10^{-5}\). The task is to determine the pH of a \(9.21 \times 10^{-1} \, M\) solution of potassium phenylacetate (\(C_6H_5CH_2CO_2^− K^+\)).

**Context:**

When salts like potassium phenylacetate are dissolved in water, they dissociate into ions. In this case, the phenylacetate ion (\(C_6H_5CH_2CO_2^−\)) acts as a conjugate base and can accept a proton (\(H^+\)) from water, forming phenylacetic acid and a hydroxide ion (\(OH^−\)). This results in a basic solution. 

The pH of such a solution can be calculated by considering the hydrolysis reaction and using the relationship between \(K_a\) and \(K_b\) (the base dissociation constant), given by:

\[ K_w = K_a \times K_b \]

where \(K_w\) is the ion-product constant of water (\(1.0 \times 10^{-14}\)). By calculating \(K_b\), we can determine the \(OH^−\) concentration and subsequently the pH.
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Transcribed Image Text:**Topic: Calculating the pH of a Weak Acid Solution** Phenylacetic acid (\(C_6H_5CH_2CO_2H\)) is a weak monoprotic acid with an acid dissociation constant (\(K_a\)) of \(4.90 \times 10^{-5}\). The task is to determine the pH of a \(9.21 \times 10^{-1} \, M\) solution of potassium phenylacetate (\(C_6H_5CH_2CO_2^− K^+\)). **Context:** When salts like potassium phenylacetate are dissolved in water, they dissociate into ions. In this case, the phenylacetate ion (\(C_6H_5CH_2CO_2^−\)) acts as a conjugate base and can accept a proton (\(H^+\)) from water, forming phenylacetic acid and a hydroxide ion (\(OH^−\)). This results in a basic solution. The pH of such a solution can be calculated by considering the hydrolysis reaction and using the relationship between \(K_a\) and \(K_b\) (the base dissociation constant), given by: \[ K_w = K_a \times K_b \] where \(K_w\) is the ion-product constant of water (\(1.0 \times 10^{-14}\)). By calculating \(K_b\), we can determine the \(OH^−\) concentration and subsequently the pH.
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