   # Acrylic acid (CH 2 9CHCO 2 H) is a precursor for many important plastics. K a for acrylic acid is 5.6 × 10 −5 . a. Calculate the pH of a 0.10- M solution of acrylic acid. b. Calculate the percent dissociation of a 0.10- M solution of acrylic acid. c. Calculate the pH of a 0.050- M solution of sodium acrylate (NaC 3 H 3 O 2 ). ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 151AE
Textbook Problem
99 views

## Acrylic acid (CH29CHCO2H) is a precursor for many important plastics. Ka for acrylic acid is 5.6 × 10−5.a. Calculate the pH of a 0.10-M solution of acrylic acid.b. Calculate the percent dissociation of a 0.10-M solution of acrylic acid.c. Calculate the pH of a 0.050-M solution of sodium acrylate (NaC3H3O2).

(a)

Interpretation Introduction

Interpretation: The acid dissociation constant Ka   and molarity of acrylic acid and sodium acrylate are given. By using these values, the pH value of solution of acrylic acid and sodium acrylate and percent dissociation of solution of acrylic acid is to be calculated.

Concept introduction: The equilibrium constant Kb is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant base.

The equilibrium constant Ka is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant acid.

To determine: The change in concentration of H+ and the pH value of acrylic acid.

### Explanation of Solution

Explanation

The dissociation reaction of acrylic acid is,

C3H3O2H(aq)C3H3O2-(aq)+H+(aq)

The concentration of H+ is 2.33×103M_ .

Given

The equilibrium constant is 5.6×105 .

The initial concentration of acrylic acid is 0.10M .

It is assumed that the change in concentration of [H+] is x .

Make the ICE table for the dissociation reaction of acrylic acid.

C3H3O2H(aq)C3H3O2(aq)H+(aq)Initial(M):0.1000Chang(M):xxxEquilibrium(M):0.10xxx

The expression for the acid dissociation constant for the given reaction is,

Ka=[C3H3O2][H+][C3H3O2H]

Substitute the value of Ka and equilibrium concentrations in the above equation.

Ka=[C3H3O2][H+][C3H3O2H]5

(b)

Interpretation Introduction

Interpretation: The acid dissociation constant Ka   and molarity of acrylic acid and sodium acrylate are given. By using these values, the pH value of solution of acrylic acid and sodium acrylate and percent dissociation of solution of acrylic acid is to be calculated.

Concept introduction: The equilibrium constant Kb is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant base.

The equilibrium constant Ka is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant acid.

To determine: The percent dissociation of solution of acrylic acid.

(c)

Interpretation Introduction

Interpretation: The acid dissociation constant Ka   and molarity of acrylic acid and sodium acrylate are given. By using these values, the pH value of solution of acrylic acid and sodium acrylate and percent dissociation of solution of acrylic acid is to be calculated.

Concept introduction: The equilibrium constant Kb is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant base.

The equilibrium constant Ka is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant acid.

To determine: The pH value of solution of sodium acrylate.

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