   Chapter 14, Problem 80E

Chapter
Section
Textbook Problem

# A typical sample of vinegar has a pH of 3.0. Assuming that vinegar is only an aqueous solution of acetic acid (Ka = 1.8 × 10−5), calculate the concentration of acetic acid in vinegar.

Interpretation Introduction

Interpretation: The pH of a vinegar solution and its Ka value is given. The concentration of acetic acid in vinegar is to be calculated, given that vinegar is only an aqueous solution of acetic acid (CH3COOH) .

Concept introduction: The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution.

The pH of a solution is calculated by the formula, pH=log[H+]

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

Explanation

Explanation

To determine: The concentration of acetic acid in vinegar.

The [H+] for acetic acid is 1.0×10-3M_ .

Given

The pH of the given concentration of acetic acid is 3.0 .

The pH of a solution is calculated by the formula,

pH=log[H+]

Rearrange the above expression to obtain the value of [H+] .

[H+]=10pH

Substitute the pH value of acetic acid in the above expression.

[H+]=103.0=1.0×10-3M_

The equilibrium constant expression for the dissociation reaction of acetic acid is,

Ka=[H+][CH3COO][CH3COOH]

CH3COOH is a comparatively stronger acid than H2O .

The dominant equilibrium reaction for the given case is,

CH3COOH(aq)H+(aq)+CH3COO(aq)

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

Ka=[H+][CH3COO][CH3COOH] (1)

The concentration of CH3COOH is 0.0565M_ .

The concentration of CH3COOH is assumed to be x .

The [H+] for acetic acid is 1.0×103M .

The ICE table for the stated reaction is,

CH3COOH(aq)H+(aq)+CH3COO(aq)Inititialconcentrationx00Change1

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