   Chapter 16.7, Problem 16.5CYU

Chapter
Section
Textbook Problem

What are the equilibrium concentrations of acetic acid, the acetate ion, and H3O+ for a 0.10 M solution of acetic acid (K2 = 1.8 × 10−5)? What is the pH of the solution?

Interpretation Introduction

Interpretation:

Equilibrium concentration of acetic acid, acetate ion, H3O+ and pH of 0.10M solution of acetic acid is to be determined

Concept introduction:

The pH of a solution of acetic acid can be calculated by using the hydronium ion concentration by using the expression, pH=log[H3O+].

Hydronium ion concentration is calculated by considering the equilibrium conditions and from the value of acid dissociation constant.

The concentration of acetic acid and acetate ion is determined by using equilibrium condition The concentration of acetate ion and hydronium ion is equal at equilibrium according to the reaction stoichiometry.

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Explanation

Calculate concentrations of all ions in the solution and pH of 0.10M solution of acetic acid from the given data.

Given data

The initial concentration of the acetic acid solution is 0.10M.

The acid dissociation constant Ka of acetic acid is 1.8×105.

The ICE table for the dissociation of acetic acid is given below,

EquilibriumCH3COOH(aq)CH3COO(aq)+H3O+(aq)Initial(M)0.1000Change(M)x+x+xEquilibrium(M)0.10xxx

From ICE table the expression for Ka;

Ka=[CH3COO][H3O+][CH3COOH]

Substituting equilibrium concentrations for acetate ion, hydronium ion, and acetic acid;

Ka=x2(0.1x)

Substituting the value of acid dissociation constant (Ka) in the above equation,

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