   Chapter 14, Problem 76E

Chapter
Section
Textbook Problem

# An acid HX is 25% dissociated in water. If the equilibrium concentration of HX is 0.30 M, calculate the Ka value for HX.

Interpretation Introduction

Interpretation: The Ka value for the given solution of HX is to be calculated.

Concept introduction: At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

The percent dissociation of an acid is calculated by the formula,

Percentdissociation=Equilibriumconcentrationof[H+]Initialconcentrationoftheacid×100

Explanation

Explanation

To determine: The Ka value for the given solution of HX .

The equilibrium constant expression for the given reaction is, Ka=[H+][X][HX]

The weak acid is assumed to be HA .

The dominant equilibrium reaction for the given case is,

HX(aq)H+(aq)+X(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+][X][HX] (1)

The initial concentration of the given acid is (0.30+x)M_ .

The initial concentration of HX is assumed to be C .

The change in concentration of HX is assumed to be x .

The ICE table for the stated reaction is,

HX(aq)H+(aq)+X(aq)InititialconcentrationC00Changex+x+xEquilibriumconcentrationCxxx

The equilibrium concentration of [HX] is (Cx)M .

The equilibrium concentration of [H+] is xM .

The equilibrium concentration of [X] is xM .

The equilibrium concentration of the acid is given to be 0.30M .

According to the ICE table formed,

The equilibrium concentration of the acid =Cx

Substitute the given value of the equilibrium concentration of the acid in the above expression.

0.30=CxC=(0.30+x)M_

The amount of the acid that dissociated is 0.10M_ .

Given

The percent dissociation is 25%

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