   Chapter 16, Problem 91GQ

Chapter
Section
Textbook Problem

A monoprotic acid HX has Ka = 1.3 × 10−3. Calculate the equilibrium concentration of HX and H3O+ and the pH for a 0.010M solution of the acid.

Interpretation Introduction

Interpretation:

The equilibrium concentration of HX and H3O+ has to be calculated.

Concept introduction:

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

The equilibrium reaction of HX with water is as follows.

HX + H2O(aq) H3O+(aq) + X-(aq)

The equilibrium expression:

Ka[X-][H3O+][HX]

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

HX          +   H2O (aq)   H3O+(aq)+ X-(aq)I     0.010M          --                      --               --C       -x                --                    + x             + xE   (0.010-x)         --                      x                x

The Ka of the HX is 1.3×10-3

Substitute the values form the ICE table.

1.3×10-3 = (x)(x)(0.010- x)1.3×10-3 = (x)2(0.010- x)1.3×10-3 = (x)2(0

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