   Chapter 14, Problem 179CP

Chapter
Section
Textbook Problem

# Consider 1000. mL of a 1.00 × 10−4-M solution of a certain acid HA that has a Ka value equal to 1.00 × 10−4. How much water was added or removed (by evaporation) so that a solution remains in which 25.0% of HA is dissociated at equilibrium? Assume that HA is nonvolatile.

Interpretation Introduction

Interpretation: The value of Ka , volume and concentration of certain acid HA is given. The volume of water added or removed is to be calculated for the given condition.

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

The number of moles is calculated using the formula,

Numberofmoles=Volume×Concentration

Explanation

Explanation

To determine: The volume of water added or evaporated.

The total number of moles of acid is 1.0×104mol_ .

Given

Volume of HA is 1000mL .

The concentration of HA is 1.0×104M .

The value of Ka is 1.0×104 .

Formula

The number of moles is calculated using the formula,

Numberofmoles=Volume×Concentration

Substitute the values of volume and concentration of HA in the above equation.

Numberofmoles=Volume×Concentration=1.0×104mol/L×1.0L=1.0×104mol_

The number of moles of H+ is 2.5×105mol_ .

The number of moles of A is 2.5×105mol_ .

The number of moles of HA is 7.50×105mol_ .

The reaction of dissociation of HA is,

HAH++A

It is given that 25% of acid is dissociated. It means that 25% of product is formed and 75% of acid is left.

The number of moles of reactant and product is calculated using the formula,

Numberofmoles=Amountofreactantorproduct×Totalmoles (1)

Substitute the percent of production of H+ and total calculated moles in the above equation.

NumberofmolesofH+=Amountofreactantorproduct×Totalmoles=25100×1.0×104mol=2.5×105mol_

Substitute the percent of production of A and total calculated moles in the equation (1).

NumberofmolesofA=Amountofreactantorproduct×Totalmoles=25100×1.0×104mol=2.5×105mol_

Substitute the percent of production of HA and total calculated moles in the equation (1).

NumberofmolesofHA=Amountofreactantorproduct×Totalmoles=75100×1.0×104mol=7

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