   Chapter 17, Problem 5PS

Chapter
Section
Textbook Problem

What is the pH of the solution that results from adding 30.0 mL of 0.015 M KOH to 50.0 mL of 0.015 M benzoic acid?

Interpretation Introduction

Interpretation:

The value of pH is to be calculated for the given mixture which contains 50mL0.015 M benzoic acid and 30mL 0.015 M potassium hydroxide solutions.

Concept introduction:

A buffer solution is defined as a solution which does not show any change in its pH values on the addition of small amount of acid or base. There are two types of buffer solutions,

(1) Acidic Buffer and (2) Basic Buffer

Acidic Buffer is a solution which has the pH value below 7. An acidic buffer is prepared by mixing a weak acid with its conjugate base. For example, a mixture of weak acid CH3COOH and its conjugate base CH3COO gives an acidic buffer.

CH3COOH(aq.)+ H2O(l)H3O+(aq.)+ CH3COO1(aq.)

Basic Buffer is a solution which has the pH value above 7. A basic buffer is prepared by mixing a weak base with its conjugate acid. For example, a mixture of weak base NH4OH and its conjugate acid NH4+ gives a basic buffer.

NH4OH(aq.)NH4+(aq.)+ OH1(aq.)

Explanation

The pH value for the resulting solution prepared by mixing 50mL0.015 M benzoic acid and 30mL 0.015 M potassium hydroxide is calculated below.

Given:

Refer to the table no. 16.2 in the textbook for the value of Ka.

The value of Ka for benzoic acid is 6.3×105.

The concentration of benzoic acid is 0.015 M.

The volume of benzoic acid is,

(50mL)(1L1000mL)=0.05L

The concentration of potassium hydroxide is 0.015 M.

The volume of potassium hydroxide solution is,

(30mL)(1L1000mL)=0.03L

Benzoic acid reacts with potassium hydroxide and the reaction involved is,

C6H5CO2H(aq)+KOH(aq)KC6H5CO2(aq)+H2O(l)

The initial number of moles of reactants benzoic acid and potassium hydroxide is calculated by using the expression,

Numberof moles=(concentration)(volume) (1)

For benzoic acid,

Substitute 0.015 M for concentration and 0.05L for volume in equation (1).

Numberof moles=(0.015 M)(0.05 L)=7.5×104mol

For potassium hydroxide,

Substitute 0.015 M for concentration and 0.03 L for volume for potassium hydroxide,

No.of moles=(0.015 M)(0.03L)=4×104mol

The ICE table (1) is as follows,

EquationC6H5CO2H(aq)+OH(aq)C6H5CO2(aq)+H2O(l)Initial(mol)7.5×1044.5×1040Change(mol)(4.5×104)(4.5×104)+(4.5×104)Afterreaction(mol)3.0×10404.5×104

After the reaction, in reaction mixture only benzoic acid and benzoate ion will be left as potassium hydroxide is limiting reagent and the total volume of the solution is equal to 0.08 L.

Therefore the remaining moles of benzoic acid will be dissolved in 0.08L and it undergoes dissociation as follows,

C6H5CO2H(aq)+H2O(aq)H3O+(aq)+C6H5COO(aq)

The concentration of benzoic acid and benzoate ion after the reaction is calculated using the following expression,

Numberof moles=(concentration)(volume)

Rearrange it for concentration.

concentration = Numberof molesvolume(molL1) (2)

For benzoic acid,

Substitute 3.0×104 for number of moles and 0.08L for volume in equation (2).

concentration=3.0×104 mol0.08L=3.75×103molL1=3.75×103M

For benzoate ion,

Substitute 4.5×104 for number of moles and 0.08L for volume in equation (2).

concentration=4.5×104 mol0.08L=5.625×103molL1=5.625×103M

There will be equilibrium between benzoic acid and potassium benzoate which will form a buffer solution. That can be represented in ICE table (2) as follows,

EquationC6H5CO2H(aq)+H2O(aq)H3O+(aq)+C6H5COO(aq)Initial(M)3

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