   # What is the pH at the equivalence point in the titration of 25.0 mL of 0.090 M phenol ( K a = 1.3 × 10 −10 ) with 0.108 M NaOH? (a) 7.00 (b) 5.65 (c) 8.98 (d) 11.29 ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17.3, Problem 2RC
Textbook Problem
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## What is the pH at the equivalence point in the titration of 25.0 mL of 0.090 M phenol (Ka = 1.3 × 10−10) with 0.108 M NaOH? (a) 7.00 (b) 5.65 (c) 8.98 (d) 11.29

Interpretation Introduction

Interpretation:

The value of pH is to be calculated at a point when 25.0mL,  0.090 M phenol is titrated against 0.108MNaOH solution the (Ka) value was given in the statement.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, HCl is titrated against NaOH .

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, HCl is titrated against NH4OH .

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, CH3COOH is titrated against NaOH .

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, CH3COOH is titrated against NH4OH .

For weak base-strong acid titration the pH value can be calculated at various points before and after equivalence point.

The equilibrium established during the titration of NH3 with HCl . The equilibrium can be represented as,

NH3(aq)+ H3O+(aq)H2O(l)+NH4+(aq)

Here, the H3O+ ion are generated from HCl , and the reaction is written as,

HCl(aq)+H2O(l)H3O+(aq)+Cl(aq)

At equivalence point all the base will be neutralized, and there will be only NH4+ ion and H3O+ . The H3O+ will be produced due to the hydrolysis of NH4+ at equivalence point. The hydrolysis equilibrium is represented as,

NH4+(aq)+H2O(l) H3O+(aq)+NH3(aq)

By using the value of Ka for the NH4+ concentration of H3O+ can be calculated.

Ka=[H3O+][H2O][NH4+] (1)

As H3O+ ions are produced due to hydrolysis of NH4+ , the value of pH will be lower than 7 at equivalence point for the weak base-strong acid titrations.

### Explanation of Solution

The pH calculation just before the equivalence point is done by using Henderson-Hasselbalch equation.

As the addition of NaOH is done there will be formation of buffer solution phenol and phenolate ions. The pOH calculation for buffer solution is done by using Henderson-Hesselbalch equation.

pOH=pKb+log[conjugateacid][base] (2)

The pH value will be calculated by using expression, pH + pOH = 14 .

Given:

Refer to table 16.2 in the textbook for the value of Ka .

Hence the net reaction bretween week acid and strong base is as fallows

C6H5OH(aq)+OH(aq)C6H5O(aq)+H2O(l)

The correcsponding reaction tells that one mole of base reacts, with one mole of phenol to produce (1 mole of ) C6H5O ions

Next we calculate the moles of phenol molecule

ThemoleC6H5OH=25.0mL×0.090M=2.50mmolmoleC6H5OH=(2.50mmol)TheequivelentpointmolC6H5OH=molOHThehydroxylions(OH)calculationmethosfallowas2.50mmol×1mL0.108mmol=20.8mL

The equvivelent pont 2.25 mmol ofphenol is converted into 2.25 mmol of C6H5O ions

So,

[C6H5O]=2.25mmol(20.8+25.0)mL=2.25mmol45.8mL=0.0491M[C6H5O]=0.0491MThepHdeterminedbyC6H5OH(aq)+OH(aq)C6H5O(aq)+H2O(l)

The equilibrium constant Kb for the given above reaction is Kw/Ka

Hence Kb

Kb=KwKaKw=1×1014Ka=1

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