Phenol, C₆H₅OH, is a weak organic acid. Suppose 0.515 g of the compound is dissolved in enough water to make 125 mL of solution. The resulting solution is titrated with 0.123 M NaOH.

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

1. (a) What is the pH of the original solution of phenol?
2. (b) What are the concentrations of all of the following ions at the equivalence point: Na⁺, H₃O⁺, OH⁻, and C₆H₅O⁻?
3. (c) What is the pH of the solution at the equivalence point?

Interpretation:

The value of pH has to be calculated at the various points during the titration between C6H5OH and NaOH. The value of pH, of the original solution of C6H5OH has to be calculated.

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 acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C6H5OH with NaOH is represented as,

C6H5OH(aq)+ OH−(aq)⇌H2O(l)+C6H5ONa(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A−](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5OH/C6H5O−. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugate base][acid] (2)

At the midpoint of the titration, when concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugate base][acid]

Substitute, [conjugate base] for[acid].

pH=pKa+log[conjugate base][conjugate base]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OH− ion and C6H5O−. The OH− will be produced due to the hydrolysis of phenoxide ion at equivalence point. The hydrolysis equilibrium is represented as,

C6H5O−(aq)+H2O(l)⇌OH−(aq)+C6H5OH(aq)

By using the value of Kb for the phenoxide ion, concentration of  OH− can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A−](eq)[HA](eq)

Given:

Refer to the Apendix H in the textbook for the value of Ka.

The value of Ka for phenol is 1.3×10−10.

The value of Kw for water is 1.0×10−14.

The pKa value is calculated as follows;

pKa=−log(Ka)

Substitute, 1.3×10−10 for Ka.

pKa=−log(1.3×10−10)=9.88

Therefore, pKa value of phenol is 9.88.

The amount of C6H5OH dissolved is 0.515 g.

Molar mass of phnol is 94 g⋅mol−1.

Number of moles of phenol are calculated by using following expression,

Number of moles = weight of solute molar mass of solute(mol)

Substitute 0.515 g for weight of solute , 94 g⋅mol−1 for molar mass of solute.

Number of moles = 0.515 g94 g⋅mol−1=0.00547 mol

Therefore Number of moles of phenol are 0.00547 mol.

Volume of the solvent is 125 mL.

Unit conversion of 125 mL into L.

(125 mL)(1 L1000 mL)=0.125 L

Therefore volume of the solvent is 0.125 L.

The initial concentration of C6H5OH is calculated by using the expression,

Molarity=Number of molesvolume of the solvent(mol⋅L−1)

Substitute 0.00547 mol for Number of moles, 0.125 L for volume of the solvent.Molarity=0.00547 mol0.125 L=0.0438 mol⋅L−1

Therefore initial concentration of C6H5OH is 0.0438 mol⋅L−1.

ICE table (1) gives the dissociation of C6H5OH.

EquationC6H5OH(aq)+H2O(l)⇌H3O+(aq)+C6H5O−(aq)Initial(mol⋅L−1)0

Interpretation:

The value of pH has to be calculated at the various points during the titration between C6H5OH and NaOH. The concentration of OH−,H3O+,C6H5O− and Na+ has to be calculated at equivalence point.

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 acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C6H5OH with NaOH is represented as,

C6H5OH(aq)+ OH−(aq)⇌H2O(l)+C6H5ONa(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A−](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5OH/C6H5O−. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugate base][acid] (2)

At the midpoint of the titration, concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugate base][acid]

Substitute, [conjugate base] for[acid].

pH=pKa+log[conjugate base][conjugate base]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OH− ion and C6H5O−. The OH− will be produced due to the hydrolysis of phenoxide ion at equivalence point. The hydrolysis equilibrium is represented as,

C6H5O−(aq)+H2O(l)⇌OH−(aq)+C6H5OH(aq)

By using the value of Kb for the phenoxide ion, concentration of  OH− can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

Interpretation:

The value of pH has to be calculated at the various points during the titration between C6H5OH and NaOH. The value of pH at equivalence point has to be calculated.

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 acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C6H5OH with NaOH is represented as,

C6H5OH(aq)+ OH−(aq)⇌H2O(l)+C6H5ONa(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A−](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5OH/C6H5O−. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugate base][acid] (2)

At the midpoint of the titration, concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugate base][acid]

Substitute, [conjugate base] for[acid].

pH=pKa+log[conjugate base][conjugate base]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OH− ion and C6H5O−. The OH− will be produced due to the hydrolysis of phenoxide ion at equivalence point. The hydrolysis equilibrium is represented as,

C6H5O−(aq)+H2O(l)⇌OH−(aq)+C6H5OH(aq)

By using the value of Kb for the phenoxide ion, concentration of  OH− can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)