Question

“When a strong base is gradually added dropwise to a weak acid, the pH changes at each addition. When the appropriate quantity of base has been added to react with all of the acid, the pH changes sharply, indicating the endpoint of the titration. A plot of pH versus volume of base added gives what is known as a titration curve”.

Consider the titration of 25.00 mL of 0.1000 M benzoic acid with 0.1000 M NaOH,.

Write a balanced chemical equation for the titration reaction using C6H5COOH for benzoic acid’s formula.

Determine the volume of NaOH solution required to reach the endpoint.

The chart on the next page has entries for several steps along the titration curve. To calculate the pH at each step, you first must understand the chemistry at that step, and then you can decide the appropriate method to calculate the pH. For each volume listed, (i) list the major species in solution, (ii) determine whether the Ka equation, Kb equation, buffer equation, or solution equilibrium equation is appropriate for the calculation of the solution [H3O+] and pH, and (iii) complete the calculations.

Volume NaOH Added (mL)	Major Species	Appropriate Equation	[H3O+]	pH
0.00
5.00
12.50
20.00
25.00
30.00

Answer 1

Titration

Balanced equation,

C6H5COOH + NaOH ---> C6H5COONa + H2O

Volume of NaOH required to reach end point = 0.1 M x 25 ml/0.1 ml = 25 ml

Table

NaOH (ml)             Major species                         [H3O+]                   pH

     0                        C6H5COOH                       2.55 x 10^-3              2.59

     5                        C6H5COOH                       2.58 x 10^-4              3.59

  12.5             C6H5COOH,C6H5COONa          6.5 x 10^-5               4.19

    20                      C6H5COONa                       1.63 x 10^-5           4.79

    25                      C6H5COONa                        3.6 x 10^-9              8.44

    30                         OH-                                   1.1 x 10^-12           11.96

Appropriate equations,

at the start :

C6H5COOH + H2O <==> C6H5COO- + H3O+

when 5 ml NaOH added

C6H5COOH + NaOH <==> C6H5COONa + H2O

Used Hendersen-Hasselbalck equation for pH calculation,

pH = pKa + log(base/acid)

when 12.5 ml NaOH added

C6H5COOH + NaOH <==> C6H5COONa + H2O

Used Hendersen-Hasselbalck equation for pH calculation,

pH = pKa + log(base/acid)

here, [C6H5COOH] left = [C6H5COO-] formed

when 20 ml NaOH added

C6H5COOH + NaOH <==> C6H5COONa + H2O

Used Hendersen-Hasselbalck equation for pH calculation,

pH = pKa + log(base/acid)

when 25 ml NaOH added

C6H5COOH + NaOH <==> C6H5COONa + H2O

C6H5COO- + H2O <==> C6H5COOH + OH-

this is equivalence point

when 30 ml NaOH added

OH- in excess. Solution basic.