Question

Consider the titration of 20.00 ml of 0.0200 M sodium benzoate, C₆H₅COO^-Na⁺, with 0.0125 M HCl.

Write the chemical reaction occurring in this titration as a net ionic equation.

Does the reaction go to completion? Show the relevant calculation.

What is the equivalence point of this titration?

Calculate the pH of the initial point, before addition of HCl (V_HCl = 0.00 mL).

Calculate the pH after the addition of 10.00 mL of HCl (V_HCl = 10.00 mL).

Calculate the pH after the addition of 16.00 mL of HCl (V_HCl = 16.00 mL).

Calculate the pH after the addition of 20.00 mL of HCl (V_HCl = 20.00 mL).

Calculate the pH after the addition of 32.00 mL of HCl (V_HCl = 32.00 mL).

Calculate the pH after the addition of 40.00 mL of HCl (V_HCl = 40.00 mL).

Answer 1

The reaction is

C6H5COONa  + HCl --------> C6H5COOH + NaCl

C6H5COO-  + H+ --------> C6H5COOH   is the net ionic reaction

1) At VHCl = 0 , the solution has only salt of a weak acid and strong base.

Its pH is given as

pH = 1/2 [pKw +pKa + log C]

      = 1/2 [14 + 4.17+ log 0.02]

     = 8.23

2) When V = 10mL

            C6H5COO-                        + H+ -------->             C6H5COOH  

     20x 0.02 = 0.4                       0.0125x10=0.125                  0              initial

     0.275                                           0                                 0.125 after

So it forms a buffer whose pH is given by Hendersen equation as

   ph = pKa + log [conjugate base]/ [acid]

         = 4.17 + log 0.275/0.125

         = 4.51

3) After 16mL added

         C6H5COO-                        + H+ -------->             C6H5COOH  

     20x 0.02 = 0.4                       0.0125x16=0.2                  0              initial

     0.2                                          0                                 0.2 after

So it forms a buffer whose pH is given by Hendersen equation as

   ph = pKa + log [conjugate base]/ [acid]

         = 4.17 + log 0.2/0.2

        = 4.17

4) after addition of 20mL

        C6H5COO-                        + H+ -------->             C6H5COOH  

     20x 0.02 = 0.4                       0.0125x20=0.25                 0              initial

     0.15                                           0                                 0.25 after

So it forms a buffer whose pH is given by Hendersen equation as

   ph = pKa + log [conjugate base]/ [acid]

         = 4.17 + log 0.15/0.25

         = 3.48

5) After 32 mL

        C6H5COO-                        + H+ -------->             C6H5COOH  

     20x 0.02 = 0.4                       0.0125x32= 0.4               0              initial

     0                                           0                                 0.4 after

So the solution now has only weak acid and this the equivalence poiint of titration and the reaction is complete,.

[acid] = mmoles / total volume

          = 0.4/ (20+32)

          =0.00769

thus pH = 1/2 [ pKa -logC]

            = 1/2 [4.17 - log 0.00769]

            = 3.14

6) after 40mL

        C6H5COO-                        + H+ -------->             C6H5COOH  

     20x 0.02 = 0.4                       0.0125x40=0.5              0              initial

           0                                           0.1                              0.4 after

the solution containsmixture   of strong acid and weak acid

So the ph is decided only strong acid

[strong acid] = 0.1/60 = 0.00167 M

pH of solution = - log 0.00167

                      = 2.777