Question

Calculate the pH during the titration of 25.0 ml of 0.150 M benzoic acid with 0.150 M NaOH after the following additions of titrant: 15.0 ml; 25.0 ml; 40.0 ml.

Answer 1

Ka of benzoic acid = 6.3 x 10–5

pKa = -logKa = -log (6.3 x 10–5)

       = 4.2

millimoles of benzoic acid = 25.0 x0.15 = 3.75

a) 15 ml NaOH

NaOH millimoles = 15 x 0.15= 2.25

C₆H₅COOH + NaOH ----------------->C₆H₅COONa + H2O

3.75                   2.25                              0                    0 --------------------initial

1.5                       0                                 2.25               2.25 ---------------after reaction

in the solution acid and salt remained so it can form buffer

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

    = 4.2 + log (2.25/1.5)

pH= 4.37

b) 25ml NaOH

NaOH millimoles = 25 x0.15 = 3.75

C₆H₅COOH + NaOH -----------------> C₆H₅COONa + H2O

3.75                  3.75                             0                      0 --------------------initial

0                          0                                3.75               3.75 ---------------after reaction

in the solution salt remained so we have to do salt hydrolysis

it is the salt of strong base and weak acid so pH should be more than 7

[salt] = salt millimoles /total volume in ml

           = 3.75/(25+25)

           = 0.075 M

pH = 7 + 1/2[Pka + logC]

      = 7 + 1/2[4.2 + log (0.075)]

pH = 8.54

c) 40ml NaOH

NaOH millimoles = 40 x 0.15= 6

C₆H₅COOH + NaOH -----------------> C₆H₅COONa + H2O

3.75                   6                              0                 0 --------------------initial

0                     2.25                                   3.75            3.75 ---------------after react

solution only contain strong base NaOH so pH decide by base only

[NaOH] = milli moles of NaOH / total volume

             = 2.25 / (25 + 40)

             = 0.035

pOH = -log[OH-]

       = -log [0.035]

       = 1.45

pH + pOH = 14

pH = 14 - pOH

     = 12.54

pH= 12.54