Question

Determine the pH during the titration of 62.1 mL of 0.445 M benzoic acid (K_a = 6.3×10^-5) by 0.445 M NaOH at the following points. (Assume the titration is done at 25 °C.)

(a) Before the addition of any NaOH

(b) After the addition of 15.0 mL of NaOH

(c) At the half-equivalence point (the titration midpoint)

(d) At the equivalence point

(e) After the addition of 93.2 mL of NaOH

Answer 1

The corresponding reaction is:

C6H5COOH + NaOH ==> H2O + C6H5COONa

Since they react in a 1:1 mole ratio,

(a) before addition of any NaOH :

We have 0.445 M C6H5COOH . For weak acids,

[H3O+] = (Ka x [HA]o)^0.5
[H3O+] = ((6.3 x 10^-5)(0.445))^0.5 = 5.29 x 10^-3 M

pH = -log[H3O+]

= -log (5.29 x 10^-3)

= 2.27

b) after addition of 15 mL of 0.445 M NaOH :

Since we have a weak acid (C6H5COOH ) and its conjugate base (C6H5COO- in the form of C6H5COONa) in the same solution, which constitutes a buffer system. So, using the Henderson-Hasselbalch equation for buffers,

pH = pKa + log ([conj. base] / [weak acid])

pH = pKa + log ([KClO] / [HClO])

Where, pKa = -log Ka = -log (6.3 x 10^-5) = 4.20

And after adding 15 mL of the NaOH, we are at the 24% mark of completion of the titration. So the ratio of [KClO] to [HClO] is

24 / 76.

So, putting the value of pKa in above equation,

pH = 4.20 + log (24 / 76) = 3.69

c) at the half equivalence point :

After adding 31.05 mL of the KOH, we are at the 50% mark of completion of the titration. So the ratio of [C6H5COONa] to [C6H5COOH ] is 50/50.

pH = 4.20 + log (50/50) = 4.20

d) at the equivalence point :

This is the equivalence point where you have reacted all of the C6H5COOH and converted it to C6H5COONa.

Initial mmoles C6H5COOH = Molarity C6H5COOH x mL C6H5COOH

= (0.445)(62.1) = 27.63 mmoles HClO = mmol of C6H5COONa

[C6H5COONa] at equivalence point = mmoles C6H5COONa / mL of solution = 27.63 / 124.2

= 0.22 M

Since C6H5COONa is made using weak acid and strong base, the resulting C6H5COONa solution will be basic (i.e. C6H5COO- on will form C6H5COOH and OH- in aqueous solution)

For solution of weak bases,

[OH-] = (Kb x [B]o)^0.5

Where, Kb = Kw / Ka

= (1 x 10^-14) / (6.3 x 10^-5)

= 1.58 x 10^-10

Putting this value in above equation :

[OH-] = ((1.58 x 10^-10)(0.22 M))^0.5

= 5.89 x 10^-6

pOH = -log [OH-]

= -log (5.89 x 10^-6)

= 5.22

pH = 14.00 - pOH

= 14.00 - 5.22

= 8.78

e) after addition of 93.2 mL of KOH :

Mol of OH- in extra 31.1 mL = 0.0311 L X 0.445 mol/L

= 0.0138 mol OH- to the solution. So,

[OH-] = 0.0138 mol / 0.1553 L = 0.088 M
pOH = - log [OH-]

pOH = - log (0.088)

pOH = 1.05
pH = 14.00 - 1.05

pH = 12.95