Consider the titration of 100.0 mL of 0.280 M benzoic acid (Ka = 6.4 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the resulting solution after each of the following volumes of NaOH has been added.

Question

Consider the titration of 100.0 mL of 0.280 M benzoic acid (K_a = 6.4 ✕ 10⁻⁵) with 0.140 M NaOH. Calculate the pH of the resulting solution after each of the following volumes of NaOH has been added. (Assume that all solutions are at 25°C.)

(a)    0.0 mL

(b)    50.0 mL

(c)    100.0 mL

(d)    150.0 mL

(e)    200.0 mL

(f)    250.0 mL

Answer 1

I'll do a few and assume that you'll be able to do the rest.

First note that you are dealing with the titration of a weak acid (benzoic acid) with a strong base (NaOH). This will result in the formation of sodium benzoate (the conjugate base) and water. Because you will form the conjugate base, you are forming a BUFFER until you reach equivalence. After equivalence, you will then have excess NaOH.

For convenience, I'll use HA as benzoic acid and A- as the conjugate base.

a) 0 ml NaOH added:

HA ==> H⁺ + A^- and Ka = 6.4x10^-5 = [H+][A-]/[HA] = (x)(x)/0.28

x² = 1.79x10^-5

x = [H+] = 4.23x10^-3

pH = -log 4.23x10-3 = 2.37

b) 50 ml 0.140 M NaOH added:

moles HA = 0.1 L x 0.28 mol/L = 0.028 moles HA

moles NaOH = 0.050 L x 0.140 mol/L = 0.007 moles NaOH

Final volume = 150 ml = 0.150 L

HA + OH- ===> A- + H₂O

0.028....0.007........0..............Initial

-0.007....-0.007.....+0.007.....Change

0.021......0.............0.007......Equilibrium

Final [ HA] = 0.21 mol/0.150 L = 0.14 M

Final [A-] = 0.007 mol/0.150 L = 0.047 M

pH = pKa + log [0.047/0.114] = 4.19 + (-0.38)

pH = 3.81

c) 100 ml 0.140 M NaOH added:

follow procedure in (b) above.

d) 150 ml of 0.140 M NaOH added:

follow procedure in (b) above.

e) 200 ml of 0.140 M NaOH (note that this is 2x the volume of HA but the concentration of NaOH is 1/2 [HA].

moles HA = 0.028

moles NaOH = 0.2 L x 0.140 mol/L = 0.028

Since moles HA = moles NaOH, you have reached equivalence and pH = pKa (plug values into HH equation if you need proof of this fact).

pH = pKa = 4.19

f) 250 ml of 0.140 M NaOH added (note that since equivalence was reached at addition of 200 ml NaOH, the [OH-] will be in excess in this addition:

moles HA = 0.028

moles NaOH = 0.250 L x 0.140 mol/L = 0.035

moles OH- in excess = 0.035 - 0.028 = 0.007 moles OH-

Final volume = 250 + 100 = 350 ml = 0.350 L

[OH-] = 0.007 mol/0.350 L = 0.02 M

pOH = 1.70

pH = 14 - 1.7 = 12.3