Calculate the pH at the halfway point and at the equivalence point for each of the following:

Question

Calculate the pH at the halfway point and at the equivalence point for each of the following:

a) 100.0 mL of 0.10 M HC₇H₅O₂ (K_a = 6.4 x 10^-5) titrated by 0.10 M NaOH

b) 11.0 mL of 0.10 C₂H₅NH₂ (K_b - 5.6 x 10^-4) titrated by 0.20 M HNO₃

I am lost with this process so please explain as simply as possible.

The answers are:

a) 4.19 , 8.45

b) 10.74 , 5.96

Please list the steps so I can follow along. Thank you!

Answer 1

This is an example of a titration of a weak acid (HC₇H₅O₂), with a strong base NaOH in part (a), and the titration of a weak base (C₂H₅NH₂) with a strong acid HNO₃ in part (b). In BOTH situations, you are forming a buffer solution. Anytime you have this situation, the pH or pOH at 1/2 the equivalence point is equal to the pKa or pKb, respectively. For part (a) pKa = -log 6.4x10^-5 = 4.19 = pH at 1/2 equivalence. For part (b) pKb = -log 5.6x10^-4 = 3.25 = pOH and pH = 14 - 3.25 = 10.75 at 1/2 equivalence

I'll work it out the long way, but this simple fact will save you time in the future.

I will only do part (a) at this time. It will be too long to do both parts. If you want me to address part (b), either message me or post part (b) separately. Or maybe another tutor will answer part (b).

Part (a): HC₇H₅O₂ + OH^- ===> C₇H₅O₂^- + H2O

Instead of typing HC₇H₅O₂ for the acid, I will simply call it HA. We then have

HA + OH^- ==> A^- + H2O

Initial moles HA = 100.0 ml x 1 L/1000 ml x 0.10 mol/L = 0.01 moles HA

Initial moles OH- to reach 1/2 equivalence = 1/2 x 0.01 moles = 0.005 moles (= 50 ml of 0.1 M NaOH)

At this point of 1/2 equivalence we now have 0.005 moles HA and 0.005 moles A-. Looking at an ICE table...

HA + OH- ===> A- + H2O

0.01....0.005............0..........0.........Initial

-0.005...-0.005......+0.005.............Change

0.005.......0.............0.005..............Equilibrium

Final volume = 100 ml HA + 50 ml OH- = 150 ml = 0.150 L

Final [HA] = 0.005 mol/0.15 L = 0.0333 M

Final [A-] = 0.005 mol/0.15 L = 0.0333 M

Using the Henderson Hasselbalch equation (where pKa = -log Ka = -log 6.4x10^-5 = 4.19)

pH = pKa + log [A-]/[HA] = 4.19 + log [0.0333]/[0.0333] = 4.19 + 0

pH = 4.19

At full equivalence point, all the HA is neutralized (none left) and it has all been converted to A-. In this case that will be 0.01 moles HA ==> 0.01 moles A^- in a final volume of 200 ml (0.2 L). The 0.2 L comes from 100 ml of HA + 100 ml of NaOH.

Final [A-] = 0.01 mol/0.2 L = 0.05 M

Now, we look at the hydrolysis of A^- as follows:

A^- + H2O ==> HA + OH^- (note that A- acts as a base in this case, so we need to use Kb)

KaKb = Kw = 1x10^-14

Kb = 1x10^-14/6.4x10^-5

Kb = 1.56x10^-10

Kb = 1.56x10^-10 = [HA][OH^-]/[HA] = (x)(x)/0.05

x² = 7.8x10^-12

x = 2.79x10^-6 = [OH^-]

pOH = -log 2.79x10^-6 = 5.55

pH = 14 - 5.55 = 8.45