Question

Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH₃CH₂)₃N (K_b = 5.2 ×10⁻⁴), with 0.1000 M HCl solution after the following additions of titrant.

(a) 11.00 mL:

pH =

(b) 20.60 mL:

pH =

(c) 26.00 mL:

pH =

Answer 1

Moles of triethylamine = 20.00 x 0.1000 = 2 mol

K_b of triethylamine K_b= 5.2 x 10^-4

K_a of triethylamine hydrochloride = K_w/K_b = 10^-14/5.2 x 10^-4 = 1.92 x 10^-11 (K_w = autoprotolysis constant of water)

pK_a = - logK_a = - log (1.92 x 10^-11) = 10.72

(a)

11.00 mL of 0.1000 M HCl = 11 x 0.1000 = 1.1 mol

1.1 mol HCl will react with 1.1 mol triethylamine to form 1.1 mol triethylamine hydrochloride

Remaining moles of triethylamine = (2.0 - 1.1) = 0.9 mol

Total volume of the solution = (20.00 + 11.00) = 31.00 mL

New concentration of triethylamine = 0.9/31.00 = 0.029 M

New concentration of triethylamine hydrochloride = 1.1/31.00 = 0.035 M

From Henderson-Hasselbalch equation

pH = pKa + log[base]/[acid]

      = 10.72 + log 0.029/0.035

      = 10.72 - 0.08

      = 10.64

-------------------------------------------

(b)

20.60 mL of 0.1000 M HCl = 20.60 x 0.1000 = 2.06 mol

Now,

2.0 mol triethylamine will react with 2.0 mol HCl to form 2.0 mol triethylamine hydrochloride

Remaining moles of HCl = (2.06 - 2.0) = 0.06 mol

Total volume of the solution = (20.00 + 20.60) = 40.60 mL

New concentration of HCl = 0.06/ 40.60 = 0.00148 M

Now this 0.00148 M HCl will completely dissociate in water.

pH = - log[H⁺]

      = - log0.00148

      = 2.83

----------------------------------------------------

(c)

26.00 mL of 0.1000 M HCl = 26.00 x 0.1000 = 2.6 mol

Now,

2.0 mol triethylamine will react with 2.0 mol HCl to form 2.0 mol triethylamine hydrochloride

Remaining moles of HCl = (2.6 - 2.0) = 0.6 mol

Total volume of the solution = (20.00 + 26.00) = 46.00 mL

New concentration of HCl = 0.6/ 46.00 = 0.01304 M

Now this 0.01304 M HCl will completely dissociate in water.

pH = - log[H⁺]

      = - log0.01304

      = 1.88