assume that the triethylamine added is completely dissolved in water.

Question

Triethylamine (C₆H₁₅N) is an organic base which is useful in pesticide production and organic synthesis. It has a pKa of 10.75 and density of 726 kgm^-3 . It will partially dissociate in water following the chemical equilibrium below after dissolution in water.

C₆H₁₅N(aq) ⇌ OH^-/(aq) +C₆H₁₅NH⁺(aq)

(c) If 5.31mL of triethylamine (C₆H₁₅N) is added into pH9 sodium hydroxide solution (NaOH), please determine the resulting pH of the solution. You may assume that the triethylamine added is completely dissolved in water.

(d) If dilute sulfuric acid (H₂SO₄) is added into the triethylamine solution, please state how it will affect the ionization of triethylamine (C₆H₁₅N) in solution and provide some reasons behind your deduction.

Answer 1

(c) pH ≈ 11.6  |  (d) The pH falls, and the solution becomes a buffer.

(c) Setting it up

First convert the volume of liquid amine to moles — that is what the density is for:

ρ = 726 kg/m³ = 0.726 g/mL (those units are the same number; kg/m³ ÷ 1000 = g/mL)

m = (5.31 mL)(0.726 g/mL) = 3.855 g
M(C₆H₁₅N) = 101.19 g/mol
n = 3.855 / 101.19 = 0.0381 mol

⚠️ The volume of the NaOH solution is not stated in the part you posted — it was presumably given earlier in the question. I will work it as 1.00 L, which is the usual basis; if yours differs, substitute your volume at this step and everything downstream follows.

[C₆H₁₅N] = 0.0381 M

Convert pK_a to K_b

The 10.75 is the pK_a of the conjugate acid, C₆H₁₅NH⁺. Since you are working with the base:

pK_b = 14.00 − 10.75 = 3.25  →  K_b = 10^−3.25 = 5.62 × 10⁻⁴

ICE on the base equilibrium

C₆H₁₅N + H₂O ⇌ C₆H₁₅NH⁺ + OH⁻

K_b = x² / (0.0381 − x) = 5.62 × 10⁻⁴

K_b is not small enough here for the shortcut — x would be 11% of the initial concentration — so solve the quadratic:

x = [OH⁻] = 4.36 × 10⁻³ M

pOH = 2.36  →  pH = 14.00 − 2.36 = 11.64

The NaOH is a red herring

pH 9 means the starting [OH⁻] was only 10⁻⁵ M. The triethylamine generates 4.36 × 10⁻³ M — more than 400 times as much. Including the original hydroxide changes the answer by well under 1%, so it is entirely swamped. Recognising that a term is negligible, and saying why, is part of the answer.

(d) Adding dilute H₂SO₄

Sulfuric acid protonates the amine:

2 C₆H₁₅N + H₂SO₄ → 2 C₆H₁₅NH⁺ + SO₄²⁻

The pH decreases, but not steeply at first. Once both C₆H₁₅N and C₆H₁₅NH⁺ are present in appreciable amounts you have a buffer, and by Henderson–Hasselbalch:

pH = 10.75 + log([C₆H₁₅N] / [C₆H₁₅NH⁺])

So the pH drifts down through the region around 10.75 — reaching exactly 10.75 at the half-equivalence point, where the two forms are equal — and only crashes once essentially all the amine has been converted. That resistance in the middle of the titration is precisely what buffering means.