Calculate [OH − ], [H+], and the pH of a 0.12 M solution of each of the following amines

Answer 1

In order to calculate the [OH-], [H+], and pH of a solution of an amine, we need to consider the basicity of the amine.

Amines are organic compounds that contain a nitrogen atom with a lone pair of electrons. Due to the presence of this lone pair, amines can act as bases and accept protons (H+) from water molecules, forming OH- ions.

Let's consider the three different types of amines separately:

1. Primary amine (RNH2):
- A primary amine has one alkyl or aryl group attached to the nitrogen atom.
- When it reacts with water, it accepts a proton (H+) from water to form OH-.
- The reaction can be represented as: RNH2 + H2O → RNH3+ + OH-

2. Secondary amine (R2NH):
- A secondary amine has two alkyl or aryl groups attached to the nitrogen atom.
- When it reacts with water, it also accepts a proton (H+) from water to form OH-.
- The reaction can be represented as: R2NH + H2O → R2NH2+ + OH-

3. Tertiary amine (R3N):
- A tertiary amine has three alkyl or aryl groups attached to the nitrogen atom.
- Tertiary amines do not have an available hydrogen atom to react with water and form OH-. Therefore, the [OH-] concentration will be very low or negligible.
- The concentration of [OH-] can be considered as approximately 0.

For a 0.12 M solution of each amine, we can calculate the [OH-] and [H+] concentrations using the given reactions.

1. Primary amine (RNH2):
- [OH-] concentration: Since the primary amine reacts with water to form OH-, the [OH-] concentration will be equal to the initial concentration of the amine, which is 0.12 M.
- [H+] concentration: [H+] = Kw/[OH-] = 1.0 x 10^-14 / 0.12 = 8.33 x 10^-14 M (using the ion product of water, Kw = 1.0 x 10^-14 at 25°C).
- pH: pH = -log[H+] = -log(8.33 x 10^-14) ≈ 13.08

2. Secondary amine (R2NH):
- [OH-] concentration: Similar to primary amines, secondary amines react with water to form OH-. Thus, [OH-] will be equal to the initial concentration of the amine, which is 0.12 M.
- [H+] concentration: [H+] = Kw/[OH-] = 1.0 x 10^-14 / 0.12 = 8.33 x 10^-14 M.
- pH: pH = -log[H+] = -log(8.33 x 10^-14) ≈ 13.08

3. Tertiary amine (R3N):
- [OH-] concentration: The [OH-] concentration for tertiary amines is very low or negligible, and can be considered as approximately 0.
- [H+] concentration: [H+] cannot be directly determined as [OH-] is negligible.
- pH: pH cannot be directly determined as [H+] is unknown.

Therefore, for a 0.12 M solution of each amine:
- Primary amine (RNH2): [OH-] = 0.12 M, [H+] = 8.33 x 10^-14 M, pH ≈ 13.08
- Secondary amine (R2NH): [OH-] = 0.12 M, [H+] = 8.33 x 10^-14 M, pH ≈ 13.08
- Tertiary amine (R3N): [OH-] ≈ 0, [H+] cannot be determined, pH cannot be determined