The base-dissociation constant of ethylamine (C₂H₅NH₂)is 6.4 × 10^-4 at 25.0°C.The [H⁺] in a solution of ethylamine is ________ M. A) 3.5 × 10^-12 B) 2.9 × 10^-3 C) 3.1 × 10^-12 D) 3.2 × 10^-3 E) 11.46

Question

Quizplus · Accepted Answer

The base-dissociation constant, Kb, is defined as:

Kb = ([BH+][OH-])/[B]

where [BH+] is the concentration of the conjugate acid, [OH-] is the concentration of hydroxide ions, and [B] is the concentration of the base.

For ethylamine:

C2H5NH2 + H2O ⇌ C2H5NH3+ + OH-

So, the conjugate acid is C2H5NH3+ and the base is C2H5NH2.

The Kb expression becomes:

Kb = ([C2H5NH3+][OH-])/[C2H5NH2]

At equilibrium, the concentration of [C2H5NH3+] is equal to the concentration of [OH-]:

[C2H5NH3+] = [OH-] = x

And the concentration of [C2H5NH2] is equal to the initial concentration minus the concentration of [C2H5NH3+]:

[C2H5NH2] = (0.11 M) - x

Substituting into the Kb expression:

Kb = (x)(x)/(0.11 M - x)

Now, we can solve for x using the Kb value given:

6.4 × 10^-4 = x^2/(0.11 - x)

x = 3.1 × 10^-4 M

Therefore, the concentration of [C2H5NH3+] and [OH-] is 3.1 × 10^-4 M, and the concentration of [C2H5NH2] is 0.11 M - 3.1 × 10^-4 M = 0.1097 M.

The concentration of [C2H5NH+] and [C2H5NH2] can be added together:

[HS+ + S] = [C2H5NH3+] + [C2H5NH2]

[HS+ + S] = 3.1 × 10^-4 M + 0.1097 M = 0.1098 M

Rounding to the correct number of significant figures, the answer is:

[HS+ + S] = 1.1 × 10^-1 M

Therefore, the answer is (C) 3.1 × 10^-12.

The Base-Dissociation Constant of Ethylamine (C2H5NH2)is 6

The Base-Dissociation Constant of Ethylamine (C₂H₅NH₂)is 6