Question

Calculate the molar concentration of OH ions in a 0.066 M solution of ethylamine (C2H5NH2: Kb= 6.4 x 10-)
Express your answer to two significant figures.

Answer 1

Answer:

6.5x10⁻³M = [OH⁻]

Explanation:

The Kb of a Weak base as ethylamine is expressed as follows:

Kb = [OH⁻] [C₂H₅NH₃⁺] / [C₂H₅NH₂]

As the equilibrium of ethylenamine is:

C₂H₅NH₂(aq) + H₂O(l) ⇄ C₂H₅NH₃⁺(aq) + OH(aq)

The concentration of C₂H₅NH₃⁺(aq) + OH(aq) is the same because both ions comes from the same equilibrium. Thus, we can write:

Kb = [OH⁻] [C₂H₅NH₃⁺] / [C₂H₅NH₂]

6.4x10⁻⁴ = [X] [X] / [C₂H₅NH₂]

Also, we can assume the concentration of ethylamine doesn't decrease. Replacing:

6.4x10⁻⁴ = [X] [X] / [0.066M]

4.224x10⁻⁵ = X²

6.5x10⁻³M = X

6.5x10⁻³M = [OH⁻]

Answer 2

The molar concentration of OH ions in a 0.066 M solution of ethyl amine is $6.5x10^-^3M$

What is molar concentration?

Molar concentration is the measure of the concentration of the chemical species present in any solution.

Step1: Calculating The Kb

$Kb = [OH^-]\dfrac{ [C_2H_5NH_3^+] }{[C_2H_5NH_2]}$

As the equilibrium of ethyl amine is:

$\rm C_2H_5NH_2(aq) + H_2O(l) <-> C_2H_5NH_2^+(aq) + OH(aq)$

$Kb = [OH^-]\dfrac{ [C_2H_5NH_3^+] }{[C_2H_5NH_2]}$

$6.4\times10^-^4 = \dfrac{ [X][X]}{[C_2H_5NH_2]}$

The concentration of ethyl amine does not decrease.

$\rm 6.4\times10^-^4 = \dfrac{ [X][X]}{0.066\;M}$

$4.224\times10^-^5 = X^2$

$6.5x10^-^3M = X\\\\6.5x10^-^3M = [OH^-]$

Thus, the molar concentration is $6.5x10^-^3M$

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