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
The molar concentration of OH ions in a 0.066 M solution of ethyl amine is [tex]6.5x10^-^3M[/tex]
Molar concentration is the measure of the concentration of the chemical species present in any solution.
Step1: Calculating The Kb
[tex]Kb = [OH^-]\dfrac{ [C_2H_5NH_3^+] }{[C_2H_5NH_2]}[/tex]
As the equilibrium of ethyl amine is:
[tex]\rm C_2H_5NH_2(aq) + H_2O(l) <-> C_2H_5NH_2^+(aq) + OH(aq)[/tex]
[tex]Kb = [OH^-]\dfrac{ [C_2H_5NH_3^+] }{[C_2H_5NH_2]}[/tex]
[tex]6.4\times10^-^4 = \dfrac{ [X][X]}{[C_2H_5NH_2]}[/tex]
The concentration of ethyl amine does not decrease.
[tex]\rm 6.4\times10^-^4 = \dfrac{ [X][X]}{0.066\;M}[/tex]
[tex]4.224\times10^-^5 = X^2[/tex]
[tex]6.5x10^-^3M = X\\\\6.5x10^-^3M = [OH^-][/tex]
Thus, the molar concentration is [tex]6.5x10^-^3M[/tex]
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