0.1 M ethylamine 용액의 pH

0.1 M ethylamine 용액의 pH

 

 

Ethylammonium (CH3CH2NH3^+)의 pKa가 10.70일 때,

0.1 M ethylamine 용액의 pH는 얼마인가?

 

 

What is the pH of a 0.1 M solution of ethylamine, given that the pKa of ethylammonium ion (CH3CH2NH3^+) is 10.70?

 

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CH3CH2NH2(aq) + H2O(l) ⇌ CH3CH2NH3^+(aq) + OH^-(aq)

Kb = [CH3CH2NH3^+][OH^-] / [CH3CH2NH2]

Kb = (x)(x) / (C-x) ... (1)

( 식 설명 https://ywpop.tistory.com/4294 )

 

 

약염기(CH3CH2NH2)와 짝산(CH3CH2NH3^+)의 관계식

Ka × Kb = Kw

( 식 설명 https://ywpop.tistory.com/2937 )

 

 

Kb = Kw / Ka = 10^(-14) / 10^(-10.70) = 0.0005012

 

 

Kb = (x)(x) / (C-x)

C-x ≒ C 라 가정하면, (1)식은

0.0005012 = x^2 / 0.1

x = (0.0005012 × 0.1)^(1/2)

= 0.007080 M = [OH^-]

 

 

0.1 × (5/100) = 0.005 > 0.007080 이므로,

C-x ≒ C 근사 ok.

 

 

pOH = -log[OH^-] = -log(0.007080) = 2.15

pH = 14 – 2.15 = 11.85

 

 

답: pH = 11.85