아닐린 용액의 pH. pH of 0.35 M aniline

아닐린 용액의 pH. pH of 0.35 M aniline

 

 

What is the pH of 0.35 M aniline solution?

(Kb of aniline = 4.0×10^-10)

 

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aniline as A:

A(aq) + H2O(l) ⇌ AH^+(aq) + OH^-(aq)

Kb = [AH+][OH-] / [A] = x^2 / (0.35-x) = 4.0×10^-10

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

 

(0.35-x) ≒ 0.35 라 가정하면,

x = [(0.35)(4.0×10^(-10))]^(1/2)

= 1.1832×10^-5 = [OH-]

 

pOH = -log[OH-] = -log(1.1832×10^(-5)) = 4.93

pH = 14 – pOH = 14 – 4.93 = 9.07

 

 

답: 9.1