0.20 M 메틸아민(CH3NH2)의 Kb

0.20 M 메틸아민(CH3NH2)의 Kb

 

 

25℃에서 0.20 M 메틸아민(CH3NH2) 용액은 5.0% 이온화된다. 메틸아민의 Kb는 얼마인가?

 

At 25°C, a 0.20 M solution of methylamine, CH3NH2, is 5.0% ionized. What is Kb for methylamine?

 

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이온화 % = 이온화 백분율

= (염기의 평형농도 / 염기의 초기농도) × 100

= ([OH^-]_평형 / [BOH]_초기) × 100

( 식 설명 http://ywpop.tistory.com/3146 )

 

5.0% = (x / 0.20) * 100

x = 5.0 * 0.20 / 100 = 0.01 M = [OH^-]

 

 

...................... CH3NH2(aq) + H2O(l) ⇌ CH3NH3^+(aq) + OH^-(aq)

평형에서, .......... C-x ........................... x .................... x

 

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

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

= (0.01)^2 / (0.20-0.01) = 0.00053

= 5.3×10^-4

 

 

답: 5.3×10^-4

 

 

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Methylamine, CH3NH2, is a weak base. At 25°C, the percent ionization in a 0.160 M CH3NH2 solution is 4.7%. Calculate the Kb for methylamine at 25°C.
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4.7% = (x / 0.160) * 100
x = 4.7 * 0.160 / 100 = 0.00752 M = [OH^-]

 

Kb = (x)(x) / C-x
= (0.00752)^2 / (0.160-0.00752) = 0.000371
= 3.71×10^-4