다이메틸암모늄이온, (CH3)2NH2^+의 pKa

다이메틸암모늄이온, (CH3)2NH2^+의 pKa

 

 

What is the pKa of the dimethylammonium ion, (CH3)2NH2^+?

 

 

Given that Kb for (CH3)2NH is 5.4×10^(-4) at 25℃, what is the value of Ka for (CH3)2NH2^+ at 25℃?

 

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dimethylamine, (CH3)2NH의 Kb = 5.4×10^(-4)

 

 

For a weak acid/conjugate base pair or

weak base/conjugate acid pair,

Kw = Ka × Kb

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

 

Ka = (10^(-14)) / (5.4×10^(-4))

= 1.9×10^(-11)

 

 

pKa = -logKa

= -log(1.9×10^(-11)) = 10.72

 

 

답: pKa = 10.72

 

 

 

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