Problem: What is the Ka of the 2-hydroxyethylammonium ion, HOCH2CH2NH3+ (pKb of HOCH2CH2NH2 = 4.49)?

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FREE Expert Solution

Recall that Ka and Kb are related as (Ka = 10-pKa):

Kw = Ka×KbKa = KwKb=1x10-1410-4.49

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Problem Details

What is the Ka of the 2-hydroxyethylammonium ion, HOCH2CH2NH3+ (pKb of HOCH2CH2NH2 = 4.49)?