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)?

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Our tutors have indicated that to solve this problem you will need to apply the Ka and Kb concept. You can view video lessons to learn Ka and Kb. Or if you need more Ka and Kb practice, you can also practice Ka and Kb practice problems.

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Based on our data, we think this problem is relevant for Professor Weng's class at YU.

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.