J.R. S. answered • 07/14/22
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What happens when we put HOCH2CH2NH2 in water?
HOCH2CH2NH2 + H2O ==> HOCH2CH2NH3+ + OH-
Kb = [HOCH2CH2NH3+][OH-] / [HOCH2CH2NH2]
3.1x10-5 = (x)(x) / 4.25x10-4 - x
x2 = 1.32x10-8 - 3.1x10-5x
x2 + 3.1x10-5x - 1.32x10-8 = 0
x = 9.96x10-5 M = [OH-]
pOH = -log 9.96x10-5
pOH = 4.00
pH = 14 - pOH = 14 - 4.00
pH = 10.00
J.R. S.
tutor
Not sure where you got 4 M. They asked for the [OH-] which I calculated to be 9.96x10-5 M. I went on to calculate the pH which wasn’t required.
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07/14/22
Katelyn A.
I put 4M into my online homework website and this was the response I received: You are asked to calculate [OH–] in a 4.25 × × 10–4 M solution of aminoethanol, given that Kb = 3.1 × × 10–5. Analyze Aminoethanol ionizes in aqueous solution to form OH– ions and its conjugate base. [OH–] can be determined from the expression for Kb: $$Kb = [OH−][HOCH2CH2NH + 3] [HOCH2CH2NH2] where HOCH2CH2NH3+ is the conjugate acid of HOCH2CH2NH2. What procedure would you follow for a percent ionization greater than 5%?07/14/22