Nicole R.
asked • 03/22/21The pOH of an aqueous solution of 0.337 M diethylamine (a weak base with the formula (C2H5)2NH) is _____.
J.R. S. answered • 03/22/21
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
(C2H5)2NH + H2O ==> (C2H5)2NH2+ + OH-
Kb = 1x10-3 (I looked it up)
Kb = 1x10-3 = [(C2H5)2NH2+] [ OH-] / [(C2H5)2NH ]
1x10-3 = (x)2 / 0.337
x2 = 3.37x10-4
x = 1.8x10-2 = [OH-]
pOH = -log [OH-]
pOH = 1.7
Param S. answered • 03/22/21
High school diploma of chemistry
pOH= -logOH
hence find the -log of 0.337
Param S.
ahh my bad but i cant seem to figure out the pOH from the OH concentration given if they had given maybe the kb or pkb we could have perhaps figured it out from the equillibrium equation03/23/21
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