Homework Answers
1) Ka value of benzoic acid = 6.3x10-5
2)concentration of benzoic acid = 0.200M
pKa of acid = -log Ka = -log 6.3x10-5 = 4.20
and pH of weak acid = 1/2[pKa-logC] = 1/2[4.2-log 0.2] =2.45
3) Kb of trimethylamine = 6.4x10-5
4)
concentration of trimethylamine= 0.200M
pKb of base = -log Ka = -log 6.4x10-5 = 4.19
and pOH of weak base = 1/2[pKa-logC] = 1/2[4.19-log 0.2] =2.44
and pH of base = 14-pOH = 14-2.44 = 11.56
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Published online 03/27/2020 Due 04/03/2020 5 points possible First name Last name 1. What is the...
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can someone please help me out with questions 1-5, please
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observed how the ph changes. Later we then used a buffer with a
weak acid to see how buffers affect ph change. These questions are
basically surrounded around those topics to help us prepare.
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