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
To add more information this was given to me for a lab that used
a weak acid and we added a strong base through titration. We just
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.
However, I'm kinda confused about answering them because weak...
please help ><
Polyprotic acids have more than one proton to donate to water, therefore have more than one equilibrium constant for proton donation. For phosphoric acid, there is a three-step equilibrium: H, PO +H,0 5 H2PO4 + H30* H,PO," +H,0 5 HPO - +H30* HPO 2- +,0 5 PO.- +,0* Kai = 7.11 x 10-3 Ka2 = 6.32 x 10-8 Ka; = 4.5 x 10-13 For all conjugate acid base pairs: K, XKK where K is for the reaction...
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