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Calculate the pH for each of the cases in the titration of 25.0 mL of 0.200 M pyridine, C5HN(aq) with 0.200 M HBr(aq). The b of pyridine is 1.7×10−9.
Before addition of any HBr, after addition of 12.5 mL of HBr, after addition of 18.0 mL of HBr,
after addition of 25.0 mL of HBr HBr, after addition of 30.0 mL of HBr HBr
Some explanation of the reasoning for the setup of the first two steps would be appreciated. From where are all of the used values obtained? It is otherwise useless to me. There is no 0.190 value in the given information.
Sagot :
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