Question

Fifteen points! Nothing like chemistry quizzes in the morning... >o<

Answer 1

Answer:

Explanation:

HA(aq)+H2O(l)⟺H3O+(aq)+A−(aq)(1)

you need to solve for the Ka value. To do that you use

Ka=[H3O+][A−][HA](2)

Another necessary value is the pKa value, and that is obtained through pKa=−logKa

The procedure is very similar for weak bases. The general equation of a weak base is

BOH⟺B++OH−(3)

Solving for the Kbvalue is the same as the Ka value. You use the formula  

Kb=[B+][OH−][BOH](4)

The pKb value is found through pKb=−logKb  

The Kw value is found withKw=[H3O+][OH−].  

Kw=1.0×10−14(5)