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)
