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(Solved): Consider the following acid in equlibrium. What is the correct expression for Ka of this acid. HA+ ...
Consider the following acid in equlibrium. What is the correct expression for Ka of this acid.
HA+H_(2)O<=>H_(3)O^(+)+A^(-)
Ka=(H3O+)/(HA)
Ka=([H_(3)O^(+)]^(2))/(HA)
Ka=(H_(3)O^(+))/(HA)[A^(-)]
Ka=[H_(3)O^(+)](A^(-))/(HA)