HCOOH ==> H^+ + HCOO^- Ka
H^+ + OH^- ==> H2O 1/Kw
Add the two.
HCOOH + H + OH => H + HCOO + H2O
H cancels to leave
HCOOH + OH => HCOO + H2O.
When adding equations one multiplies the K values.
So Keq = Ka/Kw.
Use the data in the table to calculate the equilibrium constant for the following reaction.
HCOOH(aq)+ OH −(aq) equilibrium reaction arrow HCOO−(aq)+ H2O(l)
HCOO is 5.9e-11 (Kb)
HCOOH is 1.7e-4 (Ka)
Not sure how to find the equilibrium constant given this information
1 answer
2 answers