When 0.240 mol of HCOOH is dissolved in 0.916 L of water, proton transfer occurs.
HCOOH + H2O ↔ HCOO- + H3O+
The equilibrium concentration of H3O+ ions is 0.00674 M. Evaluate Keq.
Answer:
Ka = [HCOOˉ][H3O+]/[HCOOH]
You are given [H3O+]
[HCOOˉ] = [H3O+] because they both come from the same source
[HCOOH] = (0.240/0.916) - 0.00674
Put all these figures into the first equation and you have your answer - it is not the one given in a previous answer
You've got to stop asking questions and actually do the work for yourself.
divide moles of HCOOH/liters of water to get your molarity of your reactants.
Then divide your product molarity (H3O+) / reactant molarity (HCOOH)
That will give you Keq
4.9 x 10^3
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