Let HPr = propionic acid.
HPr ==> H^ + Pr^-
Set up an ICE chart.
Ka = (H^+)(Pr^-)/(HPr)
You know (H^+) from the problem. (Pr^-) is the same as (H^+). (HPr) = moles/L and you have the information to calculate that. Solve for Ka.
A 625mL sample of an aqueous solution containing 0.275 mol propionic acid (CH3CH2CO2H),has [H30+]= 0.00239M.
What is the value of Ka for propionic acid?
3 answers
please help me to solve , I'm confused
I have get it now and the answer is
Ka =2.1×10^-5
Ka =2.1×10^-5