Asked by Olamide
0.1M solution of methanoic acid is found to have a ph of 2.4 at 25 degree celcius. Calculate the dissociation constant k at this temperature.
Answers
Answered by
DrBob222
Methanoic acid, CH3COOH, I will call HAc.
HAc ==> H^+ + Ac^-
Set up an ICE chart and substitute into Ka. Ka = (H^+)(Ac^-)/(HAc)
HAc ==> H^+ + Ac^-
Set up an ICE chart and substitute into Ka. Ka = (H^+)(Ac^-)/(HAc)
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