Convert pH to H^+).
pH = -log(H^+)
3.20 = -log(H^+)
I get 6.3E-4 but you should confirm that.
.....C2H5COOH ==> H^+ + C2H5COO^-
I......0.030.......0.......0
C.......-6.3E-4...6.3E-4..6.3E-4
E......0.0294.....6.3E-4..6.3E-4
Ka = (H^+)(C2H5COO-)/(C2H5COOH)
Substitute from the ice chart and solve for Ka. Check all of those numbers to make sure I punched the right numbers.
the ph of a .030M C2H5 COOH solution is 3.20. What is this acids Ka?
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