If that is (HA) = 0.1 before ionization, then I would have used (HA) after ionization as 0.1-0.001 = 0.099.
If Ka = 1.00E-5, then pKa = 5
pKa = -log(Ka) = -log(1E-5) = -(-5) = 5.
an acid with the equilibrium concentration is listed below:
1) HA + H2O = H3O^+ + A-
HA= 10^-1 M, H3O=10^-3M, A=10^-3M
Calculate the Ka and pKa for the acid.
Ka=[10^-3][10^-3] / [10^-1] = 1.00 X 10^-5
pKa = -log[1.00 X 10^-5] = -1.00
Did I do this correctly for the Ka and pKa?
1 answer