Posted by Steven on Friday, March 2, 2012 at 12:28am.
pH = pKa + log([A-]/[HA])
Let's call lactic acid HL.
.........HL ==> H^+ + L^-
initial.0.120....0.....0
change...-x......x.....x
equil...0.120-x...x....x
Ka = 1.4E-4 = (H^+)(L^-)/(HL)
Substitute from the ICE chart and solve for x which = (H^+). Then
% ionization = [(H^+)/(0.120)]*100 = ?
afdadf
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