Posted by Sarah on Tuesday, October 11, 2011 at 2:03pm.
Where i go from there **
You start over. Benzoic acid is an acid, not a base. Let's call benzoic acid HB.
..........HB ==> H^+ + B^-
init.....0.15.....0.....0
change....-x......x.....x
equil....0.15-x...x.....x
Ka + (H^+)(B^-)/(HB)
Substitute the equilibrium line above into the Ka expression and solve for x = (H^+) = (H3O^+).
b. %ion = [(H^+)/0.15]*100 = ?
Okay so i got
Ka = (x)(x)/0.15 = 6.5x10^-5
OK. Ka = 6.5E-5 = (x*x)/0.15; now solve for x which is (H^+) (H3O^+ to be more exact) which is what the question asked. Then use the x value to solve for percent ionization as I showed you in part b.
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