Posted by DAN on Thursday, March 1, 2012 at 4:26pm.
Didn't I show you how to do the other two of the three you posted last night? I think so. Now I've worked all three for you. This one is worked exactly like part b of your three part post.
If we call this HB (for benzoic acid), then
..........B^- + HOH ==> HB + OH^- initial.0.0444...........0....0
change....-x.............x.....x
equil..0.0444-x...........x....x
Kb = Kw/Ka = (HB)(OH^-)/(B^-)
Substitute from the ICE chart, solve for OH^- and convert to pH.
thanx
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