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.
Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH.
A) benzoic acid (C6H5COOH) Ka= 6.3*10^-5 for it
3 answers
thanx
ph = 5.57
1 answer