Posted by Confused on Tuesday, April 10, 2012 at 7:22pm.
..........OBr^- + HOH-->HOBr + OH^-
initial..0.0893..........0......0
change....-x.............x.......x
equil...0.903-x..........x.......x
Kb for OBr^- = Kw/Ka for HOBr = (HOBr)(OH^-)/(OBr^-)
Substitute and solve for x = (OH^-) and convert to pH.
Hi Dr. Bob,
When I do that, I get 11.30 as the pH, although I don't think that's it. Did I miss something?
Maybe and maybe not. That .903 is not right. I remember typing 0.0893
Using 0.0893 I get 10.79. Your Kb is ok. That 0.0893 comes from [(65mL*0.146M/(65mL+41.3mL)]. Sorry I usually proof my responses to catch those typos; I didn't proof this one.
I calculated it using 0.903 and obtained 11.3 also.
Thanks so much! I get it now!
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