Posted by Tom on Tuesday, May 27, 2008 at 9:36pm.
Solution B is 0.05 M solution of NaX.
The hydrolysis of X^- is
X^- + HOH ==> HX + OH^-
Kb = Kw/Ka = (HX)(OH^-)/(X^-)
You know pH is 10.02. Subtract from 14 to get pOH and solve for (OH^-) remembering that pOH= -log(OH^-). You know (HX) = (OH^-). YOu know Kw and (X^-) so the only unknown is Ka. Solve for Ka.
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