Posted by Kaleen on Wednesday, November 23, 2011 at 7:11pm.
..........B2NH + HOH ==> B2NH2^+ + OH^-
initial....0.053..........0.........0
change....-x..............x.........x
equil.....0.053-x.........x..........x
Kb = (B2NH^+)(OH^-)/(B2BH)
Substitute into Kb expression and solve for x (which is OH^-). Convert to pH.
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