Calculate the pH of a 0.053 M (C2H5)2NH solution (Kb = 1.3 10-3).

Answer 1

..........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.

Answer 2

2.13