What is the PH of a 0.82M Methylamine solution? given: kb=4.38x10^-4

Answer 1

..................CH3NH2 + HOH ==> [CH3NH3]^+ + OH^-
I..................0.82..............................0......................0
C..................-x.................................x.......................x
E...............0.82-x..............................x.......................x

Kb = [(CH3NH3)]^+(OH^-)/(CH3NH2)
4.38E-4 = (x)(x)/(0.82-x)
Solve for x = (OH^-), then
pOH = -log(OH^-). Next is
pH + pOH = pKw = 14. You know pOH and pKw, solve for pH.
Post your work if you get stuck.