........MeNH2 + HOH ==> MeNH3^+ + OH^-
initial.0.354M...........0.........0
change...-x..............x.........x
equil...0.354-x..........x.........x
Kb = (MeNH3)(OH^-)/(MeNH2)
Substitute and solve for OH^- and convert to pH.
I am very confused about how to start this problem:
A 29.8 mL sample of 0.354 M methylamine, CH3NH2, is titrated with 0.231 M hydrochloric acid.
The pH before the addition of any hydrochloric acid is ?.
The Kb of methylamine is 4.2x10^-4
1 answer