.........NH4^+ + H2O ==> NH3 + H3O^+
I........1E-3.............0.....0
C.........-x..............x......x
E.......1E-3-x............x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0/001-x)
Solve for x = (H3O^+).
Then (H3O^+)(OH^-) = Kw = 1E-14 and solve for OH^-
How do you find the [OH-] of a 1x10^-3 M solution of NH4Cl?
1 answer
1 answer