Let's call this BNH2 to save some typing.
..........BNH2 + HOH --> BNH3^+ + OH^-
I........0.223.............0.......0
C.........-x...............x.......x
E.......0.223-x............x.......x
But you know x = 9.7E-6. Evaluate 0.223-x and substitute into the Kb expression and solve for Kb.
Kb = (BNH3^+)(OH^-)/(BNH2)
Aniline, C6H5NH2, is a weak base. If the hydroxide ion concentration of a 0.223 M solution is 9.7 10-6 M, what is the ionization constant for the base?
1 answer
1 answer