......CH3NH2 + H2O ==> CH3NH3^+ + OH^-
I.....0.31M..............0.........0
C.......-x...............x.........x
E.....0.31-x.............x.........x
Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
Substitute the E line into the Kb expression and solve for x = (OH^-).
Then % ion = [(OH^-)/(CH3NH2)]*100
what is the percent ionization at equilibrium in a 0.31M solution of dimethylamine (Kb=4.2x10^-10) at 25 Celsius
1 answer
1 answer