I think you need to rethink b.
If pH = 11.6 then pH is 2.4 and (OH^-) must be something x E-3.
c. Call ethylamine just BNH2
..........BNH2 + HOH ==> BNH3^+ + OH^-
I..........y..............0........0
C.........-x..............x........x
E.........y-x.............x........x
You know Kb (your expression in part A is correct). You know OH^- (x in the above ICE chart), solve for y.
CH3CH2NH2(aq) + H2O(l) <-> CH3CH2NH3+(aq) + OH-(aq)
Kb = 5.0e-4
Ethylamine, CH3CH2NH, a weak base reacts with water according to the equation above. A 50.0 mL sample of a ethylamine solution is found to have a pH of the solution 11.60.
(a) Write the expression for the equilibrium constant, Kb, for ethylamine.
I got Kb = {[CH3CH2NH3+][OH-]}/[CH3CH2NH2]
(b) Calculate the molar concentration of OH- in the 50.0 mL sample of the ethylamine solution.
I got [OH-] = 3.544e-8 M
(c) Calculate the initial molar concentration of CH3CH2NH2(aq) in the solution before it reacted with water and equilibrium was established.
This one, part c, is the one I'm struggling with.
1 answer
1 answer