Diethylamine, (C2H5)2NH, is a weak base with Kb = 6.92×10−4 at 25 oC.
For a 0.133 mol L−1 solution of (C2H5)2NH at 25 oC, what is the percent ionization of (C2H5)2NH?
Let's call (C2H5)2NH simply BH.
...........BH + HOH ==> BH2 + OH^-
I..........0.133..................0...........0
C.........-x.........................x...........x
E.....0.133-x.....................x...........x
Plug the E line into the Kb expression and solve for x = (OH^-). Convert to pH. Post your work if you get stuck.
By the way, between last night and today I've worked oodles of problems for you and sarah (aka jhope). That is enough freebies. You need to show some work and thinking on your part. This is a homework help site and it's time I started HELPING and not DOING it all.
To find the percent ionization of diethylamine, we need to first calculate the concentration of diethylamine that undergoes ionization and then divide it by the initial concentration. Here's how you can calculate it step by step:
Step 1: Write the balanced equation for the ionization of diethylamine:
(C2H5)2NH + H2O ⇌ (C2H5)2NH2+ + OH-
Step 2: Let "x" represent the concentration of (C2H5)2NH2+ and OH- that forms. Since diethylamine is a weak base, we can assume that the concentration of (C2H5)2NH2+ and OH- formed is very small compared to the initial concentration of diethylamine (C2H5)2NH. Thus, we can assume that x is negligible compared to the initial concentration.
Step 3: Calculate the equilibrium concentration of (C2H5)2NH by subtracting x from the initial concentration:
Initial concentration of (C2H5)2NH = 0.133 mol/L
Equilibrium concentration of (C2H5)2NH = 0.133 - x mol/L
Step 4: Since (C2H5)2NH and OH- are produced in a 1:1 ratio, the concentration of OH- is also x mol/L.
Step 5: Write the expression for the base dissociation constant, Kb, using the equilibrium concentrations:
Kb = [ (C2H5)2NH2+ ][OH-] / [ (C2H5)2NH ]
Step 6: Substitute the equilibrium concentrations into the Kb expression:
Kb = x * x / (0.133 - x)
Step 7: Plug in the value of Kb (6.92×10−4) and solve for x:
6.92×10−4 = x^2 / (0.133 - x)
Step 8: Solve the quadratic equation for x. This can be done by rearranging the equation and using the quadratic formula. After solving, you will find x = 3.67×10−3 mol/L.
Step 9: Calculate the percent ionization:
Percent Ionization = (x / Initial concentration) * 100
Percent Ionization = (3.67×10−3 / 0.133) * 100
Percent Ionization ≈ 2.8%
Therefore, the percent ionization of diethylamine in this solution is approximately 2.8%.