posted by Lena on .
If a 0.100 mol/L solution of methyamine has a pH of 11.80, calculate the ionization constant for the weak base. What is an ionization constant. Is it just K?
This is what I did so far:
11.80 = -log (concentration of h3o+)
10^-11.80 = (concentration of h3o+)
1.58 x 10^.12 = (concentration of h3o+)
K = (1.58 x 10^-12)^2 / (0.1-1.58 x 10^-12)
Not too sure what to do from here :S
Think NH3 and yes, Ka is the ionization constant.
You know what to do with NH3.
NH3 + HOH ==> NH4^+ + OH^- so in an analogous fashion,
CH3NH2 (this is just NH3 with a CH3 in place of one of the H atoms. These are bases, too, and react the same way.
CH3NH2 + HOH ==> CH3NH3^+ + OH^-
So use an ICE chart to determine OH^-, CH3NH3^+ and CH3NH2 and calculate Ka.
I am getting something like 4 x 10^-4. Look up Kb for CH3NH2 in your text tables to see what it should be.
I don't understand how to figure out the equation. I thought it would be this:
CH3NH2 + HOH ==> CH3NH+ + H30
:S How do I know what the products will be?
NH3 + HOH ==> NH4^+ + OH^-. You KNOE NH3 is a weak base but it is a stronger base than H2O, therefore, the H^+ is pulled from the HOH to form NH4^+.
Amines are just substituted ammonia and there are number of them. When you see NH2 it is RNH2 or RNH or R3N so we are talking about three classes of amines where R may vary. In this case we have
CH3NH2 where CH3 is the R group added. And RNH2 + HOH ==> RNH3^+ + OH^-/
The reaction is strictly analogous to NH3. Do accordingly.
I understand now. Thank you :)