Methylaime, CH3NH2.

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?

You KNOW

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 :)

To calculate the ionization constant (Ka) for the weak base methylamine (CH3NH2), you need to use the pH of the solution and the concentration of the base.

First, let's understand what the ionization constant is. The ionization constant, Ka, is a measure of the extent to which a weak acid or base ionizes in water. It represents the equilibrium constant for the ionization reaction. For a weak base like methylamine, it can be represented as:

CH3NH2 + H2O ⇌ CH3NH3+ + OH-

The Ka value tells us how much of the base (CH3NH2) is ionized into its conjugate acid (CH3NH3+) and hydroxide ion (OH-) in water.

Now, let's solve the problem step by step.

1. Start with the given pH, which is 11.80. The pH is a measure of the concentration of H3O+ ions in the solution. To convert the pH value into the concentration of H3O+ ions, you need to use the formula:

[H3O+] = 10^(-pH)

In this case, [H3O+] = 10^(-11.80) = 1.58 x 10^(-12) M.

2. In an aqueous solution of the weak base methylamine, it reacts with water to form the conjugate acid (CH3NH3+) and hydroxide ion (OH-). However, since methylamine is a weak base, it does not fully ionize and only a fraction of it will form the conjugate acid.

3. Assume x is the concentration of methylamine (CH3NH2) that has ionized to form CH3NH3+ and OH-. The equilibrium can be represented with the following expression:

Kb = [CH3NH3+][OH-] / [CH3NH2]

Since Kb is the ionization constant for the base, you can use the following relationship:

Kb = Kw / Ka

Where Kw is the ion product constant of water (Kw = 1.0 x 10^(-14) at 25°C).

4. Substitute Kb = Kw / Ka and the values into the equation:

1.0 x 10^(-14) = (1.58 x 10^(-12) * x) / (0.1 - x)

5. Solve for x using the quadratic equation. However, since the value of x is expected to be small compared to 0.1, you can approximate (0.1 - x) as 0.1:

1.0 x 10^(-14) = (1.58 x 10^(-12) * x) / 0.1

Cross-multiply to get:

x = (1.0 x 10^(-14) * 0.1) / (1.58 x 10^(-12))

x = 6.33 x 10^(-4) M

6. Calculate the value of Kb using the obtained value of x:

Kb = (1.58 x 10^(-12) * 6.33 x 10^(-4)) / (0.1 - 6.33 x 10^(-4))

Kb ≈ 1.0 x 10^(-6)

So the ionization constant (Kb) for methylamine is approximately 1.0 x 10^(-6).