Methyl amine, `CH_(3)NH_(2)` is a week base. The expression for its ionization constant `K_(b)` is

Options

A. `([CH_(3) NH_(3)^(+)][OH^(-)])/([CH_(3) NH_(2)][H_(2)O])`
B. `([CH_(2)NH_(3)^(+)][OH^(-)])/([CH_(3)NH_(2)])`
C. `([CH_(3) NH^(-)][H_(3)O^(+)])/([CH_(3)NH_(2)])`
D. `([CH_(3)NH_(2)])/([CH_(3)NH_(3)^(+)][OH^(-)])`

Answer

To derive the expression for the ionization constant \( K_b \) of methyl amine (\( CH_3NH_2 \)), we can follow these steps:

### Step 1: Write the ionization reaction
Methyl amine acts as a weak base and ionizes in water as follows:
\[ CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^- \]

### Step 2: Identify the products and reactants
In this reaction:
- The reactant is \( CH_3NH_2 \) (methyl amine).
- The products are \( CH_3NH_3^+ \) (methylammonium ion) and \( OH^- \) (hydroxide ion).

### Step 3: Write the expression for \( K_b \)
The ionization constant \( K_b \) for a weak base is defined as the ratio of the concentrations of the products to the concentration of the reactants at equilibrium. Therefore, the expression for \( K_b \) is given by:
\[ K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} \]

### Step 4: Consider the concentration of water
Since water is in excess and its concentration does not change significantly during the reaction, it is often omitted from the equilibrium expression. However, in this case, we do not include water in the \( K_b \) expression because it is a pure liquid.

### Final Expression
Thus, the final expression for the ionization constant \( K_b \) of methyl amine is:
\[ K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} \]