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 (\( CH_3NH_2 \)), follow these steps:

### Step 1: Write the ionization reaction
When methyl amine is dissolved in water, it acts as a weak base and ionizes according to the following reaction:

\[ CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^- \]

### Step 2: Identify the products and reactants
In this reaction:
- The reactants are methyl amine (\( CH_3NH_2 \)) and water (\( H_2O \)).
- The products are the methylammonium ion (\( CH_3NH_3^+ \)) and hydroxide ion (\( OH^- \)).

### Step 3: Write the expression for \( K_b \)
The ionization constant \( K_b \) is defined as the ratio of the concentration of the products to the concentration of the reactants at equilibrium. Therefore, the expression for \( K_b \) can be written as:

\[ K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} \]

### Step 4: Consider the concentration of water
Since water is the solvent and its concentration remains effectively constant, it is not included in the expression for \( K_b \).

### Final Expression
Thus, the final expression for the ionization constant \( K_b \) of methyl amine is:

\[ K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} \]