What is the `K_(b)` of weak base that produce one OH-ion if a 0.05 M solution is 2.5% ionised ?

Options

A. `7.8xx10^(-8)`
B. `1.6xx10^(-6)`
C. `3.1xx10^(-5)`
D. `1.2xx10^(-3)`

Explanation

To find the \( K_b \) of the weak base that produces one OH⁻ ion when a 0.05 M solution is 2.5% ionized, we can follow these steps:

### Step 1: Calculate the concentration of OH⁻ ions produced
Given that the solution is 2.5% ionized, we can find the concentration of OH⁻ ions produced from the 0.05 M solution.

\[ \text{Ionization} = \text{Initial concentration} \times \text{Percentage ionized} \]
\[ \text{Concentration of OH}^- = 0.05 \, \text{M} \times \frac{2.5}{100} = 0.00125 \, \text{M} \]

### Step 2: Write the equilibrium expression for the base ionization
For a weak base \( B \) that ionizes in water to produce \( OH^- \):
\[ B + H_2O \rightleftharpoons BH^+ + OH^- \]

The equilibrium expression for the base ionization constant \( K_b \) is given by:
\[ K_b = \frac{[BH^+][OH^-]}{[B]} \]

### Step 3: Determine the concentrations at equilibrium
- The initial concentration of the base \( [B] \) is 0.05 M.
- At equilibrium, the concentration of \( OH^- \) is 0.00125 M (from Step 1).
- Since the base produces one \( OH^- \) ion for every molecule of base that ionizes, the concentration of \( BH^+ \) will also be 0.00125 M.
- The concentration of the un-ionized base \( [B] \) at equilibrium will be:
\[ [B] = 0.05 \, \text{M} - 0.00125 \, \text{M} = 0.04875 \, \text{M} \]

### Step 4: Substitute the values into the \( K_b \) expression
Now we can substitute the equilibrium concentrations into the \( K_b \) expression:
\[ K_b = \frac{[BH^+][OH^-]}{[B]} = \frac{(0.00125)(0.00125)}{0.04875} \]

### Step 5: Calculate \( K_b \)
\[ K_b = \frac{0.0000015625}{0.04875} \approx 0.0000321 \]

Thus, the \( K_b \) of the weak base is approximately \( 3.21 \times 10^{-5} \).

### Final Answer:
\[ K_b \approx 3.21 \times 10^{-5} \]

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