The pH of 0.04 M hydrazine solution is 9.7 . Calculate its ionization constant `K_(b) and pK_(b)`.

Explanation

To solve the problem of calculating the ionization constant \( K_b \) and \( pK_b \) for a 0.04 M hydrazine solution with a pH of 9.7, we can follow these steps:

### Step 1: Calculate the concentration of \( H^+ \) ions
The pH of the solution is given as 9.7. We can find the concentration of \( H^+ \) ions using the formula:
\[ [H^+] = 10^{-\text{pH}} \]
Substituting the value of pH:
\[ [H^+] = 10^{-9.7} \approx 1.67 \times 10^{-10} \, \text{M} \]

### Step 2: Calculate the concentration of \( OH^- \) ions
Using the relationship between \( H^+ \) and \( OH^- \) ions in water at 25°C, we have:
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]
Rearranging for \( [OH^-] \):
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{1.67 \times 10^{-10}} \approx 5.98 \times 10^{-5} \, \text{M} \]

### Step 3: Set up the expression for \( K_b \)
The ionization of hydrazine (\( NH_2NH_2 \)) can be represented as:
\[ NH_2NH_2 + H_2O \rightleftharpoons NH_2NH_3^+ + OH^- \]
The expression for the base ionization constant \( K_b \) is:
\[ K_b = \frac{[NH_2NH_3^+][OH^-]}{[NH_2NH_2]} \]
Assuming that the initial concentration of hydrazine is 0.04 M and that \( x \) is the change in concentration due to ionization, we can express:
- \( [NH_2NH_3^+] = x \)
- \( [OH^-] = x \)
- \( [NH_2NH_2] = 0.04 - x \approx 0.04 \) (since \( x \) is small)

Substituting into the \( K_b \) expression:
\[ K_b = \frac{x \cdot x}{0.04} = \frac{x^2}{0.04} \]

### Step 4: Substitute \( [OH^-] \) into the \( K_b \) expression
From Step 2, we found that \( [OH^-] \approx 5.98 \times 10^{-5} \, \text{M} \), which is equal to \( x \):
\[ K_b = \frac{(5.98 \times 10^{-5})^2}{0.04} \]

### Step 5: Calculate \( K_b \)
Calculating \( K_b \):
\[ K_b = \frac{(5.98 \times 10^{-5})^2}{0.04} = \frac{3.57 \times 10^{-9}}{0.04} \approx 8.93 \times 10^{-8} \]

### Step 6: Calculate \( pK_b \)
To find \( pK_b \), we use the formula:
\[ pK_b = -\log(K_b) \]
Substituting the value of \( K_b \):
\[ pK_b = -\log(8.93 \times 10^{-8}) \approx 7.05 \]

### Final Results
Thus, the ionization constant \( K_b \) is approximately \( 8.93 \times 10^{-8} \) and \( pK_b \) is approximately \( 7.05 \).

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