The pH of 0.004 M hydrazine solution is 9.7. its ionisation constant `(K_(b))` is

Options

A. `7.79xx10^(-8)`
B. `4.49xx10^(-9)`
C. `1.67xx10^(-10)`
D. `6.25xx10^(-7)`

Answer

To find the ionization constant \( K_b \) of hydrazine from the given pH and concentration, we can follow these steps:

### Step 1: Calculate the concentration of hydroxide ions \([OH^-]\)

Given the pH of the solution is 9.7, we can find the pOH using the formula:

\[ pOH = 14 - pH \]

Substituting the given pH:

\[ pOH = 14 - 9.7 = 4.3 \]

Next, we can find the concentration of hydroxide ions using the formula:

\[ [OH^-] = 10^{-pOH} \]

Calculating this gives:

\[ [OH^-] = 10^{-4.3} \approx 5.01 \times 10^{-5} \, \text{M} \]

### Step 2: Set up the equilibrium expression

For the weak base hydrazine (\( N_2H_4 \)), the ionization can be represented as:

\[ N_2H_4 + H_2O \rightleftharpoons N_2H_5^+ + OH^- \]

The equilibrium expression for \( K_b \) is given by:

\[ K_b = \frac{[N_2H_5^+][OH^-]}{[N_2H_4]} \]

### Step 3: Determine the concentrations at equilibrium

Let \( C \) be the initial concentration of hydrazine, which is 0.004 M. At equilibrium, the change in concentration due to ionization will be \( x \), where \( x \) is the concentration of \( OH^- \) produced.

From the previous calculation, we know:

\[ [OH^-] = x = 5.01 \times 10^{-5} \, \text{M} \]

At equilibrium:

- \([N_2H_5^+] = x = 5.01 \times 10^{-5} \, \text{M}\)
- \([OH^-] = x = 5.01 \times 10^{-5} \, \text{M}\)
- \([N_2H_4] = C - x \approx 0.004 - 5.01 \times 10^{-5} \approx 0.004 \, \text{M}\) (since \( x \) is very small compared to \( C \))

### Step 4: Substitute into the \( K_b \) expression

Now we can substitute these values into the \( K_b \) expression:

\[ K_b = \frac{(5.01 \times 10^{-5})(5.01 \times 10^{-5})}{0.004} \]

Calculating this gives:

\[ K_b = \frac{(5.01 \times 10^{-5})^2}{0.004} = \frac{2.51 \times 10^{-9}}{0.004} = 6.275 \times 10^{-7} \]

### Final Answer

Thus, the ionization constant \( K_b \) for hydrazine is approximately:

\[ K_b \approx 6.31 \times 10^{-7} \]