The ionisation constant `K_(b)` for hydrazine `(N_(2)H_(4))" is " 9.6 xx 10^(-7)`. What would be the % age hydrolysis of `0.001 M N_(2)H_(5)Cl`, a salt containing acid ion conjugate to hydrazine base.

Answer

To solve the problem of calculating the percentage hydrolysis of 0.001 M N2H5Cl, we will follow these steps:

### Step 1: Understand the Hydrolysis Reaction
The salt N2H5Cl dissociates in water to give hydrazine (N2H4) and H⁺ ions:
\[ \text{N}_2\text{H}_5\text{Cl} \rightleftharpoons \text{N}_2\text{H}_4 + \text{H}^+ \]

### Step 2: Define Initial Concentrations
Let the initial concentration of N2H5Cl be \( C = 0.001 \, \text{M} \). When hydrolysis occurs, let \( x \) be the degree of hydrolysis (the concentration of H⁺ produced).

### Step 3: Write the Equilibrium Expression
At equilibrium, the concentration of N2H4 will be \( x \) and the concentration of H⁺ will also be \( x \). The concentration of N2H5Cl will be \( C - x \).

### Step 4: Use the Ionization Constant
The ionization constant \( K_b \) for hydrazine is given as \( 9.6 \times 10^{-7} \). The expression for \( K_b \) is:
\[ K_b = \frac{[\text{N}_2\text{H}_4][\text{H}^+]}{[\text{N}_2\text{H}_5^+]} \]

Substituting the equilibrium concentrations:
\[ K_b = \frac{x \cdot x}{C - x} \]
\[ K_b = \frac{x^2}{0.001 - x} \]

### Step 5: Assume \( x \) is Small
Since \( K_b \) is small, we can assume \( x \) is much smaller than \( C \). Thus, \( 0.001 - x \approx 0.001 \):
\[ K_b \approx \frac{x^2}{0.001} \]
\[ x^2 = K_b \cdot 0.001 \]
\[ x^2 = 9.6 \times 10^{-7} \cdot 0.001 \]
\[ x^2 = 9.6 \times 10^{-10} \]

### Step 6: Solve for \( x \)
Taking the square root of both sides:
\[ x = \sqrt{9.6 \times 10^{-10}} \]
\[ x \approx 3.1 \times 10^{-5} \, \text{M} \]

### Step 7: Calculate Percentage Hydrolysis
The percentage hydrolysis is given by:
\[ \text{Percentage Hydrolysis} = \left( \frac{x}{C} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Hydrolysis} = \left( \frac{3.1 \times 10^{-5}}{0.001} \right) \times 100 \]
\[ \text{Percentage Hydrolysis} = 0.031 \times 100 \]
\[ \text{Percentage Hydrolysis} = 3.1\% \]

### Final Answer
The percentage hydrolysis of 0.001 M N2H5Cl is approximately **3.1%**.

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