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, we need to calculate the percentage hydrolysis of the salt N2H5Cl, which is derived from the weak base hydrazine (N2H4) and the strong acid HCl. The steps for the solution are as follows:

### Step 1: Understand the Reaction
N2H5Cl dissociates in water to form hydrazine (N2H4) and hydrochloric acid (HCl):
\[ \text{N}_2\text{H}_5\text{Cl} \rightleftharpoons \text{N}_2\text{H}_4 + \text{HCl} \]

### Step 2: Identify the Given Values
- Ionization constant \( K_b \) for hydrazine, \( K_b = 9.6 \times 10^{-7} \)
- Concentration of N2H5Cl, \( C = 0.001 \, M \)

### Step 3: Calculate \( pK_b \)
First, we calculate \( pK_b \):
\[ pK_b = -\log(K_b) = -\log(9.6 \times 10^{-7}) \approx 6.02 \]

### Step 4: Calculate \( pK_w \)
The value of \( pK_w \) at 25°C is:
\[ pK_w = 14 \]

### Step 5: Calculate \( pK_a \)
Using the relationship \( pK_a + pK_b = pK_w \):
\[ pK_a = pK_w - pK_b = 14 - 6.02 = 7.98 \]

### Step 6: Calculate the Concentration of \( H^+ \)
Using the formula for the pH of a weak base in the presence of a strong acid:
\[ pH = \frac{1}{2} pK_w - \frac{1}{2} pK_b - \log C \]
Substituting the values:
\[ pH = \frac{1}{2} \times 14 - \frac{1}{2} \times 6.02 - \log(0.001) \]
Calculating:
\[ pH = 7 - 3.01 + 3 = 6.99 \]

### Step 7: Calculate \( [H^+] \)
Now, we find the concentration of \( H^+ \):
\[ [H^+] = 10^{-pH} = 10^{-6.99} \approx 1.02 \times 10^{-7} \, M \]

### Step 8: Determine the Degree of Hydrolysis \( h \)
Let \( h \) be the degree of hydrolysis. The concentration of \( H^+ \) produced will be equal to \( h \) times the initial concentration of the salt:
\[ h = \frac{[H^+]}{C} = \frac{1.02 \times 10^{-7}}{0.001} = 1.02 \times 10^{-4} \]

### Step 9: Calculate Percentage Hydrolysis
The percentage hydrolysis is given by:
\[ \text{Percentage Hydrolysis} = h \times 100 = 1.02 \times 10^{-4} \times 100 = 0.0102\% \]

### Final Answer
The percentage hydrolysis of 0.001 M N2H5Cl is approximately **0.0102%**.

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