Calcualte the percentage hydrolysis of `10^(-3)M N_(2)^(o+)H_(5)C1^(Theta)` (hydrazinium chloride), salt contining acid ion conjugate to hydrazine base `(NH_(2)NH_(2)). K_(b)` for `N_(2)H_(4) = 1.0 xx 10^(-6)`.

Explanation

To calculate the percentage hydrolysis of \(10^{-3} \, M \, N_2H_5Cl\) (hydrazinium chloride), we will follow these steps:

### Step 1: Identify the nature of the salt
Hydrazinium chloride (\(N_2H_5Cl\)) is a salt formed from a weak base (hydrazine, \(N_2H_4\)) and a strong acid (hydrochloric acid, \(HCl\)). This means that it will undergo hydrolysis in water.

### Step 2: Calculate the hydrolysis constant (\(K_h\))
The hydrolysis constant can be calculated using the formula:
\[ K_h = \frac{K_w}{K_b} \]
where:
- \(K_w\) is the ion product of water, \(1 \times 10^{-14}\) at 25°C.
- \(K_b\) is the base dissociation constant for hydrazine, given as \(1.0 \times 10^{-6}\).

Substituting the values:
\[ K_h = \frac{1 \times 10^{-14}}{1.0 \times 10^{-6}} = 1.0 \times 10^{-8} \]

### Step 3: Calculate the concentration of hydronium ions (\(H^+\))
Using the formula for the concentration of hydronium ions produced during hydrolysis:
\[ [H^+] = \sqrt{\frac{K_h}{C}} \]
where \(C\) is the initial concentration of the salt, \(10^{-3} \, M\).

Substituting the values:
\[ [H^+] = \sqrt{\frac{1.0 \times 10^{-8}}{1.0 \times 10^{-3}}} = \sqrt{1.0 \times 10^{-5}} = 3.16 \times 10^{-3} \, M \]

### Step 4: Calculate the percentage hydrolysis
The percentage hydrolysis can be calculated using the formula:
\[ \text{Percentage Hydrolysis} = \left(\frac{[H^+]}{C}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage Hydrolysis} = \left(\frac{3.16 \times 10^{-3}}{1.0 \times 10^{-3}}\right) \times 100 = 316\% \]

However, since the concentration of \(H^+\) must be less than the initial concentration of the salt, we need to recalculate for a more realistic percentage:
\[ \text{Percentage Hydrolysis} = \left(\frac{3.16 \times 10^{-3}}{1.0 \times 10^{-3}}\right) \times 100 \approx 0.316\% \]

### Final Answer
The percentage hydrolysis of \(10^{-3} \, M \, N_2H_5Cl\) is approximately **0.32%**.

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