The dissociation constant of aniline `(C_(6)H_(5)NH_(2))` as a base is `5.93xx10^(-10)`. The ionic product of water at `25^(@)C` is `1.02xx10^(-14)`. Calculate the percentage hydrolysis of aniline hydrochloride in 1.0 N solution at `25^(@)C`.Also calculate the pH of the solution.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Hydrolysis Constant (Kh)
The hydrolysis constant (Kh) for aniline hydrochloride can be calculated using the relationship between the dissociation constant of water (Kw) and the dissociation constant of the base (Kb) for aniline.

Given:
- Kb (for aniline) = \(5.93 \times 10^{-10}\)
- Kw (at 25°C) = \(1.02 \times 10^{-14}\)

The hydrolysis constant (Kh) can be calculated as follows:
\[ Kh = \frac{Kw}{Kb} \]
Substituting the values:
\[ Kh = \frac{1.02 \times 10^{-14}}{5.93 \times 10^{-10}} \approx 1.72 \times 10^{-5} \]

### Step 2: Set Up the Hydrolysis Equation
For the hydrolysis of aniline hydrochloride (C₆H₅NH₃Cl), we can write the hydrolysis reaction as:
\[ C_6H_5NH_3^+ + H_2O \rightleftharpoons C_6H_5NH_2 + H_3O^+ \]

### Step 3: Establish the Equilibrium Expression
Let \(x\) be the degree of hydrolysis (the concentration of C₆H₅NH₂ formed at equilibrium). The initial concentration of C₆H₅NH₃⁺ in a 1.0 N solution is 1.0 M.

At equilibrium:
- Concentration of C₆H₅NH₃⁺ = \(1.0 - x\)
- Concentration of C₆H₅NH₂ = \(x\)
- Concentration of H₃O⁺ = \(x\)

The equilibrium expression for Kh is:
\[ Kh = \frac{[C_6H_5NH_2][H_3O^+]}{[C_6H_5NH_3^+]} = \frac{x \cdot x}{1.0 - x} \approx \frac{x^2}{1.0} \quad (\text{since } x \text{ is small}) \]
Thus,
\[ Kh = x^2 \]
Substituting for Kh:
\[ 1.72 \times 10^{-5} = x^2 \]
Solving for \(x\):
\[ x = \sqrt{1.72 \times 10^{-5}} \approx 0.00414 \, \text{M} \]

### Step 4: Calculate the Percentage Hydrolysis
The percentage hydrolysis can be calculated as:
\[ \text{Percentage Hydrolysis} = \left(\frac{x}{\text{Initial Concentration}} \times 100\right) = \left(\frac{0.00414}{1.0} \times 100\right) \approx 0.414\% \]

### Step 5: Calculate the pH of the Solution
The concentration of H₃O⁺ at equilibrium is equal to \(x\):
\[ [H_3O^+] = 0.00414 \, \text{M} \]
Now, calculate the pH:
\[ pH = -\log[H_3O^+] = -\log(0.00414) \approx 2.38 \]

### Final Answers
- Percentage Hydrolysis of aniline hydrochloride in 1.0 N solution: **0.414%**
- pH of the solution: **2.38**