Calculate:
(a) the hydrolysis constant
(b) degree of hydrolysis, and
(c) the `pH` of `0.01 M` `NH_4Cl` at `25^@C`.
`K_b(NH_4OH)=1.81xx10^(-5)`

Explanation

To solve the problem, we need to calculate the hydrolysis constant, degree of hydrolysis, and pH of a 0.01 M solution of NH4Cl at 25°C. Here’s the step-by-step solution:

### Step 1: Calculate the Hydrolysis Constant (K_h)

The hydrolysis constant (K_h) can be calculated using the formula:

\[ K_h = \frac{K_w}{K_b} \]

Where:
- \( K_w \) is the ion product of water at 25°C, which is \( 1.0 \times 10^{-14} \).
- \( K_b \) is the base dissociation constant of NH4OH, given as \( 1.81 \times 10^{-5} \).

Now substituting the values:

\[ K_h = \frac{1.0 \times 10^{-14}}{1.81 \times 10^{-5}} \approx 5.52 \times 10^{-10} \]

### Step 2: Calculate the Degree of Hydrolysis (h)

The degree of hydrolysis (h) can be calculated using the formula:

\[ h = \sqrt{\frac{K_w}{K_b \cdot C}} \]

Where:
- \( C \) is the concentration of the salt, which is \( 0.01 \, M \).

Substituting the values:

\[ h = \sqrt{\frac{1.0 \times 10^{-14}}{1.81 \times 10^{-5} \cdot 0.01}} \]

Calculating the denominator:

\[ h = \sqrt{\frac{1.0 \times 10^{-14}}{1.81 \times 10^{-7}}} \]

Now calculating:

\[ h = \sqrt{5.52 \times 10^{-8}} \approx 2.35 \times 10^{-4} \]

### Step 3: Calculate the pH of the Solution

To find the pH, we first need to find the concentration of \( H^+ \) ions in the solution. The concentration of \( H^+ \) ions can be calculated as:

\[ [H^+] = C \cdot h \]

Substituting the values:

\[ [H^+] = 0.01 \cdot 2.35 \times 10^{-4} = 2.35 \times 10^{-6} \, M \]

Now, we can calculate the pH using the formula:

\[ pH = -\log[H^+] \]

Substituting the value:

\[ pH = -\log(2.35 \times 10^{-6}) \approx 5.63 \]

### Final Answers:
(a) Hydrolysis constant \( K_h \approx 5.52 \times 10^{-10} \)

(b) Degree of hydrolysis \( h \approx 2.35 \times 10^{-4} \)

(c) pH of the solution \( \approx 5.63 \)