Calculate the hydrolysis constant of `NH_4CI,` determine the degree of hydrolysis of this salt in 0.01 M solution and the pH of the solution. `K_b(NH_4OH)= 1.8 xx 10^(-5)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Hydrolysis Constant (K_h) of NH4Cl
NH4Cl is a salt formed from a weak base (NH4OH) and a strong acid (HCl). The hydrolysis constant (K_h) can be calculated using the relationship:

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

Where:
- \( K_w \) is the ion product of water, \( 1.0 \times 10^{-14} \) at 25°C.
- \( K_b \) is the base dissociation constant of NH4OH, given as \( 1.8 \times 10^{-5} \).

Substituting the values:

\[ K_h = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \]

### Step 2: Determine the Degree of Hydrolysis (h) of NH4Cl in 0.01 M Solution
The degree of hydrolysis (h) can be calculated using the formula:

\[ h = \sqrt{\frac{K_h}{C}} \]

Where:
- \( C \) is the concentration of the salt solution, which is \( 0.01 \, M \) or \( 1.0 \times 10^{-2} \, M \).

Substituting the values:

\[ h = \sqrt{\frac{5.56 \times 10^{-10}}{1.0 \times 10^{-2}}} = \sqrt{5.56 \times 10^{-8}} \approx 7.45 \times 10^{-4} \]

### Step 3: Calculate the pH of the Solution
To find the pH, we first need to find the concentration of \( H^+ \) ions produced due to hydrolysis. The concentration of \( H^+ \) ions can be approximated as:

\[ [H^+] \approx h \cdot C = (7.45 \times 10^{-4}) \cdot (1.0 \times 10^{-2}) = 7.45 \times 10^{-6} \, M \]

Now, we can calculate the pH using the formula:

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

Substituting the value:

\[ pH = -\log(7.45 \times 10^{-6}) \approx 5.13 \]

### Final Results
1. Hydrolysis constant \( K_h \) of NH4Cl: \( 5.56 \times 10^{-10} \)
2. Degree of hydrolysis \( h \) in 0.01 M solution: \( 7.45 \times 10^{-4} \)
3. pH of the solution: \( 5.13 \)