Calculate the hydrolysis constant of `NH_4Cl`. Determine the degree of hydrolysis of this salt in 0.01 M solution and the pH of the solution. Kb(`NH_4OH`)=1.8×`10^(−5)`

Explanation

To solve the problem step by step, we will calculate the hydrolysis constant of `NH4Cl`, determine the degree of hydrolysis in a 0.01 M solution, and find the pH of the solution.

### Step 1: Calculate the Hydrolysis Constant (Kh) of NH4Cl

The hydrolysis constant (Kh) can be calculated using the relationship:
\[ K_h = \frac{K_w}{K_b} \]

Where:
- \( K_w \) is the ion product of water, which is \( 1.0 \times 10^{-14} \) at 25°C.
- \( K_b \) is the base dissociation constant of \( NH_4OH \), given as \( 1.8 \times 10^{-5} \).

Substituting the values:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \]

Calculating this gives:
\[ K_h = 5.55 \times 10^{-10} \]

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

The degree of hydrolysis (α) can be calculated using the formula:
\[ K_h = C \alpha^2 \]

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

Rearranging the formula to find α:
\[ \alpha = \sqrt{\frac{K_h}{C}} \]

Substituting the values:
\[ \alpha = \sqrt{\frac{5.55 \times 10^{-10}}{0.01}} \]

Calculating this gives:
\[ \alpha = \sqrt{5.55 \times 10^{-8}} \]
\[ \alpha \approx 7.45 \times 10^{-4} \]

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

To find the pH, we can use the formula:
\[ \text{pH} = \frac{1}{2} \text{pK}_w - \frac{1}{2} \log C + \frac{1}{2} \text{pK}_b \]

Where:
- \( \text{pK}_w = -\log K_w = -\log(1.0 \times 10^{-14}) = 14 \)
- \( \text{pK}_b = -\log K_b = -\log(1.8 \times 10^{-5}) \)

Calculating \( \text{pK}_b \):
\[ \text{pK}_b \approx 4.74 \]

Now substituting into the pH formula:
\[ \text{pH} = \frac{1}{2} (14) - \frac{1}{2} \log(0.01) + \frac{1}{2} (4.74) \]

Calculating each term:
1. \( \frac{1}{2} (14) = 7 \)
2. \( \frac{1}{2} \log(0.01) = \frac{1}{2} (-2) = -1 \)
3. \( \frac{1}{2} (4.74) = 2.37 \)

Putting it all together:
\[ \text{pH} = 7 + 1 + 2.37 = 10.37 \]

### Final Answers:
- Hydrolysis constant \( K_h \) of \( NH_4Cl \) = \( 5.55 \times 10^{-10} \)
- Degree of hydrolysis \( \alpha \) = \( 7.45 \times 10^{-4} \)
- pH of the solution = \( 10.37 \)