Calculate the hydrolysis constant. Degree of hydrolysis and pH of 0.5 M solution of `NH_4Cl.`
( Given `: K_a = 1.78 xx 10 ^(-5) ,K_b= 1.8 xx 10 ^(-5) and K_w = 1.8 xx 10 ^(-14) ) `

Answer

To solve the problem of calculating the hydrolysis constant, degree of hydrolysis, and pH of a 0.5 M solution of ammonium chloride (NH₄Cl), we can follow these steps:

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

The hydrolysis constant \( K_h \) can be calculated using the relationship between \( K_w \), \( K_a \), and \( K_b \):

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

Given:
- \( K_w = 1.8 \times 10^{-14} \)
- \( K_b = 1.8 \times 10^{-5} \)

Substituting the values:

\[ K_h = \frac{1.8 \times 10^{-14}}{1.8 \times 10^{-5}} = 10^{-9} \]

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

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

\[ K_h = \frac{C \cdot H^2}{1 - H} \]

Assuming \( H \) is small, we can neglect \( H \) in the denominator:

\[ K_h \approx C \cdot H^2 \]

Rearranging gives:

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

Where \( C = 0.5 \, \text{M} \):

\[ H = \sqrt{\frac{10^{-9}}{0.5}} = \sqrt{2 \times 10^{-9}} = 4.47 \times 10^{-5} \]

### Step 3: Calculate the pH

To find the pH of the solution, we can use the formula for weak acids derived from the hydrolysis of ammonium chloride:

\[ \text{pH} = 7 - \frac{1}{2} \log C + \text{pK}_b \]

Where \( \text{pK}_b = -\log K_b \).

Calculating \( \text{pK}_b \):

\[ \text{pK}_b = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

Now substituting \( C = 0.5 \):

\[ \text{pH} = 7 - \frac{1}{2} \log(0.5) + 4.74 \]

Calculating \( \log(0.5) \):

\[ \log(0.5) \approx -0.301 \]

So:

\[ \text{pH} = 7 - \frac{1}{2}(-0.301) + 4.74 = 7 + 0.1505 + 4.74 = 4.627 \]

### Final Results

- Hydrolysis constant \( K_h = 10^{-9} \)
- Degree of hydrolysis \( H = 4.47 \times 10^{-5} \)
- pH of the solution \( \text{pH} = 4.627 \)

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