Calculate the degree of hydrolysi and the pH of 0.02 M ammonium cyanide solution 298 K. `K_b(NH_4OH)=1.77xx10^(-5) , K_a(HCN)=4.99xx10^(-10)`

Explanation

To solve the problem of calculating the degree of hydrolysis and the pH of a 0.02 M ammonium cyanide solution at 298 K, we can follow these steps:

### Step 1: Understand the Components
Ammonium cyanide (NH4CN) is a salt formed from a weak base (NH4OH) and a weak acid (HCN). The hydrolysis of this salt will lead to the formation of NH4+ and CN- ions in solution.

### Step 2: Use the Hydrolysis Degree Formula
The degree of hydrolysis (H) can be calculated using the formula:
\[ H = \sqrt{\frac{K_w}{K_a \cdot K_b}} \]
where:
- \( K_w = 1.0 \times 10^{-14} \) at 298 K,
- \( K_a \) for HCN = \( 4.99 \times 10^{-10} \),
- \( K_b \) for NH4OH = \( 1.77 \times 10^{-5} \).

### Step 3: Calculate \( K_w \), \( K_a \), and \( K_b \)
Substituting the values into the formula:
\[ H = \sqrt{\frac{1.0 \times 10^{-14}}{(4.99 \times 10^{-10}) \cdot (1.77 \times 10^{-5})}} \]

### Step 4: Calculate the Denominator
First, calculate the product of \( K_a \) and \( K_b \):
\[ K_a \cdot K_b = (4.99 \times 10^{-10}) \cdot (1.77 \times 10^{-5}) = 8.8263 \times 10^{-15} \]

### Step 5: Calculate \( H \)
Now substitute back into the formula:
\[ H = \sqrt{\frac{1.0 \times 10^{-14}}{8.8263 \times 10^{-15}}} \]
\[ H = \sqrt{1.134} \approx 1.06 \]

### Step 6: Calculate pH
To find the pH, we can use the formula:
\[ pH = \frac{1}{2} (pK_w + pK_a - pK_b) \]

### Step 7: Calculate \( pK_w \), \( pK_a \), and \( pK_b \)
1. Calculate \( pK_w \):
\[ pK_w = -\log(1.0 \times 10^{-14}) = 14 \]
2. Calculate \( pK_a \):
\[ pK_a = -\log(4.99 \times 10^{-10}) \approx 9.30 \]
3. Calculate \( pK_b \):
\[ pK_b = -\log(1.77 \times 10^{-5}) \approx 4.75 \]

### Step 8: Substitute into the pH Formula
Now substitute these values into the pH formula:
\[ pH = \frac{1}{2} (14 + 9.30 - 4.75) \]
\[ pH = \frac{1}{2} (18.55) \approx 9.275 \]

### Final Results
- Degree of hydrolysis (H) ≈ 1.06
- pH ≈ 9.275