Calculate the degree of hydrolysis of `CH_3COOK` in 0.1 M and the pH of the solution. `K_a(CH_3COOH) =1.8 xx 10^(-5)`.

Answer

To solve the problem of calculating the degree of hydrolysis of `CH₃COOK` in a 0.1 M solution and the pH of the solution, we will follow these steps:

### Step 1: Understand the Hydrolysis of the Salt
`CH₃COOK` is a salt formed from the weak acid `CH₃COOH` and the strong base `KOH`. In solution, it undergoes hydrolysis to produce `CH₃COO⁻` ions and `OH⁻` ions.

### Step 2: Write the Hydrolysis Reaction
The hydrolysis reaction can be represented as:
\[ CH₃COOK \rightleftharpoons CH₃COO⁻ + K^+ \]
\[ CH₃COO⁻ + H₂O \rightleftharpoons CH₃COOH + OH⁻ \]

### Step 3: Calculate the Hydrolysis Constant (K_h)
The hydrolysis constant \( K_h \) can be calculated using the relation:
\[ K_h = \frac{K_w}{K_a} \]
Where:
- \( K_w = 1.0 \times 10^{-14} \) (ion product of water)
- \( K_a = 1.8 \times 10^{-5} \) (acid dissociation constant of acetic acid)

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

### Step 4: Set Up the Hydrolysis Equation
Let \( x \) be the degree of hydrolysis. The concentration of `CH₃COO⁻` at equilibrium will be \( 0.1 - x \) and the concentration of `OH⁻` will be \( x \).

Using the hydrolysis constant:
\[ K_h = \frac{[CH₃COOH][OH⁻]}{[CH₃COO⁻]} = \frac{x \cdot x}{0.1 - x} \approx \frac{x^2}{0.1} \quad (\text{since } x \text{ is small}) \]

### Step 5: Solve for x
Setting up the equation:
\[ 5.56 \times 10^{-10} = \frac{x^2}{0.1} \]
\[ x^2 = 5.56 \times 10^{-10} \times 0.1 = 5.56 \times 10^{-11} \]
\[ x = \sqrt{5.56 \times 10^{-11}} \approx 7.45 \times 10^{-6} \]

### Step 6: Calculate the Degree of Hydrolysis
The degree of hydrolysis \( \alpha \) is given by:
\[ \alpha = \frac{x}{C} = \frac{7.45 \times 10^{-6}}{0.1} = 7.45 \times 10^{-5} \]

### Step 7: Calculate the pH of the Solution
To find the pOH:
\[ pOH = -\log[OH⁻] = -\log(7.45 \times 10^{-6}) \approx 5.13 \]
Now, using the relation \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 5.13 = 8.87 \]

### Final Results
- Degree of hydrolysis \( \alpha \approx 7.45 \times 10^{-5} \)
- pH of the solution \( \approx 8.87 \)