`A0.1M` sodium acetate solution was prepared. The `K_(h)=5.6xx10^(-10)`

Options

A. The degree of hydrolysis is `7.48xx10^(-5)`
B. The `[OH^(-)]` concentration is `7.48xx10^(-3)M`
C. The `[OH^(-)]` concentration is `7.48xx10^(-6)M`
D. The pH is approximately `8.88`

Explanation

To solve the problem regarding the sodium acetate solution and its hydrolysis, we can follow these steps:

### Step 1: Understanding the Hydrolysis of Sodium Acetate
Sodium acetate (CH₃COONa) is a salt derived from a weak acid (acetic acid, CH₃COOH) and a strong base (sodium hydroxide, NaOH). In solution, it undergoes hydrolysis:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

### Step 2: Calculate the Degree of Hydrolysis
The degree of hydrolysis (h) can be calculated using the formula:
\[ K_h = \frac{h^2 C}{1 - h} \]
Where:
- \( K_h \) is the hydrolysis constant (given as \( 5.6 \times 10^{-10} \))
- \( C \) is the concentration of the salt (given as \( 0.1 \, M \))

Assuming \( h \) is small, we can simplify the equation to:
\[ K_h \approx h^2 C \]
Thus,
\[ h^2 = \frac{K_h}{C} \]
Substituting the values:
\[ h^2 = \frac{5.6 \times 10^{-10}}{0.1} = 5.6 \times 10^{-9} \]
Taking the square root:
\[ h = \sqrt{5.6 \times 10^{-9}} \approx 7.48 \times 10^{-5} \]

### Step 3: Calculate the Concentration of OH⁻
The concentration of OH⁻ ions produced during hydrolysis is equal to the degree of hydrolysis multiplied by the initial concentration:
\[ [OH^-] = h \times C = 7.48 \times 10^{-5} \times 0.1 = 7.48 \times 10^{-6} \, M \]

### Step 4: Calculate the pH of the Solution
To find the pH, we first calculate the pOH:
\[ pOH = -\log[OH^-] \]
Substituting the value of [OH⁻]:
\[ pOH = -\log(7.48 \times 10^{-6}) \approx 5.13 \]
Now, using the relation \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 5.13 = 8.87 \]

### Conclusion
The degree of hydrolysis is \( 7.48 \times 10^{-5} \), the concentration of OH⁻ is \( 7.48 \times 10^{-6} \, M \), and the pH of the solution is approximately \( 8.87 \).

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