0.1 M solution of a salt of a weak acid a strong base hydrolysis. If `K_(a)` of the weak acid is `1 xx 10^(-5)`, the percent hydrolysis of the salt is

Options

A. 0.01
B. 0.0001
C. 0.001
D. 0.1

Answer

To solve the problem of finding the percent hydrolysis of a salt of a weak acid and a strong base, we can follow these steps:

### Step 1: Identify the salt and its components
Given that we have a salt from a weak acid and a strong base, we can assume the salt is sodium acetate (CH3COONa), which is derived from acetic acid (CH3COOH) and sodium hydroxide (NaOH).

### Step 2: Write the hydrolysis reaction
The hydrolysis of the acetate ion (CH3COO-) in water can be represented as:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

### Step 3: Define the hydrolysis constant (K_H)
The hydrolysis constant (K_H) can be expressed as:
\[ K_H = \frac{[\text{CH}_3\text{COOH}][\text{OH}^-]}{[\text{CH}_3\text{COO}^-]} \]

### Step 4: Relate K_H to K_a and K_w
Using the relationship between K_H, K_a (the dissociation constant of the weak acid), and K_w (the ion product of water), we have:
\[ K_H = \frac{K_w}{K_a} \]
Where:
- \( K_w = 1 \times 10^{-14} \) (at 25°C)
- \( K_a = 1 \times 10^{-5} \)

### Step 5: Calculate K_H
Substituting the values:
\[ K_H = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9} \]

### Step 6: Set up the equilibrium expression
Let the initial concentration of the acetate ion be \( C = 0.1 \, \text{M} \). If \( h \) is the degree of hydrolysis, then at equilibrium:
- Concentration of CH3COOH = \( hC \)
- Concentration of OH- = \( hC \)
- Concentration of CH3COO- = \( C - hC \approx C \) (since \( h \) is small)

Thus, we can write:
\[ K_H = \frac{(hC)(hC)}{C} = h^2C \]

### Step 7: Solve for h
Substituting \( K_H \) into the equation:
\[ 1 \times 10^{-9} = h^2 \times 0.1 \]
\[ h^2 = \frac{1 \times 10^{-9}}{0.1} = 1 \times 10^{-8} \]
\[ h = \sqrt{1 \times 10^{-8}} = 1 \times 10^{-4} \]

### Step 8: Calculate percent hydrolysis
Percent hydrolysis is given by:
\[ \text{Percent Hydrolysis} = h \times 100 = (1 \times 10^{-4}) \times 100 = 0.01\% \]

### Final Answer
The percent hydrolysis of the salt is **0.01%**.

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