The sodium salt of a weak acid is hydrolysed to the extent of 3% in 0.1 M solution in water at `25^@C` . If `K_a` for weak acid is `1.3xx10^(-10)` . The ionic product of water is

Options

A. `1.17xx10^(-14)`
B. `1.17xx10^(14)`
C. `11.7xx10^(-14)`
D. `11.7xx10^(-10)`

Answer

To solve the problem, we need to determine the ionic product of water (K_w) given the hydrolysis of a sodium salt of a weak acid. Here's a step-by-step solution:

### Step 1: Understanding Hydrolysis
The sodium salt of a weak acid (let's denote it as CH₃COONa) undergoes hydrolysis in water. The hydrolysis reaction can be represented as:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

### Step 2: Setting Up the Equilibrium Expression
For the hydrolysis reaction, we can define the equilibrium constant (K_h) as:
\[ K_h = \frac{[\text{CH}_3\text{COOH}][\text{OH}^-]}{[\text{CH}_3\text{COO}^-]} \]

### Step 3: Initial Concentrations
Let the initial concentration of the salt (CH₃COONa) be \( C = 0.1 \, \text{M} \). The extent of hydrolysis is given as 3%, which means:
- The concentration of CH₃COO⁻ that hydrolyzes is \( 0.03 \times C = 0.03 \times 0.1 = 0.003 \, \text{M} \).

### Step 4: Change in Concentrations
At equilibrium:
- The concentration of CH₃COO⁻ will be \( [\text{CH}_3\text{COO}^-] = C - 0.003 = 0.1 - 0.003 = 0.097 \, \text{M} \).
- The concentrations of CH₃COOH and OH⁻ will both be \( 0.003 \, \text{M} \).

### Step 5: Substituting into the Equilibrium Expression
Now substituting the equilibrium concentrations into the expression for K_h:
\[ K_h = \frac{(0.003)(0.003)}{0.097} \]

### Step 6: Calculating K_h
Calculating K_h:
\[ K_h = \frac{0.000009}{0.097} \approx 9.27 \times 10^{-5} \]

### Step 7: Relating K_h to K_w
We know that:
\[ K_h \cdot K_a = K_w \]
Where \( K_a \) for the weak acid is given as \( 1.3 \times 10^{-10} \).

### Step 8: Calculating K_w
Now we can calculate K_w:
\[ K_w = K_h \cdot K_a = (9.27 \times 10^{-5}) \cdot (1.3 \times 10^{-10}) \]

Calculating K_w:
\[ K_w \approx 1.2051 \times 10^{-14} \]

### Final Answer
Thus, the ionic product of water (K_w) at 25°C is approximately:
\[ K_w \approx 1.21 \times 10^{-14} \]