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 will follow these steps:

### Step 1: Understand the given data
We are given:
- The extent of hydrolysis of the sodium salt of a weak acid = 3% = 0.03
- The concentration of the solution (C) = 0.1 M
- The dissociation constant of the weak acid (K_a) = 1.3 x 10^(-10)

### Step 2: Calculate the hydrolysis constant (K_h)
The hydrolysis constant (K_h) can be calculated using the formula:
\[ K_h = C \cdot (h^2) \]
where \( h \) is the extent of hydrolysis (in decimal form).

Substituting the values:
\[ K_h = 0.1 \cdot (0.03)^2 \]
\[ K_h = 0.1 \cdot 0.0009 \]
\[ K_h = 9 \times 10^{-5} \]

### Step 3: Use the relationship between K_h, K_a, and K_w
The relationship between the hydrolysis constant (K_h), the dissociation constant of the weak acid (K_a), and the ionic product of water (K_w) is given by:
\[ K_h \cdot K_a = K_w \]

Rearranging this gives:
\[ K_w = K_h \cdot K_a \]

### Step 4: Substitute the values to find K_w
Now, substituting the values of K_h and K_a:
\[ K_w = (9 \times 10^{-5}) \cdot (1.3 \times 10^{-10}) \]
\[ K_w = 1.17 \times 10^{-14} \]

### Conclusion
Thus, the ionic product of water (K_w) at 25°C is:
\[ K_w = 1.17 \times 10^{-14} \]

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