The dissociation constant of HA
is 2×10−8
. The hydrolysis constant of NaA
is
A. 1.
2×10−8

B. 2.
4×10−5

C. 3.
5×10−7

D. 4.
7×10−3

To find the hydrolysis constant (\( K_h \)) of \( \text{NaA} \), given the dissociation constant (\( K_a \)) of the weak acid \( \text{HA} \), we can use the relationship between the dissociation constant of the weak acid and the hydrolysis constant of its salt.

Here’s a step-by-step solution:

1. **Understand the given data**:
- \( K_a \) (dissociation constant of HA) = \( 2 \times 10^{-8} \)

2. **Recall the relationship**:
For a salt \( \text{NaA} \) derived from a weak acid \( \text{HA} \), the hydrolysis constant \( K_h \) of \( \text{NaA} \) is related to the dissociation constant of the weak acid by the equation:

\[
K_h = \frac{K_w}{K_a}
\]

where \( K_w \) is the ion-product constant of water, which is \( 1 \times 10^{-14} \) at 25°C.

3. **Calculate \( K_h \)**:
- Substitute \( K_a \) and \( K_w \) into the equation:

\[
K_h = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{2 \times 10^{-8}}
\]

- Perform the division:

\[
K_h = \frac{1 \times 10^{-14}}{2 \times 10^{-8}} = \frac{1}{2} \times 10^{-6} = 5 \times 10^{-7}
\]

4. **Select the correct option**:
The hydrolysis constant \( K_h \) is \( 5 \times 10^{-7} \), which matches option C.

So, the hydrolysis constant of \( \text{NaA} \) is:

**C. \( 5 \times 10^{-7} \)**