What is the hydrolysis constant of the `OCl^(-)` ion? The ionization constant of `HOCl` is `3.0xx10^(-8).`

Options

A. `3.33xx10^(-8)`
B. `3.33xx10^(-7)`
C. `3.0xx10^(-7`
D. `3.33xx10^(-6)`

Answer

To find the hydrolysis constant of the `OCl^(-)` ion, we can follow these steps:

### Step 1: Understand the relationship between hydrolysis constant and ionization constant
The hydrolysis constant (Kh) for the `OCl^(-)` ion can be calculated using the formula:
\[ K_h = \frac{K_w}{K_a} \]
where:
- \( K_h \) is the hydrolysis constant of the `OCl^(-)` ion.
- \( K_w \) is the ionic product of water, which is \( 1.0 \times 10^{-14} \) at 25°C.
- \( K_a \) is the ionization constant of the weak acid \( HOCl \), which is given as \( 3.0 \times 10^{-8} \).

### Step 2: Substitute the values into the formula
Now, we can substitute the known values into the formula:
\[ K_h = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-8}} \]

### Step 3: Perform the calculation
To perform the division:
\[ K_h = \frac{1.0}{3.0} \times \frac{10^{-14}}{10^{-8}} \]
\[ K_h = \frac{1}{3} \times 10^{-6} \]

### Step 4: Simplify the expression
Calculating \( \frac{1}{3} \):
\[ K_h \approx 0.333 \times 10^{-6} \]

### Step 5: Convert to scientific notation
To express this in proper scientific notation:
\[ K_h = 3.33 \times 10^{-7} \]

### Conclusion
Thus, the hydrolysis constant of the `OCl^(-)` ion is:
\[ K_h = 3.33 \times 10^{-7} \]