The ionization constant of water is `1.1 xx 10^(-16)`. What is its ionic product of water?

Options

A. `1.1 xx 10^(-16) xx 6 xx 10^(23)`
B. `(1.1 xx 10^(-16))/(6.023 xx 10^(23))`
C. `1.1 xx 10^(-16) xx 55.4`
D. `(1.1 xx 10^(-16))/(10^(-14))`

Answer

To find the ionic product of water (Kw), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The ionization constant of water (Kw) is given as \(1.1 \times 10^{-16}\).
- The ionic product of water is defined as the product of the concentrations of hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\) in pure water.

2. **Write the Expression for Ionization Constant**:
- The ionization constant of water can be expressed as:
\[ K_w = \frac{[H^+][OH^-]}{[H_2O]} \]
- In pure water, the concentration of water \([H_2O]\) is approximately constant and can be considered as 55.5 mol/L.

3. **Rearranging the Equation**:
- To find the ionic product of water, we can rearrange the equation:
\[ [H^+][OH^-] = K_w \times [H_2O] \]

4. **Substituting the Values**:
- Substitute the values into the equation:
\[ [H^+][OH^-] = (1.1 \times 10^{-16}) \times (55.5) \]

5. **Calculating the Ionic Product**:
- Now, perform the multiplication:
\[ [H^+][OH^-] = 1.1 \times 55.5 \times 10^{-16} \]
- Calculate \(1.1 \times 55.5\):
\[ 1.1 \times 55.5 = 61.05 \]
- Therefore, the ionic product of water is:
\[ [H^+][OH^-] = 61.05 \times 10^{-16} = 6.105 \times 10^{-15} \]

### Final Answer:
The ionic product of water (Kw) is approximately \(6.1 \times 10^{-15}\).

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