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. **Understanding Ionization of Water**:
Water (H₂O) can dissociate into hydrogen ions (H⁺) and hydroxide ions (OH⁻). The ionization reaction can be represented as:
\[ H_2O \rightleftharpoons H^+ + OH^- \]

2. **Defining the Ionization Constant of Water**:
The ionization constant of water (Kw) is defined as the product of the concentrations of the ions produced from the dissociation of water:
\[ K_w = [H^+][OH^-] \]
Given in the question, \( K_w = 1.1 \times 10^{-16} \).

3. **Concentration of Water**:
The concentration of pure water is approximately 55.5 moles per liter. This is derived from the density of water and its molar mass:
\[ \text{Concentration of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.5 \, \text{mol/L} \]

4. **Calculating the Ionic Product of Water**:
The ionic product of water can be calculated using the formula:
\[ K_w = K \times [H_2O] \]
Where \( K \) is the ionization constant given as \( 1.1 \times 10^{-16} \) and \([H_2O] \approx 55.5 \, \text{mol/L}\).

Therefore, substituting the values:
\[ K_w = (1.1 \times 10^{-16}) \times (55.5) \]

5. **Performing the Calculation**:
\[ K_w = 1.1 \times 55.5 \times 10^{-16} = 61.05 \times 10^{-16} \approx 6.1 \times 10^{-15} \]

6. **Final Answer**:
Thus, the ionic product of water (Kw) is approximately:
\[ K_w \approx 6.1 \times 10^{-15} \]