One litre of water contains `10^(-7)` mole `H^(+)` ions. Degree of ionisation of water is:

Options

A. `1.8 xx 10^(-7)%`
B. `1.8 xx 10^(-9)%`
C. `3.6 xx 10^(-7)%`
D. `3.6 xx 10^(-9)%`

Answer

To find the degree of ionization of water, we can follow these steps:

### Step-by-Step Solution:

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

2. **Given Information**:
We are given that 1 liter of water contains \(10^{-7}\) moles of \(H^+\) ions. This means at equilibrium:
\[ [H^+] = 10^{-7} \text{ moles/L} \]

3. **Calculating the Concentration of Water**:
The concentration of pure water can be calculated using its molar mass. The molar mass of water (H₂O) is approximately 18 g/mol. Since 1 liter of water weighs about 1000 g, the concentration (C) of water in moles per liter is:
\[ C = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.5 \text{ moles/L} \]

4. **Using the Degree of Ionization Formula**:
Let α be the degree of ionization. The concentration of \(H^+\) ions at equilibrium can be expressed as:
\[ C \cdot \alpha = [H^+] = 10^{-7} \]
Substituting the value of C:
\[ 55.5 \cdot \alpha = 10^{-7} \]

5. **Solving for α**:
Rearranging the equation gives:
\[ \alpha = \frac{10^{-7}}{55.5} \]
Performing the calculation:
\[ \alpha \approx 1.8 \times 10^{-9} \]

6. **Calculating Degree of Ionization in Percentage**:
To convert α into a percentage, we multiply by 100:
\[ \text{Degree of Ionization} = \alpha \times 100 \approx 1.8 \times 10^{-9} \times 100 = 1.8 \times 10^{-7} \% \]

### Final Answer:
The degree of ionization of water is approximately \(1.8 \times 10^{-7} \%\).

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