One litre of water contains `10^(−8)` mole of `H^+` ions. Degree of ionisation of water is:

Options

A. 1.8 x `10^(-5)`
B. 1.8 x `10^(-7)`
C. 1.8 x `10^(-9)`
D. 1.8 x `10^(-8)`

Explanation

To find the degree of ionization of water given that one liter of water contains \(10^{-8}\) moles of \(H^+\) ions, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dissociation of Water**:
The dissociation of water can be represented as:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
At equilibrium, the concentration of \(H^+\) ions is equal to the concentration of \(OH^-\) ions.

2. **Given Information**:
We know that:
- The concentration of \(H^+\) ions in water is \(10^{-8}\) moles per liter.
- The molarity of water is approximately \(55.5 \, M\) (since the density of water is about \(1 \, g/mL\) and the molar mass of water is \(18 \, g/mol\)).

3. **Defining Degree of Ionization**:
Let \( \alpha \) be the degree of ionization. The degree of ionization is defined as the fraction of the total number of moles that dissociates into ions.

4. **Setting Up the Equation**:
The concentration of \(H^+\) ions can be expressed as:
\[ [H^+] = C \cdot \alpha \]
where \(C\) is the concentration of water (55.5 M). Therefore:
\[ [H^+] = 55.5 \cdot \alpha \]
We know that \([H^+] = 10^{-8}\) M.

5. **Solving for \(\alpha\)**:
Setting the two expressions for \([H^+]\) equal gives:
\[ 55.5 \cdot \alpha = 10^{-8} \]
Solving for \(\alpha\):
\[ \alpha = \frac{10^{-8}}{55.5} \]
\[ \alpha \approx 1.8 \times 10^{-10} \]

6. **Calculating Percentage of Ionization**:
To find the percentage of ionization, we multiply \(\alpha\) by 100:
\[ \text{Percentage of ionization} = \alpha \times 100 = 1.8 \times 10^{-10} \times 100 = 1.8 \times 10^{-8} \% \]

### Final Answer:
The degree of ionization of water is approximately \(1.8 \times 10^{-8} \%\).