One litre of water contains `10^(−4)` 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^(-4)`
D. 1.8 x `10^(-9)`

Explanation

To find the degree of ionization of water given that one litre of water contains \(10^{-4}\) moles of \(H^+\) ions, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Ionization of Water**:
The ionization of water can be represented by the following equilibrium reaction:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
Here, water dissociates into hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)).

2. **Define Degree of Ionization**:
Let \(x\) be the degree of ionization. This means that \(x\) fraction of the water molecules dissociate into \(H^+\) and \(OH^-\) ions.

3. **Calculate the Molarity of Water**:
The molarity of pure water is approximately \(55.5 \, \text{mol/L}\).

4. **Using the Degree of Ionization Formula**:
The concentration of \(H^+\) ions produced can be expressed as:
\[ C \cdot \alpha = [H^+] \]
where \(C\) is the concentration of water (in molarity), \(\alpha\) is the degree of ionization (which is \(x\) in this case), and \([H^+]\) is the concentration of \(H^+\) ions.

5. **Substituting Known Values**:
We know:
\[ C = 55.5 \, \text{mol/L} \quad \text{and} \quad [H^+] = 10^{-4} \, \text{mol/L} \]
Thus, we can write:
\[ 55.5 \cdot x = 10^{-4} \]

6. **Solving for \(x\)**:
Rearranging the equation gives:
\[ x = \frac{10^{-4}}{55.5} \]
Performing the calculation:
\[ x \approx 1.8 \times 10^{-6} \]

7. **Calculating the Percentage of Ionization**:
To express the degree of ionization as a percentage, we multiply \(x\) by 100:
\[ \text{Percentage of Ionization} = x \cdot 100 \approx 1.8 \times 10^{-6} \cdot 100 = 1.8 \times 10^{-4} \% \]

### Final Answer:
The degree of ionization of water is approximately \(1.8 \times 10^{-4} \%\).