One litre of water contains `10^(−9)` 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 1 liter of water contains \(10^{-9}\) moles of \(H^+\) ions, we can follow these steps:

### Step 1: Understand the Ionization of Water
Water ionizes according to the following equilibrium reaction:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
In pure water, the concentration of \(H^+\) ions and \(OH^-\) ions is equal.

### Step 2: Determine the Molarity of Water
The molarity of water is approximately \(55.5 \, M\) (moles per liter). This is derived from the fact that the molar mass of water is about \(18 \, g/mol\) and there are \(1000 \, g\) in 1 liter of water:
\[ \text{Molarity of water} = \frac{1000 \, g}{18 \, g/mol} \approx 55.5 \, M \]

### Step 3: Set Up the Equation for Degree of Ionization
Let \( \alpha \) be the degree of ionization. The concentration of \(H^+\) ions produced from the ionization of water can be expressed as:
\[ \text{Concentration of } H^+ = \text{Molarity of water} \times \alpha \]
Thus, we have:
\[ 55.5 \, \alpha = 10^{-9} \, \text{moles/L} \]

### Step 4: Solve for \( \alpha \)
Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{10^{-9}}{55.5} \]
Calculating this gives:
\[ \alpha \approx 1.8 \times 10^{-11} \]

### Step 5: Convert to Percentage
To find the percentage of the degree of ionization, we multiply by 100:
\[ \text{Percentage of degree of ionization} = \alpha \times 100 = 1.8 \times 10^{-11} \times 100 = 1.8 \times 10^{-9} \% \]

### Final Answer
The degree of ionization of water is approximately \(1.8 \times 10^{-9} \%\).

---