One litre of water contains `10^(−5)` mole of `H^+` ions. Degree of ionisation of water is:

Options

A. 1.8 x `10^(-5)`
B. 1.8 x `10^(-7)`
C. 36 x `10^(-5)`
D. 36 x `10^(-7)`

Explanation

To find the degree of ionization of water given that 1 liter of water contains \(10^{-5}\) moles of \(H^+\) ions, we can follow these steps:

### Step 1: Understand the dissociation of water
The dissociation of water can be represented by the equation:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
In pure water, the concentration of \(H^+\) ions is equal to the concentration of \(OH^-\) ions.

### Step 2: Define the degree of ionization
Let \(X\) be the degree of ionization of water. The degree of ionization is defined as the fraction of the original substance that has dissociated. In this case, we can express it as:
\[ X = \frac{\text{moles of } H^+ \text{ ions produced}}{\text{initial moles of water}} \]

### Step 3: Calculate the initial concentration of water
The molarity of water is approximately \(55.5 \, \text{mol/L}\). Therefore, in 1 liter of water, the initial moles of water is:
\[ \text{Initial moles of water} = 55.5 \, \text{mol} \]

### Step 4: Set up the equation for ionization
From the information provided, we know that the concentration of \(H^+\) ions is \(10^{-5}\) moles. Thus, we can set up the equation:
\[ X \times 55.5 = 10^{-5} \]

### Step 5: Solve for \(X\)
Rearranging the equation gives:
\[ X = \frac{10^{-5}}{55.5} \]

Calculating this gives:
\[ X \approx 1.8 \times 10^{-7} \]

### Step 6: Convert to percentage
To express the degree of ionization as a percentage, we multiply by 100:
\[ \text{Degree of ionization (\%)} = X \times 100 = (1.8 \times 10^{-7}) \times 100 = 1.8 \times 10^{-5} \% \]

### Final Answer
The degree of ionization of water is:
\[ 1.8 \times 10^{-5} \% \]