One litre of water contains `10^(−6)` 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^(-6)`
D. 1.8 x `10^(-8)`

Explanation

To find the degree of ionization of water given that one liter of water contains \(10^{-6}\) moles of \(H^+\) ions, we can follow these steps:

### Step 1: Understand the Ionization of Water
Water (\(H_2O\)) can dissociate into \(H^+\) and \(OH^-\) ions:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
At equilibrium, the concentration of \(H^+\) ions is equal to the concentration of \(OH^-\) ions.

### Step 2: Define the Degree of Ionization
Let \( \alpha \) be the degree of ionization. The degree of ionization is defined as the fraction of the original substance that has ionized.

### Step 3: Calculate the Concentration of Water
The molarity of water is approximately \(55.5 \, \text{mol/L}\) since the density of water is about \(1 \, \text{g/mL}\) and the molar mass of water is \(18 \, \text{g/mol}\).

### Step 4: Relate the Degree of Ionization to Ion Concentration
Using the formula:
\[ C \cdot \alpha = [H^+] \]
Where:
- \(C\) is the concentration of water in moles per liter (which is \(55.5 \, \text{mol/L}\)),
- \([H^+]\) is the concentration of \(H^+\) ions given as \(10^{-6} \, \text{mol/L}\).

### Step 5: Substitute the Values
Substituting the known values into the equation:
\[ 55.5 \cdot \alpha = 10^{-6} \]

### Step 6: Solve for \(\alpha\)
Rearranging the equation gives:
\[ \alpha = \frac{10^{-6}}{55.5} \]
Calculating this:
\[ \alpha \approx 1.8 \times 10^{-8} \]

### Step 7: Convert to Percentage
To express the degree of ionization as a percentage:
\[ \text{Degree of ionization} = \alpha \times 100 = 1.8 \times 10^{-8} \times 100 = 1.8 \times 10^{-6} \% \]

### Final Answer
The degree of ionization of water is \(1.8 \times 10^{-6} \%\).

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