One litre of water contains 10 (-7)moles of `H^(+)` ions. Degree of ionisation of water (in percentage) is

Options

A. `1.8xx10^(-7)`
B. `1.8xx10^(-9)`
C. `3.6xx10^(-7)`
D. `3.6xx10^(-9)`

Answer

To find the degree of ionization of water in percentage, we can follow these steps:

### Step 1: Understand the Ionization of Water
Water ionizes according to the following reaction:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]

### Step 2: Identify the Concentration of Ions
From the question, we know that in 1 liter of water, there are \(10^{-7}\) moles of \(H^+\) ions. Therefore, the concentration of \(H^+\) ions in water is:
\[ [H^+] = 10^{-7} \text{ moles/L} \]

### Step 3: Calculate the Concentration of Water
The concentration of pure water can be calculated using its density and molar mass. The density of water is approximately \(1 \text{ g/mL}\), which means:
\[ 1 \text{ L of water} \approx 1000 \text{ g} \]
The molar mass of water (H₂O) is approximately \(18 \text{ g/mol}\). Therefore, the concentration of water is:
\[ [H_2O] = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.55 \text{ mol/L} \]

### Step 4: Set Up the Expression for Degree of Ionization
Let \(C\) be the initial concentration of water, which is approximately \(55.55 \text{ mol/L}\). The degree of ionization (\(\alpha\)) can be defined as the fraction of the original substance that has dissociated.

At equilibrium:
- The concentration of \(H^+\) ions is \(C \cdot \alpha\).
- The concentration of \(OH^-\) ions is also \(C \cdot \alpha\).

Given that the concentration of \(H^+\) ions is \(10^{-7} \text{ mol/L}\), we can write:
\[ C \cdot \alpha = 10^{-7} \]

### Step 5: Solve for \(\alpha\)
Substituting the value of \(C\):
\[ 55.55 \cdot \alpha = 10^{-7} \]
\[ \alpha = \frac{10^{-7}}{55.55} \]
\[ \alpha \approx 1.8 \times 10^{-9} \]

### Step 6: Convert to Percentage
To find the degree of ionization in percentage, we multiply by 100:
\[ \text{Degree of ionization} = \alpha \times 100 \]
\[ \text{Degree of ionization} \approx 1.8 \times 10^{-9} \times 100 \]
\[ \text{Degree of ionization} \approx 1.8 \times 10^{-7} \% \]

### Conclusion
The degree of ionization of water in percentage is approximately \(1.8 \times 10^{-7} \%\).