In aqueous solution concentration of `H^(+)` ions is `10^(-4)` mol/1. If equal volume of water is added to the solution, then the concentration of `OH^(-)` ion in mol/`dm^(3)` is

Options

A. `0.5 xx 10^(-10)`
B. `2 xx 10^(-10)`
C. `10^(-6)`
D. `10^(-8)`

Answer

To solve the problem step-by-step, we will follow these instructions:

### Step 1: Understand the initial concentration of H⁺ ions
The concentration of H⁺ ions in the solution is given as:
\[ [H^+] = 10^{-4} \, \text{mol/L} \]

### Step 2: Determine the effect of diluting the solution
When we add an equal volume of water to the solution, the total volume of the solution doubles. Therefore, if the initial volume of the solution is \( V \), the new volume \( V_2 \) after adding water becomes:
\[ V_2 = V + V = 2V \]

### Step 3: Use the dilution formula
The dilution formula states that:
\[ M_1 V_1 = M_2 V_2 \]
Where:
- \( M_1 \) is the initial concentration of H⁺ ions
- \( V_1 \) is the initial volume of the solution
- \( M_2 \) is the new concentration of H⁺ ions after dilution
- \( V_2 \) is the new volume of the solution

Substituting the known values:
\[ M_1 = 10^{-4} \, \text{mol/L} \]
\[ V_1 = V \]
\[ V_2 = 2V \]

We can rearrange the equation to solve for \( M_2 \):
\[ 10^{-4} \cdot V = M_2 \cdot 2V \]

### Step 4: Solve for M₂
Dividing both sides by \( V \):
\[ 10^{-4} = M_2 \cdot 2 \]

Now, solving for \( M_2 \):
\[ M_2 = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{mol/L} \]

### Step 5: Calculate the concentration of OH⁻ ions
We know that in pure water at 25°C, the product of the concentrations of H⁺ and OH⁻ ions is given by:
\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]

Now, substituting the value of \( [H^+] \):
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{1 \times 10^{-14}}{5 \times 10^{-5}} \]

### Step 6: Calculate [OH⁻]
Calculating the above expression:
\[ [OH^-] = \frac{1 \times 10^{-14}}{5 \times 10^{-5}} = 2 \times 10^{-10} \, \text{mol/L} \]

### Final Answer
The concentration of OH⁻ ions in the solution after dilution is:
\[ [OH^-] = 2 \times 10^{-10} \, \text{mol/L} \]

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