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, we need to find the concentration of hydroxide ions (OH⁻) after diluting a solution with a known concentration of hydrogen ions (H⁺). Here’s a step-by-step solution:

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

### Step 2: Determine the Effect of Dilution
When equal volumes of water are added to the solution, the total volume of the solution doubles. This means the concentration of H⁺ ions will be halved.

### Step 3: Calculate the New Concentration of H⁺
Since the volume is doubled, the new concentration of H⁺ ions will be:
\[ [H^+]_{\text{new}} = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Use the Ion Product of Water
At 25°C, the product of the concentrations of H⁺ and OH⁻ ions in water is constant:
\[ K_w = [H^+][OH^-] = 10^{-14} \]

### Step 5: Calculate the Concentration of OH⁻
Now, we can find the concentration of OH⁻ ions using the ion product of water:
\[ [OH^-] = \frac{K_w}{[H^+]} \]
Substituting the values:
\[ [OH^-] = \frac{10^{-14}}{5 \times 10^{-5}} \]

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

### Final Answer
Thus, the concentration of OH⁻ ions in the solution after dilution is:
\[ [OH^-] = 2 \times 10^{-10} \, \text{mol/dm}^3 \]

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