Hydrogen ion concentration of an aqueous solution is `1 xx 10^(-4) M`. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol `dm^(-3)` is

Options

A. `1 xx 10^(-8)`
B. `1 xx 10^(-6)`
C. `2 xx 10^(-10)`
D. `0.5 xx 10^(-10)`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Determine the initial hydrogen ion concentration
The initial hydrogen ion concentration \([H^+]\) is given as:
\[ [H^+] = 1 \times 10^{-4} \, \text{M} \]

### Step 2: Calculate the new hydrogen ion concentration after dilution
Since the solution is diluted with an equal volume of water, the volume doubles. When the volume doubles, the concentration is halved. Therefore, the new concentration of hydrogen ions \([H^+]_{new}\) will be:
\[ [H^+]_{new} = \frac{[H^+]}{2} = \frac{1 \times 10^{-4}}{2} = 0.5 \times 10^{-4} \, \text{M} = 5 \times 10^{-5} \, \text{M} \]

### Step 3: Use the ion product of water to find hydroxyl ion concentration
The ion product of water (\(K_w\)) at 25°C is:
\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]
Now, we can find the hydroxyl ion concentration \([OH^-]\) using the new hydrogen ion concentration:
\[ [OH^-] = \frac{K_w}{[H^+]_{new}} = \frac{1 \times 10^{-14}}{5 \times 10^{-5}} \]

### Step 4: Calculate the hydroxyl ion concentration
Now, we perform the division:
\[ [OH^-] = \frac{1 \times 10^{-14}}{5 \times 10^{-5}} = 2 \times 10^{-10} \, \text{M} \]

### Final Answer
The hydroxyl ion concentration of the resultant solution is:
\[ [OH^-] = 2 \times 10^{-10} \, \text{mol/dm}^3 \]

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