At `25^(0)`C, the hydroxyl ion concentration of a basic solution is `6.75 xx 10^(-3)` M.Then the value of `K_(w)` is

Options

A. `13.5xx10^(-6) M^(2)`
B. `13.5xx10^(-12)M^(2)`
C. `13.5xx10^(-8) M^(2)`
D. `10^(-14) M^(2)`

Answer

To solve the question, we need to find the value of \( K_w \) given the hydroxyl ion concentration \([OH^-]\) of a basic solution at 25°C, which is \( 6.75 \times 10^{-3} \) M.

### Step-by-Step Solution:

1. **Understanding the relationship between \( K_w \), \([H^+]\), and \([OH^-]\)**:
The ionic product of water (\( K_w \)) at any temperature is given by the formula:
\[ K_w = [H^+][OH^-] \]
At 25°C, \( K_w \) is a constant value, specifically \( 1.0 \times 10^{-14} \) M².

2. **Finding \([H^+]\)**:
Since we know the concentration of hydroxyl ions \([OH^-]\), we can find the concentration of hydrogen ions \([H^+]\) using the relationship:
\[ [H^+] = \frac{K_w}{[OH^-]} \]
Substituting the known values:
\[ [H^+] = \frac{1.0 \times 10^{-14}}{6.75 \times 10^{-3}} \]

3. **Calculating \([H^+]\)**:
Performing the calculation:
\[ [H^+] = \frac{1.0 \times 10^{-14}}{6.75 \times 10^{-3}} \approx 1.48 \times 10^{-12} \text{ M} \]

4. **Confirming the value of \( K_w \)**:
Now that we have both \([H^+]\) and \([OH^-]\), we can confirm the value of \( K_w \):
\[ K_w = [H^+][OH^-] = (1.48 \times 10^{-12})(6.75 \times 10^{-3}) \]
Calculating this gives:
\[ K_w \approx 1.0 \times 10^{-14} \text{ M}^2 \]

5. **Final Answer**:
Therefore, the value of \( K_w \) at 25°C is:
\[ K_w = 1.0 \times 10^{-14} \text{ M}^2 \]