Find out the `OH^(-)` ion concentration in `100 mL` of `0.015 M HCl` is

Options

A. `2.0 xx 10^(-9) M`
B. `6.7 xx 10^(-13) M`
C. `3 xx 10^(-10) M`
D. `5 xx 10^(-12) M`

Explanation

To find the concentration of \( OH^- \) ions in a \( 100 \, mL \) solution of \( 0.015 \, M \) HCl, we can follow these steps:

### Step 1: Understand the dissociation of HCl
HCl is a strong acid and dissociates completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Since the concentration of HCl is \( 0.015 \, M \), the concentration of \( H^+ \) ions produced will also be \( 0.015 \, M \).

### Step 2: Use the ion product of water
The ion product of water (\( K_w \)) at \( 25^\circ C \) is given by:
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]

### Step 3: Substitute the concentration of \( H^+ \)
From the dissociation, we know:
\[ [H^+] = 0.015 \, M \]
Now we can substitute this value into the \( K_w \) expression to find \( [OH^-] \):
\[ 1.0 \times 10^{-14} = (0.015)[OH^-] \]

### Step 4: Solve for \( [OH^-] \)
Rearranging the equation to solve for \( [OH^-] \):
\[ [OH^-] = \frac{1.0 \times 10^{-14}}{0.015} \]
Calculating this gives:
\[ [OH^-] = \frac{1.0 \times 10^{-14}}{0.015} \approx 6.67 \times 10^{-13} \, M \]

### Step 5: Conclusion
Thus, the concentration of \( OH^- \) ions in the \( 100 \, mL \) of \( 0.015 \, M \) HCl solution is approximately:
\[ [OH^-] \approx 6.67 \times 10^{-13} \, M \]