Calculate the `(H_3O^+)` and `[OH^+]` of (i)0.01 M NaOH , (ii) 0.1 M `HNO_3`

Explanation

To solve the problem of calculating the concentrations of \( [H_3O^+] \) and \( [OH^-] \) for the given solutions of NaOH and HNO3, we will follow these steps:

### Step 1: Calculate \( [OH^-] \) for 0.01 M NaOH
- **NaOH** is a strong base and dissociates completely in water:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
- Therefore, the concentration of \( [OH^-] \) is equal to the concentration of NaOH:
\[ [OH^-] = 0.01 \, \text{M} \]

### Step 2: Calculate \( [H_3O^+] \) using the ion product of water
- The ion product of water (\( K_w \)) at 25°C is:
\[ K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} \]
- We can rearrange this to find \( [H_3O^+] \):
\[ [H_3O^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{0.01} = 1.0 \times 10^{-12} \, \text{M} \]

### Step 3: Calculate \( [H_3O^+] \) for 0.1 M HNO3
- **HNO3** is also a strong acid and dissociates completely:
\[ \text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^- \]
- Thus, the concentration of \( [H_3O^+] \) is equal to the concentration of HNO3:
\[ [H_3O^+] = 0.1 \, \text{M} \]

### Step 4: Calculate \( [OH^-] \) using the ion product of water
- Again using the ion product of water:
\[ [OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{0.1} = 1.0 \times 10^{-13} \, \text{M} \]

### Summary of Results:
- For 0.01 M NaOH:
- \( [OH^-] = 0.01 \, \text{M} \)
- \( [H_3O^+] = 1.0 \times 10^{-12} \, \text{M} \)

- For 0.1 M HNO3:
- \( [H_3O^+] = 0.1 \, \text{M} \)
- \( [OH^-] = 1.0 \times 10^{-13} \, \text{M} \)