`100 `ml of 0.1 M HCl and 100 ml of 0.1 M HOCN are mixed then ` (K_a= 1.2 xx 10^(-6))`

Options

A. `OCN^(-)` concentration in the solution is `1.2 xx 10^(-6)`
B. pH of the solution is 1.3
C. solution is a buffer
D. `H^(+)` in the solution is `10^(-6)`

Answer

To solve the problem of mixing 100 mL of 0.1 M HCl and 100 mL of 0.1 M HOCN, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- HCl is a strong acid and will completely dissociate in solution.
- HOCN is a weak acid with a dissociation constant \( K_a = 1.2 \times 10^{-6} \).

2. **Calculate the Total Volume**:
- When mixing 100 mL of HCl and 100 mL of HOCN, the total volume of the solution becomes:
\[ V_{total} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} \]

3. **Calculate the Moles of Each Acid**:
- Moles of HCl:
\[ \text{Moles of HCl} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]
- Moles of HOCN:
\[ \text{Moles of HOCN} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]

4. **Determine the Concentration of H\(^+\) Ions from HCl**:
- Since HCl is a strong acid, it will dissociate completely:
\[ [H^+]_{HCl} = \frac{0.01 \, \text{moles}}{0.2 \, \text{L}} = 0.05 \, \text{M} \]

5. **Calculate the pH from HCl**:
- The pH can be calculated using the concentration of H\(^+\):
\[ pH = -\log[H^+] = -\log(0.05) \approx 1.3 \]

6. **Determine the pH Contribution from HOCN**:
- The dissociation of HOCN can be represented as:
\[ HOCN \rightleftharpoons H^+ + OCN^- \]
- The equilibrium expression is given by:
\[ K_a = \frac{[H^+][OCN^-]}{[HOCN]} \]
- At equilibrium, let \( x \) be the amount that dissociates:
\[ K_a = \frac{(0.05)(x)}{(0.1 - x)} \approx \frac{(0.05)(x)}{0.1} \quad \text{(assuming } x \text{ is small)} \]
- Setting this equal to \( 1.2 \times 10^{-6} \):
\[ 1.2 \times 10^{-6} = \frac{(0.05)(x)}{0.1} \]
- Solving for \( x \):
\[ x = \frac{1.2 \times 10^{-6} \times 0.1}{0.05} = 2.4 \times 10^{-6} \]

7. **Final Concentration of OCN\(^-\)**:
- The concentration of OCN\(^-\) is equal to \( x \):
\[ [OCN^-] = 2.4 \times 10^{-6} \, \text{M} \]

### Summary of Results:
- The concentration of OCN\(^-\) in the solution is approximately \( 2.4 \times 10^{-6} \, \text{M} \).
- The pH of the solution is approximately \( 1.3 \).