100 mL of 0.5 M hydrazoic acid `(N_3H, K_a=3.6xx10^(-4))` and 400 mL of 0.1 M cyanic acid `(HOCN, K_a=8xx10^(-4))` are mixed .Which of the following is true for final solution ?

Options

A. `[H^+]=2xx10^(-2) M`
B. `[N^(3-)]=3.6xx10^(-2) M`
C. `[OCN^-]=4.571xx10^(-3) M`
D. `[OCN^-]=6.4xx10^(-3) M`

Answer

To solve the problem, we need to find the concentrations of \( H^+ \), \( N_3^- \), and \( OCN^- \) in the final solution after mixing 100 mL of 0.5 M hydrazoic acid (N₃H) and 400 mL of 0.1 M cyanic acid (HOCN).

### Step 1: Calculate the number of moles of each acid before mixing.

1. **For hydrazoic acid (N₃H)**:
- Volume = 100 mL = 0.1 L
- Concentration = 0.5 M
- Moles of N₃H = Volume × Concentration = \( 0.1 \, \text{L} \times 0.5 \, \text{mol/L} = 0.05 \, \text{mol} \)

2. **For cyanic acid (HOCN)**:
- Volume = 400 mL = 0.4 L
- Concentration = 0.1 M
- Moles of HOCN = Volume × Concentration = \( 0.4 \, \text{L} \times 0.1 \, \text{mol/L} = 0.04 \, \text{mol} \)

### Step 2: Calculate the total volume of the mixed solution.

- Total Volume = Volume of N₃H + Volume of HOCN = 100 mL + 400 mL = 500 mL = 0.5 L

### Step 3: Calculate the concentrations of the acids in the final solution.

1. **Concentration of N₃H in the final solution**:
- Concentration of N₃H = Moles / Total Volume = \( \frac{0.05 \, \text{mol}}{0.5 \, \text{L}} = 0.1 \, \text{M} \)

2. **Concentration of HOCN in the final solution**:
- Concentration of HOCN = Moles / Total Volume = \( \frac{0.04 \, \text{mol}}{0.5 \, \text{L}} = 0.08 \, \text{M} \)

### Step 4: Calculate the \( H^+ \) concentration from each acid.

1. **For N₃H**:
- The dissociation of N₃H is:
\[ N_3H \rightleftharpoons N_3^- + H^+ \]
- Let \( \alpha \) be the degree of dissociation.
- At equilibrium:
- \( [N_3H] = 0.1 - \alpha \)
- \( [N_3^-] = \alpha \)
- \( [H^+] = \alpha \)

- The equilibrium expression is:
\[ K_a = \frac{[N_3^-][H^+]}{[N_3H]} = \frac{\alpha^2}{0.1 - \alpha} \]
- Given \( K_a = 3.6 \times 10^{-4} \), assuming \( \alpha \) is small, we can approximate:
\[ K_a \approx \frac{\alpha^2}{0.1} \]
- Solving for \( \alpha \):
\[ \alpha^2 = 3.6 \times 10^{-4} \times 0.1 = 3.6 \times 10^{-5} \]
\[ \alpha = \sqrt{3.6 \times 10^{-5}} \approx 6 \times 10^{-3} \]

- Therefore, the concentration of \( H^+ \) from N₃H:
\[ [H^+]_{N_3H} = \alpha \approx 6 \times 10^{-3} \, \text{M} \]

2. **For HOCN**:
- The dissociation of HOCN is:
\[ HOCN \rightleftharpoons H^+ + OCN^- \]
- Let \( \beta \) be the degree of dissociation.
- At equilibrium:
- \( [HOCN] = 0.08 - \beta \)
- \( [H^+] = \beta \)
- \( [OCN^-] = \beta \)

- The equilibrium expression is:
\[ K_a = \frac{[H^+][OCN^-]}{[HOCN]} = \frac{\beta^2}{0.08 - \beta} \]
- Given \( K_a = 8 \times 10^{-4} \), assuming \( \beta \) is small, we can approximate:
\[ K_a \approx \frac{\beta^2}{0.08} \]
- Solving for \( \beta \):
\[ \beta^2 = 8 \times 10^{-4} \times 0.08 = 6.4 \times 10^{-5} \]
\[ \beta = \sqrt{6.4 \times 10^{-5}} \approx 8 \times 10^{-3} \]

- Therefore, the concentration of \( H^+ \) from HOCN:
\[ [H^+]_{HOCN} = \beta \approx 8 \times 10^{-3} \, \text{M} \]

### Step 5: Calculate the total \( H^+ \) concentration in the final solution.

- Total \( [H^+] \) = \( [H^+]_{N_3H} + [H^+]_{HOCN} \)
\[ [H^+] = 6 \times 10^{-3} + 8 \times 10^{-3} = 14 \times 10^{-3} \, \text{M} = 1.4 \times 10^{-2} \, \text{M} \]

### Step 6: Determine the concentrations of \( N_3^- \) and \( OCN^- \).

- From N₃H:
\[ [N_3^-] = 6 \times 10^{-3} \, \text{M} \]

- From HOCN:
\[ [OCN^-] = 8 \times 10^{-3} \, \text{M} \]

### Conclusion:

The final concentrations in the solution are:
- \( [H^+] \approx 1.4 \times 10^{-2} \, \text{M} \)
- \( [N_3^-] \approx 6 \times 10^{-3} \, \text{M} \)
- \( [OCN^-] \approx 8 \times 10^{-3} \, \text{M} \)