100 ml aqueous 0.1 molar `M(CN)_(2)`, (80% ionized) solution is mixed with 100 ml of 0.05 molar `H_(2),SO_(2)` solution (80% ionized). `(K_(b)`, of CN- = `10^(-6)`)

Options

A. pH of the solution is 8
B. the resultant solution is acidic buffer
C. the resultant solution is basic
D. the resultant solution is acidic in nature

Answer

To solve the problem, we need to analyze the mixing of two solutions: a cyanide solution and a sulfuric acid solution. We will calculate the concentrations of the ions in the solution after mixing and then determine the pH of the resulting solution.

### Step-by-Step Solution:

1. **Calculate the moles of `M(CN)₂`:**
- Volume of `M(CN)₂` solution = 100 mL = 0.1 L
- Molarity of `M(CN)₂` = 0.1 M
- Moles of `M(CN)₂` = Molarity × Volume = 0.1 mol/L × 0.1 L = 0.01 moles

2. **Calculate the moles of `CN⁻`:**
- Given that `M(CN)₂` is 80% ionized, the moles of `CN⁻` produced = 0.01 moles × 2 (since 1 mole of `M(CN)₂` produces 2 moles of `CN⁻`) × 0.8 (80% ionized) = 0.016 moles of `CN⁻`.

3. **Calculate the moles of `H₂SO₄`:**
- Volume of `H₂SO₄` solution = 100 mL = 0.1 L
- Molarity of `H₂SO₄` = 0.05 M
- Moles of `H₂SO₄` = Molarity × Volume = 0.05 mol/L × 0.1 L = 0.005 moles

4. **Calculate the moles of `H⁺`:**
- Since `H₂SO₄` is a strong acid and is 80% ionized, the moles of `H⁺` produced = 0.005 moles × 2 (since 1 mole of `H₂SO₄` produces 2 moles of `H⁺`) × 0.8 (80% ionized) = 0.008 moles of `H⁺`.

5. **Determine the reaction between `H⁺` and `CN⁻`:**
- The reaction is: `H⁺ + CN⁻ ⇌ HCN`
- Initial moles before reaction:
- `H⁺`: 0.008 moles
- `CN⁻`: 0.016 moles
- After the reaction:
- `H⁺` remaining = 0.008 - 0.008 = 0 moles
- `CN⁻` remaining = 0.016 - 0.008 = 0.008 moles
- `HCN` formed = 0.008 moles

6. **Calculate the total volume of the mixed solution:**
- Total volume = 100 mL + 100 mL = 200 mL = 0.2 L

7. **Calculate the concentrations of `CN⁻` and `HCN`:**
- Concentration of `CN⁻` = moles/volume = 0.008 moles / 0.2 L = 0.04 M
- Concentration of `HCN` = moles/volume = 0.008 moles / 0.2 L = 0.04 M

8. **Use the Henderson-Hasselbalch equation to find pH:**
- The pKa of `HCN` can be calculated from `Kb` of `CN⁻`:
- \( K_w = K_a \times K_b \)
- \( K_a = \frac{K_w}{K_b} = \frac{1 \times 10^{-14}}{10^{-6}} = 1 \times 10^{-8} \)
- \( pK_a = -\log(1 \times 10^{-8}) = 8 \)

- Now, using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
\[ pH = 8 + \log\left(\frac{0.04}{0}\right) \]
Since the concentration of acid is effectively zero due to complete reaction, we can say the solution is basic, and thus we can approximate the pH to be around 8.

### Final Answer:
The pH of the resulting solution is approximately 8.