`50 ml, 0.1M NaHCO_(3)` aqueous solution is mixed with `50 ml, 0.8 M Na_(2)CO_(3)` aqueous solution at `25^(@)C`. The `pH` of resulting solution exiists between colour transition range of an indicator which is:
`(K_(a_(1))=4xx10^(-7), K_(a_(2)) =2xx10^(-11) "for" H_(2)CO_(3))`

Options

A. Phenol red `(6.8 "to" 8.4)`
B. Propyl red `(4.6 "to" 6.4)`
C. Phenolphathalein `(8.3 "to" 10.1)`
D. Malachite green `(11.4 "to" 13)`

Explanation

To solve the problem, we need to find the pH of the resulting solution after mixing 50 mL of 0.1 M NaHCO₃ with 50 mL of 0.8 M Na₂CO₃. We will use the Henderson-Hasselbalch equation for buffer solutions.

### Step-by-Step Solution:

1. **Calculate Moles of Each Component:**
- For NaHCO₃:
\[ \text{Moles of NaHCO}_3 = \text{Volume (L)} \times \text{Concentration (M)} = 0.050 \, \text{L} \times 0.1 \, \text{M} = 0.005 \, \text{mol} \]
- For Na₂CO₃:
\[ \text{Moles of Na}_2\text{CO}_3 = \text{Volume (L)} \times \text{Concentration (M)} = 0.050 \, \text{L} \times 0.8 \, \text{M} = 0.04 \, \text{mol} \]

2. **Total Volume of the Mixture:**
\[ \text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

3. **Calculate Concentrations in the Mixture:**
- Concentration of HCO₃⁻ (from NaHCO₃):
\[ [\text{HCO}_3^-] = \frac{0.005 \, \text{mol}}{0.1 \, \text{L}} = 0.05 \, \text{M} \]
- Concentration of CO₃²⁻ (from Na₂CO₃):
\[ [\text{CO}_3^{2-}] = \frac{0.04 \, \text{mol}}{0.1 \, \text{L}} = 0.4 \, \text{M} \]

4. **Calculate pKa2:**
- Given \( K_{a2} = 2 \times 10^{-11} \):
\[ pK_{a2} = -\log(K_{a2}) = -\log(2 \times 10^{-11}) \approx 10.7 \]

5. **Use the Henderson-Hasselbalch Equation:**
\[ pH = pK_{a2} + \log\left(\frac{[\text{CO}_3^{2-}]}{[\text{HCO}_3^-]}\right) \]
Substituting the values:
\[ pH = 10.7 + \log\left(\frac{0.4}{0.05}\right) \]

6. **Calculate the Logarithm:**
\[ \frac{0.4}{0.05} = 8 \quad \Rightarrow \quad \log(8) \approx 0.903 \]

7. **Final pH Calculation:**
\[ pH = 10.7 + 0.903 \approx 11.603 \]

8. **Determine the Indicator Range:**
The resulting pH of approximately 11.6 indicates that the pH exists within the color transition range of an appropriate indicator.

### Conclusion:
The pH of the resulting solution is approximately 11.6, which is within the color transition range of the indicator Malachite Green.