100 mL of 0.1 M ` Na_2 CO _ 3` solution is titrated with `0.1 M ` HCl solution. If reaction taking placing during titrationis
` Na _ 2 CO _ 3 + HCl to Na HCO _ 3 + NaCl`
then find pH at half equivalent point. for `H_2 CO _ 3 , Ka_ 1 = 10 ^(-7)` and `Ka _ 2 = 10 ^( -11)`.

Explanation

To find the pH at the half-equivalent point during the titration of Na2CO3 with HCl, we can follow these steps:

### Step 1: Understand the Reaction
The reaction taking place is:
\[ \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaHCO}_3 + \text{NaCl} \]

At the half-equivalent point, half of the Na2CO3 has been converted to NaHCO3.

### Step 2: Calculate the Initial Moles of Na2CO3
We have 100 mL of 0.1 M Na2CO3 solution:
\[ \text{Moles of Na}_2\text{CO}_3 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} = 10 \, \text{mmol} \]

### Step 3: Determine Moles at Half-Equivalent Point
At the half-equivalent point, half of the moles of Na2CO3 will have reacted:
\[ \text{Moles of NaHCO}_3 = \text{Moles of Na}_2\text{CO}_3 \text{ at half point} = \frac{10 \, \text{mmol}}{2} = 5 \, \text{mmol} \]
\[ \text{Moles of Na}_2\text{CO}_3 \text{ remaining} = 10 \, \text{mmol} - 5 \, \text{mmol} = 5 \, \text{mmol} \]

### Step 4: Calculate Total Volume at Half-Equivalent Point
The total volume after adding HCl will be:
\[ \text{Total Volume} = 100 \, \text{mL} + 50 \, \text{mL} = 150 \, \text{mL} \]

### Step 5: Calculate Concentrations of NaHCO3 and Na2CO3
\[ \text{Concentration of NaHCO}_3 = \frac{5 \, \text{mmol}}{150 \, \text{mL}} = \frac{5}{150} \, \text{M} = \frac{1}{30} \, \text{M} \]
\[ \text{Concentration of Na}_2\text{CO}_3 = \frac{5 \, \text{mmol}}{150 \, \text{mL}} = \frac{5}{150} \, \text{M} = \frac{1}{30} \, \text{M} \]

### Step 6: Use the Henderson-Hasselbalch Equation
At the half-equivalent point, the pH can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa}_2 + \log\left(\frac{[\text{CO}_3^{2-}]}{[\text{HCO}_3^{-}]}\right) \]
Since at the half-equivalent point, \([\text{CO}_3^{2-}] = [\text{HCO}_3^{-}]\), the log term becomes zero:
\[ \text{pH} = \text{pKa}_2 \]

### Step 7: Calculate pKa2
Given \( K_{a2} = 10^{-11} \):
\[ \text{pKa}_2 = -\log(10^{-11}) = 11 \]

### Final Answer
Thus, the pH at the half-equivalent point is:
\[ \text{pH} = 11 \]