A weak acid `(HA)` is titrated with `N//100 NaOH`. What will be the `pH` when `50%` of titration is completed. Given `K_(a)=1^(-4)` & concentration of `HA=0.1M`

Options

A. 4
B. 8
C. 6.9
D. 10

Explanation

To find the pH when 50% of the titration of a weak acid (HA) with NaOH is completed, we can follow these steps:

### Step 1: Understand the Titration Process
When a weak acid (HA) is titrated with a strong base (NaOH), the reaction can be represented as:
\[ HA + NaOH \rightarrow NaA + H_2O \]
At 50% completion, half of the weak acid has been converted to its conjugate base (A⁻).

### Step 2: Calculate the Concentrations at 50% Titration
Given:
- Initial concentration of HA = 0.1 M
- Volume of NaOH used = \( \frac{N}{100} \) (which is equivalent to 0.01 N)

At 50% titration:
- Concentration of HA remaining = \( \frac{0.1}{2} = 0.05 \, M \)
- Concentration of A⁻ formed = \( \frac{0.1}{2} = 0.05 \, M \)

### Step 3: Use the Henderson-Hasselbalch Equation
At this point, we can use the Henderson-Hasselbalch equation to find the pH:
\[ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
Where:
- \( pK_a = -\log(K_a) \)
- Given \( K_a = 1 \times 10^{-4} \)

### Step 4: Calculate pK_a
\[ pK_a = -\log(1 \times 10^{-4}) = 4 \]

### Step 5: Substitute Values into the Henderson-Hasselbalch Equation
At 50% titration:
\[ pH = pK_a + \log \left( \frac{0.05}{0.05} \right) \]
\[ pH = 4 + \log(1) = 4 + 0 = 4 \]

### Final Answer
The pH when 50% of the titration is completed is **4**.

---