pH of a strong diprotic acid `(H_(2)A)` at concentrations:
(i) `10^(-4)` M, (ii) `10^(-4)` N
are respectively:

Options

A. `3.7` and `4.0`
B. `4` and `3.7`
C. `4` and `4`
D. `3.7` and `3.7`

Answer

To find the pH of a strong diprotic acid \( (H_2A) \) at the given concentrations, we will follow these steps:

### Step 1: Understand the nature of the diprotic acid
A diprotic acid can donate two protons (H⁺ ions) per molecule when dissolved in water. Therefore, for every mole of \( H_2A \), two moles of H⁺ will be produced.

### Step 2: Calculate pH for \( 10^{-4} \) M concentration
1. **Dissociation of \( H_2A \)**:
\[ H_2A \rightarrow 2H^+ + A^{2-} \]
For \( 10^{-4} \) M concentration of \( H_2A \), the concentration of H⁺ ions will be:
\[ [H^+] = 2 \times 10^{-4} \, \text{M} \]

2. **Calculate pH**:
\[ \text{pH} = -\log[H^+] = -\log(2 \times 10^{-4}) \]
Using the logarithmic property:
\[ \text{pH} = -\log(2) - \log(10^{-4}) = -\log(2) + 4 \]
The value of \( \log(2) \) is approximately \( 0.301 \):
\[ \text{pH} = 4 - 0.301 = 3.699 \approx 3.7 \]

### Step 3: Calculate pH for \( 10^{-4} \) N concentration
1. **Normality to Molarity**:
Since the diprotic acid can donate 2 protons, the n-factor is 2. Thus, the molarity can be calculated from normality:
\[ \text{Normality} = n \times \text{Molarity} \Rightarrow \text{Molarity} = \frac{\text{Normality}}{n} = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{M} \]

2. **Dissociation of \( H_2A \)**:
The concentration of H⁺ ions will be:
\[ [H^+] = 2 \times 5 \times 10^{-5} = 10^{-4} \, \text{M} \]

3. **Calculate pH**:
\[ \text{pH} = -\log[H^+] = -\log(10^{-4}) = 4 \]

### Final Results
- The pH of \( H_2A \) at \( 10^{-4} \) M is approximately **3.7**.
- The pH of \( H_2A \) at \( 10^{-4} \) N is **4**.

### Summary
The pH values for the strong diprotic acid \( H_2A \) at the specified concentrations are:
- (i) \( 3.7 \)
- (ii) \( 4 \)