Consider the following statements:
(a) The pH of a mixture containing 400 mL of 0.1 `MH_(2)SO_(4)` and 400 mL of 0.1 M NaOH will be approximately 1.3.
(b) Ionic product of water is termperature dependent.
( c) A monobasic acid with `K_(a) = 10^(-5)` has a pH =5. The degree of dissociation of this acid is `50%`
(d) The Le Chatelier's principle is not applicable to common-ion effect.

Options

A. a and b
B. a,b and c
C. b and c
D. a,b and d

Explanation

To solve the question, we need to analyze each statement one by one and determine their correctness.

### Statement (a):
**The pH of a mixture containing 400 mL of 0.1 M H₂SO₄ and 400 mL of 0.1 M NaOH will be approximately 1.3.**

1. **Calculate the moles of H₂SO₄:**
- Volume = 400 mL = 0.4 L
- Molarity = 0.1 M
- Moles of H₂SO₄ = Molarity × Volume = 0.1 mol/L × 0.4 L = 0.04 moles
- Since H₂SO₄ is a diprotic acid, it will produce 2 moles of H⁺ for every mole of H₂SO₄.
- Moles of H⁺ = 0.04 moles × 2 = 0.08 moles.

2. **Calculate the moles of NaOH:**
- Volume = 400 mL = 0.4 L
- Molarity = 0.1 M
- Moles of NaOH = Molarity × Volume = 0.1 mol/L × 0.4 L = 0.04 moles.

3. **Determine the limiting reagent:**
- H₂SO₄ provides 0.08 moles of H⁺ and NaOH provides 0.04 moles of OH⁻.
- The reaction will consume all of the NaOH, which means:
- Moles of H⁺ remaining = 0.08 - 0.04 = 0.04 moles.

4. **Calculate the total volume of the solution:**
- Total volume = 400 mL + 400 mL = 800 mL = 0.8 L.

5. **Calculate the concentration of H⁺:**
- Concentration of H⁺ = Moles of H⁺ / Total volume = 0.04 moles / 0.8 L = 0.05 M.

6. **Calculate the pH:**
- pH = -log[H⁺] = -log(0.05) ≈ 1.3.

**Conclusion for Statement (a):** This statement is **correct**.

### Statement (b):
**Ionic product of water is temperature dependent.**

1. **Understanding the ionic product of water (K_w):**
- K_w = [H⁺][OH⁻] and it varies with temperature.
- At 25°C, K_w = 1.0 × 10⁻¹⁴, but this value changes with temperature.

**Conclusion for Statement (b):** This statement is **correct**.

### Statement (c):
**A monobasic acid with Kₐ = 10⁻⁵ has a pH = 5. The degree of dissociation of this acid is 50%.**

1. **Given pH = 5, calculate [H⁺]:**
- [H⁺] = 10⁻⁵ M.

2. **Using the formula for Kₐ:**
- For a weak acid HA, Kₐ = [H⁺][A⁻] / [HA].
- Let the initial concentration of HA be C.
- At equilibrium, [H⁺] = [A⁻] = x and [HA] = C - x.
- Kₐ = x² / (C - x).

3. **Assuming 50% dissociation:**
- If α = 0.5, then x = 0.5C.
- Substitute into Kₐ: Kₐ = (0.5C)² / (C - 0.5C) = 0.25C / (0.5C) = 0.5.

4. **Setting Kₐ = 10⁻⁵:**
- 0.5 ≠ 10⁻⁵, hence the degree of dissociation cannot be 50%.

**Conclusion for Statement (c):** This statement is **incorrect**.

### Statement (d):
**The Le Chatelier's principle is not applicable to common-ion effect.**

1. **Understanding the common-ion effect:**
- The common-ion effect is a specific application of Le Chatelier's principle.
- When a common ion is added to a solution at equilibrium, the equilibrium shifts to counteract the change, which is exactly what Le Chatelier's principle states.

**Conclusion for Statement (d):** This statement is **incorrect**.

### Final Summary:
- Statement (a): Correct
- Statement (b): Correct
- Statement (c): Incorrect
- Statement (d): Incorrect