Match List I with List II and select the correct options given below `:`
List -I List -II
( P ) 10 ml 0.1 M NaOH aq. Solution is added to 10 0.1 M HCl aqueous solution ( 1) Salt of weak acid and strong base
(Q) 10 ml, 0.1 M NaOH aq. solution is added to 10 ml 0.1 M `CH_(3) COOH` aq. solution ( 2) Salt of weak base and strong acid
( R ) 10 ml 0.1 M HCl aq. solution is added to 10 ml 0.1 M `NH_(3)` aq. solution ( 3) pH of the solution will be 7
( s) 10 ml, 0.2 M HCl aq. solution to 10ml, 0.1 M KOH aq. solution ( 4) pH of the solution will be 1.33
Codes `:`

Options

A. `{:(P,Q,R,S),(3,1,2,4):}`
B. `{:(P,Q,R,S),(2,1,4,3):}`
C. `{:(P,Q,R,S),(4,1,3,2):}`
D. `{:(P,Q,R,S),(4,1,2,3):}`

Answer

To solve the problem, we need to match the items in List I with the corresponding items in List II based on the acid-base reactions described. Let's analyze each item step by step.

### Step 1: Analyze Item P
**Item P:** 10 ml 0.1 M NaOH solution is added to 10 ml 0.1 M HCl solution.

- **Reaction:** NaOH (strong base) + HCl (strong acid) → NaCl + H2O
- Since both the acid and base are strong and are present in equal volumes and concentrations, they will completely neutralize each other.
- **Result:** The pH of the solution will be 7 (neutral).

**Match:** P → 3 (pH of the solution will be 7)

### Step 2: Analyze Item Q
**Item Q:** 10 ml 0.1 M NaOH solution is added to 10 ml 0.1 M CH3COOH solution.

- **Reaction:** NaOH (strong base) + CH3COOH (weak acid) → CH3COONa + H2O
- The reaction produces a salt (sodium acetate) from a weak acid and a strong base.
- **Result:** The solution will contain the salt of a weak acid and a strong base.

**Match:** Q → 1 (Salt of weak acid and strong base)

### Step 3: Analyze Item R
**Item R:** 10 ml 0.1 M HCl solution is added to 10 ml 0.1 M NH3 solution.

- **Reaction:** HCl (strong acid) + NH3 (weak base) → NH4Cl
- The reaction produces a salt (ammonium chloride) from a strong acid and a weak base.
- **Result:** The solution will contain the salt of a strong acid and a weak base.

**Match:** R → 2 (Salt of weak base and strong acid)

### Step 4: Analyze Item S
**Item S:** 10 ml 0.2 M HCl solution is added to 10 ml 0.1 M KOH solution.

- **Calculating moles:**
- Moles of HCl = 0.2 M × 0.01 L = 0.002 moles
- Moles of KOH = 0.1 M × 0.01 L = 0.001 moles
- **Reaction:** HCl (strong acid) + KOH (strong base) → KCl + H2O
- The moles of HCl are greater than KOH, so HCl will be in excess.
- **Remaining moles of HCl:** 0.002 - 0.001 = 0.001 moles
- **Total volume of the solution:** 10 ml + 10 ml = 20 ml
- **Concentration of remaining HCl:** 0.001 moles / 0.02 L = 0.05 M
- **Calculating pH:** pH = -log[H+] = -log(0.05) ≈ 1.33

**Match:** S → 4 (pH of the solution will be 1.33)

### Final Matches:
- P → 3
- Q → 1
- R → 2
- S → 4

### Summary of Matches:
- P matches with 3
- Q matches with 1
- R matches with 2
- S matches with 4