Consider the following reaction equation: 2HCl + Ca(OH) 2 CaCl 2 + H 2 O . What is the volume of 0.1 moldm^ -3…

Analyze the question: We need the volume of 0.1 mol/dm³ HCl that will exactly neutralize 25 cm³ of 0.3 mol/dm³ Ca(OH)₂.

Concept: For a titration, find the moles of the known solution, then use the mole ratio from the balanced equation. The equation as written is not balanced (water should be 2H₂O), but the acid-to-base ratio is still 2 HCl : 1 Ca(OH)₂.

Solution:

1. Moles of Ca(OH)₂ = concentration × volume (in dm³) = 0.3 × (25/1000) = 0.0075 mol

2. From the equation, 1 mol Ca(OH)₂ reacts with 2 mol HCl: Moles of HCl = 2 × 0.0075 = 0.015 mol

3. Volume of HCl = moles ÷ concentration = 0.015 ÷ 0.1 = 0.15 dm³ = 150 cm³

Check using the titration formula: (C_a V_a)/(C_b V_b) = n_a/n_b (0.1 × V_a)/(0.3 × 25) = 2/1 0.1 V_a = 15 V_a = 150 cm³ ✓

Answer: A (150 cm³).

Tip: Always check the mole ratio from the balanced equation. If you forget the factor of 2 (the ratio of HCl to Ca(OH)₂), you would get 75 cm³ (option B), which is the common trap. Also, Ca(OH)₂ gives 2 OH⁻ ions per formula unit, so it is effectively "twice as strong" as its molarity suggests: 0.3 M Ca(OH)₂ supplies 0.6 M of OH⁻.