JEE Main 2023 · 6 April · evening · Single correct · +4 / −1
The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)₂ is (Assume complete neutralization)
- (a) 7.5 mL
- (b) 5.0 mL
- (c) 10.0 mL
- (d) 2.5 mL
Answer and solution
Answer (c) 10.0 mL
Solution
Chemistry ma’am · 3 steps
Each Ba(OH)₂ gives two OH⁻, so count hydroxide millimoles before you match the acid.
- Count the base. Millimoles of Ba(OH)₂ = 0.01 × 10.0 = 0.1. Each formula unit releases two OH⁻ ions, so there are 0.2 millimoles of OH⁻.
2HBr + Ba(OH)₂ → BaBr₂ + 2H₂O
- Match the acid. HBr gives one H⁺ each, so you need 0.2 millimoles of HBr. Volume = 0.2/0.02 = 10.0 mL.
- So. The volume of HBr needed is 10.0 mL, which is option (c).
The idea to keep. In neutralisation, match millimoles of H⁺ to millimoles of OH⁻, not moles of acid to moles of base. Always multiply by how many H⁺ or OH⁻ each unit gives.
Answer from the NTA final answer key. Solution by PrepWiser.