What volume of a 0.1 M H₃PO₄ will be required to neutralize 45.0 cm³ of a 0.2 M NaOH?

JAMB Chemistry 2024 234 views
Source: JAMB · 2024

What volume of a 0.1 M H₃PO₄ will be required to neutralize 45.0 cm³ of a 0.2 M NaOH?

  1. 10.0 cm³
  2. 20.0 cm³
  3. 27.0 cm³
  4. 30.0 cm³ ✓
Explanation

Complete neutralisation: H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O. One mole of acid needs three moles of base.

Moles of NaOH = 0.2 × 45.0 ÷ 1000 = 0.009 mol.

Moles of H₃PO₄ = 0.009 ÷ 3 = 0.003 mol.

Volume = 0.003 ÷ 0.1 = 0.03 dm³ = 30.0 cm³.

Was this explanation helpful?

Something wrong or missing? Tell us — we read every report and fix issues fast.

pastQuestions.ng pastquestions.ng