What volume of 0.2 M KOH is required to fully neutralize 20 ml of 0.1 M H₂SO₄?

  1. 20 ml
  2. 10 ml
  3. 40 ml
  4. 5 ml

📝 Batafsil tushuntirish

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O.
n(H₂SO₄) = 0.02 L * 0.1 M = 0.002 mol.
n(KOH) needed = 0.002 * 2 = 0.004 mol.
Volume KOH = 0.004 / 0.2 = 0.02 L = 20 ml.
25 / 30 (83%)
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