0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm3. If 15.0 cm3 of the solution requires 12.50 cm3 of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution

  • A 0.30 mol dm3
  • B 0.40 mol dm3
  • C 0.50 mol dm3
  • D 0.60 mol dm3
jamb 1999chemistry

The correct answer is A. 0.30 mol dm3

M1 V1 = M2 V2

∴ Mn Vn represent volume and molar concentration of HCL also MyVy represent volume and molar concentration of Na2CO3

MnVn = MyVy

0.25 * 15 = 12.50 * My

My = (0.25 * 15)/12.50

My = 0.3m/dm3

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