Moderate

20 ml of a H3PO4 solution needs 40 ml of 0.1 M NaOH to convert it into sodium dihydrogen phosphate. How much volume of 0.1 M Ca(OH)2 is needed to neutralize the same volume of the same H3PO4 completely?

Correct answer: D. 60ml

  • A. 120ml
  • B. 20ml
  • C. 40ml
  • D. 60ml

Explanation

To determine the volume of 0.1 M Ca(OH)2 needed to neutralize the same volume of the given H3PO4 solution, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction.The balanced chemical equation for the neutralization reaction between H3PO4 and Ca(OH)2 is as follows:H3PO4 + 3 Ca(OH)2 → Ca(H2PO4)2 + 2 H2OFrom the equation, we can see that 1 mole of H3PO4 reacts with 3 moles of Ca(OH)2 to form 1 mole of Ca(H2PO4)2 and 2 moles of water.Given that 20 ml of H3PO4 solution reacts with 40 ml of 0.1 M NaOH, we can calculate the number of moles of H3PO4 reacting:Number of moles of H3PO4 = Volume of H3PO4 solution (in L) × Molarity of H3PO4Number of moles of H3PO4 = 20 ml × (1 L / 1000 ml) × 0.1 mol/LNumber of moles of H3PO4 = 0.002 molesSince 1 mole of H3PO4 reacts with 3 moles of Ca(OH)2, the number of moles of Ca(OH)2 needed to react completely with the given H3PO4 is:Number of moles of Ca(OH)2 = Number of moles of H3PO4 × (3 moles of Ca(OH)2 / 1 mole of H3PO4)Number of moles of Ca(OH)2 = 0.002 moles × 3Number of moles of Ca(OH)2 = 0.006 molesNow, let's calculate the volume of 0.1 M Ca(OH)2 required to provide 0.006 moles:Volume of Ca(OH)2 solution (in L) = Number of moles of Ca(OH)2 / Molarity of Ca(OH)2Volume of Ca(OH)2 solution (in L) = 0.006 moles / 0.1 mol/LVolume of Ca(OH)2 solution (in L) = 0.06 LFinally, convert the volume to milliliters (ml):Volume of Ca(OH)2 solution = 0.06 L × 1000 ml/LVolume of Ca(OH)2 solution = 60 mlSo, 60 ml of 0.1 M Ca(OH)2 is needed to completely neutralize the same volume of the given H3PO4 solution.

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