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If 25.0 mL of 0.100 M Ca(OH)2 is titrated with 0.200 M HNO3, what volume of nitric acid is required to neutralize the base?

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Answer:

25 mL

Explanation:

The reaction that takes place is:

  • Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O

First we calculate how many Ca(OH)₂ moles were spent in the titration:

  • 25.0 mL * 0.100 M = 2.5 mmol Ca(OH)₂

Then we convert Ca(OH)₂ moles into HNO₃ moles, using the stoichiometric ratio:

  • 2.5 mmol Ca(OH)₂ * [tex]\frac{2mmolHNO_3}{1mmolCa(OH)_2}[/tex] = 5.0 mmol HNO₃

Finally we calculate the volume of required nitric acid solution, using the concentration:

  • 5.0 mmol ÷ 0.200 mmol/mL = 25 mL

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