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5.00 ml of an H2SO4 solution with unknown molarity was diluted with 25.00 ml. This was tirated with 15.00ml of .1000 M NaOH to reach the equivalence point. What is the molarity of the original unknown H2SO4 solution? Also, what would the balanced chemical equation for this reaction be?

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Answer:

0.1500 M

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Explanation:

The balanced chemical equation is:

  • H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

First we calculate how many NaOH moles reacted, using the volume and concentration:

  • 15.00 mL * 0.100 M = 1.5 mmol NaOH

Then we convert NaOH moles into H₂SO₄ moles:

  • 1.5 mmol NaOH * [tex]\frac{1mmolH_2SO_4}{2mmolNaOH}[/tex] = 0.75 mmol H₂SO₄

This amount of H₂SO₄ moles was originally contained within 5.00 mL of solution, thus the molarity is:

  • 0.75 mmol H₂SO₄ / 5.00 mL = 0.1500 M

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