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A 10.0 mL sample of H2SO4 was exactly neutralized by 23.5 mL of 1.00 M KOH. Calculate the molarity of the acid. H2SO4(aq) + KOH(aq) —> K2SO4(aq) + H2O(l)

Answer :

MAACASTROBRGO

Answer:

1.175 M

Explanation:

The balanced reaction is:

  • H₂SO₄(aq) + 2KOH(aq) —> K₂SO₄(aq) + 2H₂O(l)

First we calculate how many moles of KOH reacted, using the given concentration and volume:

  • 1.00 M * 23.5 mL = 23.5 mmol KOH

Then we convert moles of KOH to moles of H₂SO₄, using the stoichiometric coefficients of the balanced reaction:

  • 23.5 mmol KOH * [tex]\frac{1molH_2SO_4}{2molKOH}[/tex] = 11.75 mmol H₂SO₄

Finally we calculate the molarity of the acid:

  • 11.75 mmol H₂SO₄ / 10.0 mL = 1.175 M

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