Question

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 1

1.175 M

Step-by-step 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 *
  `(1molH_2SO_4)/(2molKOH)` = 11.75 mmol H₂SO₄

Finally we calculate the molarity of the acid:

• 11.75 mmol H₂SO₄ / 10.0 mL = 1.175 M