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During a titration a student neutralizes 50.0 ml of sulfuric acid with 75.0 ml of 0.25 m sodium hydroxide, as shown below. H2SO4(aq) + 2naoh (aq) → Na2SO4 (aq) + 2h2o (l), What was the molarity of the H2SO4?
Accepted Answer
From the balanced chemical equation: H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l), we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. So, the number of moles of NaOH used in the titration must first be calculated:
Number of moles of NaOH = Molarity of NaOH × Volume of NaOH in liters
Number of moles of NaOH = 0.25 M × 75.0 mL / 1000
Number of moles of NaOH = 0.01875 moles
Since 2 moles of NaOH react with 1 mole of H2SO4, the number of moles of H2SO4 must be:
Number of moles of H2SO4 = Number of moles of NaOH / 2
Number of moles of H2SO4 = 0.01875 moles / 2
Number of moles of H2SO4 = 0.009375 moles
Finally, the molarity of the H2SO4 can be calculated:
Molarity of H2SO4 = Number of moles of H2SO4 / Volume of H2SO4 in liters
Molarity of H2SO4 = 0.009375 moles / 50.0 mL / 1000
Molarity of H2SO4 = 0.1875 M
Therefore, the molarity of the H2SO4 was 0.1875 M.
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