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Suppose 0.800 M NaOH (aq) is used to titrate 50.00 mL of an unknown H2SO4
(aq) solution. If it takes 15.00 mL of the NaOH (aq) to reach the end point, how
many moles of H2SO4 were present in the unknown solution? Enter your
unrounded answer.

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Eduard22sly

The number of mole of H₂SO₄ present in the unknown solution is 0.006 mole

Balanced equation

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, H₂SO₄ (nA) = 1

The mole ratio of the base, NaOH (nB) = 2

How to determine the molarity of H₂SO₄

  • Molarity of base, NaOH (Mb) = 0.8 M
  • Volume of base, NaOH (Vb) = 15 mL
  • Volume of acid, H₂SO₄ (Va) = 50 mL
  • Molarity of acid, H₂SO₄ (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 50) / (0.8 × 15) = 1 / 2

(Ma × 50) / 12  = 1 / 2

Cross multiply

Ma × 50 × 2 = 12

Ma × 100 = 12

Divide both side by 40

Ma = 12 / 100

Ma = 0.12 M

How to determine the mole of H₂SO₄

  • Molarity of H₂SO₄ = 0.12 M
  • Volume = 50 mL = 50 / 1000 = 0.05 L
  • Mole of H₂SO₄ =?

Mole = Molarity x Volume

Mole of H₂SO₄ = 0.12 × 0.05

Mole of H₂SO₄ = 0.006 mole

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