In the titration of a 50.0 mL solution of H₂SO₄ of unknown concentration, it takes 30.2 mL of a 2.00 molar solution of NaOH to reach the end point. What is the molarity of the H₂SO₄ solution?

Question

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First, we can write a balanced chemical equation for the reaction that occurs during the titration: H₂SO₄ + NaOH → Na₂SO₄ + H2O From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of H₂SO₄. Therefore, the number of moles of NaOH used in the titration is equal to the number of moles of H₂SO₄ in the original solution. We can use the following equation to calculate the number of moles of NaOH used: moles NaOH = M NaOH x V NaOH where M NaOH is the molarity of the NaOH solution and V NaOH is the volume of NaOH added. Substituting the given values, we get: moles NaOH = 2.00 M x 0.0302 L = 0.0604 moles NaOH Since the number of moles of NaOH used is equal to the number of moles of H₂SO₄ in the original solution, we can calculate the molarity of the H₂SO₄ solution as follows: M H₂SO₄ = moles H₂SO₄ / V solution where V solution is the volume of the original H₂SO₄ solution, which is given as 50.0 mL. Substituting the calculated and given values, we get: M H₂SO₄ = 0.0604 moles / 0.0500 L = 1.21 M The molarity of the H₂SO₄ solution is therefore 1.21 M, which is closest to option D, 0.604 molar.

In the Titration of a 50