You are given 5.00 mL of an H₂SO₄ solution of unknown concentration. You divide the 5.00-mL sample into five 1.00-mL samples and titrate each separately with 0.1000 M NaOH. In each titration, the H₂SO₄ is completely neutralized. The average volume of NaOH solution used to reach the endpoint is 15.3 mL. What was the concentration of H₂SO₄ in the 5.00-mL sample? A) 0.306 M B) 0.765 M C) 3.06 M D) 1.53 × 10^-3 M E) 1.53 M

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In each of the five titrations, 0.1000 M NaOH was used to neutralize the H₂SO₄ in 1.00 mL of the unknown solution. This means that the number of moles of NaOH used in each titration is equal to the number of moles of H₂SO₄ in the 1.00 mL sample. Therefore, the concentration of the unknown solution can be determined as follows: Concentration of H₂SO₄ = moles of H₂SO₄ / volume of solution (in L) Since 0.1000 M NaOH was used to neutralize the H₂SO₄ in 1.00 mL of the unknown solution, the number of moles of H₂SO₄ in each titration is: moles of H₂SO₄ = (0.1000 mol/L) x (volume of NaOH used, in L) The average volume of NaOH solution used in the five titrations is 15.3 mL, which is equivalent to 0.0153 L. Therefore, the average number of moles of H₂SO₄ in the 1.00 mL samples is: average moles of H₂SO₄ = (0.1000 mol/L) x (0.0153 L) = 0.00153 mol To determine the concentration of H₂SO₄ in the original 5.00 mL sample, we need to convert the average number of moles of H₂SO₄ in a 1.00 mL sample to the number of moles in the entire 5.00 mL sample: moles of H₂SO₄ in 5.00 mL = (average moles of H₂SO₄ / 1.00 mL) x (5.00 mL) = 0.00765 mol Finally, we can calculate the concentration of H₂SO₄ in the original 5.00 mL sample: Concentration of H₂SO₄ = moles of H₂SO₄ / volume of solution (in L) = 0.00765 mol / 0.00500 L = 1.53 M Therefore, the answer is B, 0.765 M (since 5.00 mL is only one-third of 15.00 mL, the concentration would be one-third of 1.53 M).

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