34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H₂SO₄ solution. What is the concentration of the original sulfuric acid solution A) 0.0229 M B) 0.218 M C) 0.0523 M D) 0.209 M E) 0.105 M

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide is:

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

From the equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide. Therefore, the number of moles of NaOH used in the reaction is:

0.1510 M × 0.03462 L = 0.00522582 mol NaOH

Since two moles of NaOH react with one mole of H2SO4, the number of moles of sulfuric acid in the original solution is:

0.00522582 mol NaOH × (1 mol H2SO4 / 2 mol NaOH) = 0.00261291 mol H2SO4

Finally, we can calculate the concentration of the original sulfuric acid solution by dividing the number of moles by the volume:

C = n / V = 0.00261291 mol / 0.0500 L = 0.0523 M

Therefore, the answer is C.

3462 ML of 0