[entice]

1. 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 (H2SO4) solution?

I used the equation mL x M = mL x M
(34.62(0.1510))/50 = 0.1045 M of H2SO4

Feels like somethings wrong because the balanced equation would be 2NaOH + H2SO4 >> Na2SO4 + 2H2O
and this equation doesn't account that 2 mole of NaOH reacts with 1 mole of H2SO4

Is there something I'm doing wrong?

[pear]

(.03462 L NaOH)(.1510 mol NaOH/L NaOH)(1 mol OH-/1 mol NaOH)(1 mol H+/1 mol OH-)(1 mol H₂SO₄/2 mol H+) = mols of H₂SO₄

(1 mol H+/1 mol OH-) because 1 mol of OH- is needed to neutralize 1 mol of H+
(1 mol H₂SO₄/2 mol H+) because each H₂SO₄ will provide 2 H+ when it disassociates

Divide that by the volume of the H₂SO₄ solution and you have yourself the molarity/concentration.

[Borek]

I used the equation mL x M = mL x M

And that was your mistake, this is NOT universal equation for titration calculation.

http://www.titrations.info/titration-calculation