If 40.06mL of a standard 0.3814 M solution of NaOH is required to neutralize 48.70mL of H2SO4, what is the molarity of the acid solution?

When a solution is neutral moles of H+ and moles of OH- are equal. NaOH produces one OH- ion per molecule and H2SO4 produces two H+ ions per molecule. 

moles of OH- = (0.3814 moles/L)(40.06 ml)(1 L/1000 ml) = 0.01528 moles

moles H+ = moles OH- = 0.01528 moles

moles H2SO4 = (0.01528 moles H+)(1 mole H2SO4/2 moles H+) =0.007639 

molarity of H2SO4 = (0.007639 moles)/0.0487 L) = 0.1569 M

Here's a shortcut that's useful for neutralization problems:

Since moles acid (H+) = moles base (OH-),

(molarity of acid)(volume of acid)(n) = (molarity of base)(volume of base)(n) 

(n is the number of H+ ions produced per molecule of acid or OH- ions per molecule of base.) 

(M of acid)(48.70 ml)(2) = (0.3814 M)(40.06 ml)(1),

molarity of H2SO4 = 0.1569 M

When using this formula it's not necessary to convert volume to liters; the volumes just need to be in the same units.