If 26.42 mL of a 0.1650 M NaOH solution is required to neutralize 30.40 mL of H2SO4, what is the molarity of?

Question

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If 26.42 mL of a standard 0.1650 M NaOH solution is required to neutralize 30.40 mL of H2SO4, what is the molarity of the acid solution?

Answer 1

Remember, titration is #moles of base reacting with the same number of moles of acid (H+).

First, figure out the #moles of NaOH you've added. M = (moles/L), so 0.1650 M = moles/(0.02642 L) or (0.1650*0.02642) = moles.

Now, that should be the same number of moles of the acid, provided the acid isn't multi-protic. Here it is (H2SO4 has two H+'s for every mole!). So it takes *twice as much base* to neutralize the acid. You need to divide the #moles NaOH by two to get the moles of H2SO4.

Now just divide #moles H2SO4 by the number of liters (0.03040) and it gives molarity.

Answer 2

moles of sulfuric acid = 1/2 moles of sodium hydroxide

moles of sodium hydroxide = molarity of s.h. x volume of s.h.
(Volume must be liters)

then:

molarity of sulfuric acid = moles of sulfuric acid / volume of sulfuric acid

(Volume must be in liters, combine equations and make the proper units arangement)

good luck.