Question

One milliliter (1.00 mL) of acid taken from a lead storage battery is pipetted into a flask. Water and phenolphthalein indicator are added, and the solution is titrated with 0.44 M NaOH until a pink color appears; 13.6 mL is required. Find, to within 5%, the number of grams of H2SO4 (formula weight = 98) present in 1 L of the battery acid.

Answer 1

During the titration process involving an acid titrated against a base, when the "end point" is reached, essentially all the acid and base have reacted with one another.

So find out first how many moles of NaOH was used.

Moles NaOH = molarity NaOH x volume (in L)

Moles NaOH = 0.44 M x 0.0136 liters = 0.006 moles of NaOH used,

The reaction in the titration is

H₂SO₄ + 2NaOH = Na₂SO₄ + 2 H₂O

The balanced equation shows that two moles of NaOH are required to neutralize one mole of H₂SO₄

so if we used 0.006 moles of NaOH we must have neutralized 0.003 moles of H₂SO₄

Now if 1 mL of battery acid contains 0.003 moles of H₂SO₄, then for 1 L (or l000 mL) of battery acid must contain l000 times 0.003 moles or 3 moles of H₂SO₄.

Molar mass of H₂SO₄ = 98 g/mol. That is 1 mole of H₂SO₄ weighs 98g.

so 3 moles of H₂SO₄ weighs 3 x 98 g = 294 grams