Question

# One milliliter (1.00 mL) of acid taken from a lead storage battery is pipetted into a...

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.

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

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