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 0.38 M NaOH until a pink color appears; 15.8 mL is required. Find, to within 5%, the number of grams of H₂SO₄ (formula weight = 98) present in 1 L of the battery acid.

Question

Quizplus · Accepted Answer

The correct formula for the acid in question is H2SO4 (sulfuric acid), with a formula weight of 98. The titration process indicates the neutralization of H2SO4 by NaOH. The equation for the reaction is: H2SO4 + 2NaOH → Na2SO4 + 2H2O. From the given data, 15.8 mL of 0.38 M NaOH is used to neutralize 1 mL of the acid. This means (15.8 mL) * (0.38 mol/L) = 6.004 moles of NaOH are used. Since it takes 2 moles of NaOH to neutralize 1 mole of H2SO4, the moles of H2SO4 in 1 mL are half of that, which is 3.002 moles in 1000 mL (or 1 L). The mass of H2SO4 in 1 L is then (3.002 moles) * (98 g/mole) = 294.196 g, which is approximately 290 g, considering a 5% error margin.

One Milliliter (1