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 0.55 M NaOH until a pink color appears;12.0 mL are required.The number of grams of H₂SO₄ (formula weight = 98)present in one liter of the battery acid is:

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of NaOH used in the titration: moles NaOH = Molarity x Volume = 0.55 mol/L x 0.012 L = 0.0066 mol Since the stoichiometric ratio of NaOH to H₂SO₄ is 1:1, there must have been 0.0066 mol of H₂SO₄ in the original sample. Next, we can calculate the concentration of H₂SO₄ in the original lead acid solution: concentration = (moles / volume) x formula weight = (0.0066 mol / 0.001 L) x 98 g/mol = 64.68 g/L However, the original sample was only 1 mL, so we need to adjust the concentration accordingly: concentration = 64.68 g/L x (1/0.001 L) = 64,680 g/L Finally, we convert to grams per liter: grams/L = 64,680 g/L x (1/1000) = 64.68 g/L Therefore, the answer is B) 323.

One Milliliter (1