Question

Scientists determine the quantity of atmospheric sulfur dioxide (SO2), a large contributor to acid rain, by an indirect titration. First, a scientist collects a sample of air and reacts it with hydrogen peroxide (H,O,) to form a solution of sulfuric acid ( H, SO2). Then, the amount of H, SO, produced is determined by titration with a sodium hydroxide solution (NaOH) of known concentration. H,02(aq) + SO2(g) → H, 50,(aq) H, SO (aq) + 2NaOH(aq) + 2 H2O(l) + Na, So (aq) Suppose Lane collects a 879.0 g sample of air that is known to contain So, and reacts it with excess H, O,. A volume of 21.2 mL of 0.050 M NaOH is required to neutralize the H, SO, produced. Calculate the mass percent of So, in the air sample. mass percent: 3.86 x10-7 9% 50,

Answer 1

H₂O2 + 50 H₂SO4 + 2 Naon H₂SO4 -1) ango+ Nasou-lil - on moles of Noon used to neutralize Hasoy = M XV = 0.05X812mL = 1:06 m moles & moles of Naoh reacts with = I mole of H2SO4 = 1 mole of Masoy = 1X106 mmoles of Hasoy. 1:06 m moles = 0.53 mmoles of H2SO4. So, we have 0.53 mmoles of H₂SO4. from is equation I mole of H₂SO4 is produce by Imole of 50 2. o. Somolle . " " 0.53 mmoles o soa. (0.53 mmoles) - not mole normal -3 - because inmoles. mas g 0.83 mmoles of soa =lox0.5364 [now? = 33.92g x 10g We nam mass percent of soa in air = 33.92 x 100xio n adas 879 - 3.86x103 moles man sorry for the mistakes (erross)! we have calculated immoles that's why we are multiplying with 103