00 L sample of urban air was bubbled through a solution containing 50.0 mL of 0.0116 M Ba(OH)2, which caused the CO2 in the sample to precipitate as...

A 3.00 L sample of urban air was bubbled through a solution containing 50.0 mL of 0.0116 M Ba(OH)2, which caused the CO2 in the sample to precipitate as BaCO3. The excess base was back-titrated to a phenolphthalein endpoint with 23.6 mL of 0.0108 M HCl. Calculate the parts-per-million of CO2 in the air (i.e. mL CO2/10^6 mL air); use 1.98 g/L as the density of CO2.

1A]3.00 L sample of urban air was bubbled through a solution containing 50.0 mL of 0.0116M Ba(OH)2, which caused the CO2 in the sample to precipitate as BaCO3. The excess basewas back-titrated...