Question

The concentration of CO2 in air is determined by an indirect acid– base titration. A sample of air is bubbled through a solution containing an excess of Ba(OH)2, precipitating BaCO3. The excess Ba(OH)2 is back titrated with HCl. In a typical analysis a 3.5-L sample of air was bubbled through 50.00 mL of 0.0200 M Ba(OH)2. Back titrating with 0.0316 M HCl required 38.58 mL to reach the end point. Determine the ppm CO2 in the sample of air given that the density of CO2 at the temperature of the sample is 1.98 g/L.

Answer 1

Vair = 3.5 L

mol of Ba(OH)2 = MV = 50*0.02 = 1 mmol of Ba(OH)2

mmol of HCl used = MV = 0.0316*38.58 = 1.2191 mmol

so... Ba(OH)2 neutralized = 1.2191/2 = 0.60955 mmol neutralized

so

Ba(OH)2 used in other BaCO3 reaction = 1-0.60955 = 0.39045 mmol of BaCO3

mol of CO2 = 1:1 ratio = 0.39045mmol = 0.39045*10^-3 mol of CO2

MW CO2 = 44 g/mol

mass of CO2 = mol *MW = 0.39045*10^-3)(44) = 0.0171798 grams

volume = mass/D = 0.0171798 / 1.98 = 0.008676 mL of CO2

ppm = mL/L = (0.008676)/3.5 = 0.00247885714 ppm of CO2 in air