A 3.00 L sample of air from a big city was bubbled through a solution containing 50.0 mL of 0.0116 M Ba(OH)2?

Question

a 3.00 L sample of air from a big city 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

Answer 1

find moles:

0.0500 L of 0.0116 mol/litre Ba(OH)2 = 0.00058 moles Ba(OH)2

0.0236 L of 0.0108 mol/litre HCl = 0.0002549 moles HCl

since 2 moles HCl react with 1 mole Ba(OH)2,

the 0.0002549 moles HCl found 0.0001274 moles of Ba(OH)2 unreacted

this means:

(0.00058 mol)  -  (0.0001274 mol)  = 0.0004526 moles of Ba(OH)2 must have reacted with the CO2

since Ba(OH)2 reacts with CO2 on a 1:1 ratio:

1 Ba(OH)2  &  1 CO2 -->  1 BaCO3  &  1 H2O ,

implies that there was 0.0004526 moles of CO2 in the 3.00 liters of air

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let's find the grams of CO2:

0.0004526 mol CO2 @ 44.01 g / mol = 0.01992 grams of CO2

let's find the ml's of CO2,(by the way, your density is off, CO2 Is 1.96g/liter at STP, it should be less than 1.96g/l in your city):

0.01992 g CO2 @ (-:-) 1.98 g/1000ml = 10.06 ml CO2

lets find ppm:

? ml CO2 in 1e6ml air @ 10.06 ml CO2 / 3000 ml air = 3353 ppm

your answer is(3 sig figs): 3350 ppm CO2