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

$CO_2+Ba(OH)_2\to BaCO_3+H_2O$$Ba(OH)_2+2HCl\to BaCl_2+2H_2O$$MolesofHCladded=molarity ×volume=0.0316M×38.58×10^{-3}=1.219×10^{-3}mol$$MolesofBa(OH)_2added =50×10^{-3}×0.02=10^{-3}mol$mol

Excess Ba(OH)2​ added

1$Ba(OH)_2$ requires $2HCl$

$6.095×10^{-4} Ba(OH)_2$ require$1.219×10^{-3}HCl$

$Ba(OH)_2$ reacted$=10^{-3}-(6.095×10^{-4})=3.9×10^{-4} mol$mol

$CO_2$ Concentration $=\frac{3.9×10^{-4}mol}{Volume of CO_2}$

$=\frac{3.9×10^{-4}mol}{3.5}=1.157×10^{-4}mol/L$

$=4.9×10^{-3}g/l=4.9mg/l$

$=4.9ppm$