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10.0 L of air of STP was slowly bubbled ...

`10.0 L` of air of `STP` was slowly bubbled through `50 mL` of `N//25Ba(OH)_(2)` solution and the final solution rendered red with phenoophthalein. After filtering the solution from the precipitated `BaCO_(3)`, the filtrate requried `22.5 mL` of `N//12.5 HCl` to becomes just colourless. Calculate the % age by volume of `CO_(2)` in the air.

Text Solution

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Air contains `CO_(2)` (acidic oxide) which reacts with `Ba(OH)_(2)` as: `Ba(OH)_(2) + CO_(2) rarr BaCO_(3) + H_(2) O`
It is clear from the data the `CO_(2)` in the air is not sufficient to neutrialise whole of hydroxide, so `HCl` is used to completely neutralise the solution.
mEq of `Ba(OH)_(2) = (1)/(12.5) xx 22.5 = 1.8`
Initial mEq of `Ba (OH)_(2) = (1)/(25) xx 50 = 2`
mEq of `Ba(OH)_(2)` neutralised by `CO_(2)` in the air `= 2 - 1.8 = 0.2`
As `Ba(OH)_(2)` is diacidic base:
1 mEq of `Ba(OH)_(2) = 1//2` millimoles of `Ba(OH)_(2)`
`implies` Millimoles of `Ba(OH)_(2) = 0.2//2 = 0.1`
`implies` Moles of `Ba(OH)_(2) = 0.1 xx 10^(-3) = 10^(-4)` moles
From stoichiometry, we have:
1 mol of `Ba(OH)_(2) -=` 1 mole of `CO_(2)`
`implies` Moles of `CO_(2) = 10^(4)`
`implies` volume at `STP = 10^(-4) xx 22.4 L`
Percentage of `CO_(2) = (22.4 xx 10^(-4))/(10) xx 100 = 0.0224 %`
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