The concentration of `CO_(2)` in atmosphere is 88ppm. If all of the `CO_(2)` present in `10^(5)mL` of air is dissolved in `1dm^(3)` water, then approximate `pOH` of solution at `27^(@)C` will be `(K_(a_(1)) = 10^(-7), K_(a_(2)) = 10^(-11)` for `H_(2)CO_(3)]`

Options

A. `3.2`
B. `3.85`
C. `10.15`
D. None

Answer

C

Explanation

`88 "ppm"` means:
`10^(6) mL` of air contains `= 88 of CO_(2)`
`= (88)/(44) = 2 mol CO_(2)`
Moles of `CO_(2)` is `10^(5) mL = (2)/(10^(6)) xx 10^(5) = 0.2mol`
`[CO_(2)] = (0.2mol)/(1dm^(3)) = 0.2M`
For `H_(2)CO_(3)`, consider only first dissociation constant `(K_(a_(1))` to calculate `pH` of solution.
`H_(2)CO_(3) hArr H^(o+) + HCO_(3)^(Theta)`
`[H^(o+)] = sqrt(K_(a_(1))C) = sqrt(10^(-7)xx0.2) = sqrt(2) xx 10^(-4)`
`pH =- log2^((1)/(2)) + 4`
`= -(1)/(2) xx 0.3 + 4 = 3.85`
`pOH = 14 - 3.85 = 10.15`