Question

Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.53 × 10⁵ Pa at the same temperature.

Answer 1

Given: K_H = 1.67 × 10⁸ Pa

`P_(CO_2)` = 2.5 atm = 2.53 × 10⁵ Pa

According to Henry’s law,

`P_(CO_2) = K_H xx x_(CO_2)`

∴ `x_(CO_2) = P_(CO_2)/K_H`

= `(2.53 xx 10^5  Pa)/(1.67 xx 10^8  Pa)`

= 1.515 × 10⁻³

We can write, `x_(CO_2) = (n_(CO_2))/(n_(CO_2) + n_(H_2O)) = (n_(CO_2))/(n_(H_2O))`

= `1.515 × 10^(−3)  ...["Since", n_(CO_2)  "is negligible as compared to"  n_(H_2O)]`

For 500 mL of soda water, the volume of water = 500 mL; the mass of water = 500 g,

We can write:

= `500/18  "mol of water"`

= 27.78 mol of water

i.e., `n_(H_2O)` = 27.78

∴ `(n_(CO_2))/n_(H_2O) = x_(CO_2)`

`(n_(CO_2))/(27.78) = 1.515 xx 10^(−3)`

= 1.515 × 10⁻³ × 27.78

`n_(CO_2)` = 42.08 × 10⁻³ mol

= 0.042 mol

∴ Mass of CO₂ = `n_(CO_2) xx "Molar mass"`

= 0.042 × 10⁻³ × 44 g

= 1.848 g