Explain the equilibrium in soda water and explain it using Henry's law.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) In a closed soda water bottle,a saturated solution of $CO_{2}$ gas is maintained under high pressure. An equilibrium exists between the $CO_{2}$ molecules dissolved in water and those in the gaseous state:
$CO_{2(g)} \rightleftharpoons CO_{2(aq)}$ ... $(I)$
(At constant pressure and temperature)
Henry's law states that the mass of a gas dissolved in a given volume of a solvent at a constant temperature is directly proportional to the partial pressure of the gas present above the surface of the solution.
Mathematically,$m \propto p$ or $p = K_{H} \cdot x$,where $p$ is the partial pressure of the gas,$x$ is the mole fraction of the gas in the solution,and $K_{H}$ is Henry's law constant.
As the pressure of $CO_{2}$ above the liquid is high in a sealed bottle,the solubility of $CO_{2}$ in water is high. When the bottle is opened,the pressure of $CO_{2}$ above the liquid drops to the atmospheric pressure. According to Henry's law,the solubility of the gas decreases as the pressure decreases. Consequently,the excess dissolved $CO_{2}$ escapes from the solution to reach a new equilibrium,causing the soda water to become 'flat'.

Explore More

Similar Questions

If $O_2$ gas is bubbled through water at $303 \; K$,the number of millimoles of $O_2$ gas that dissolve in $1 \; L$ of water is (Nearest Integer). (Given: Henry's Law constant for $O_2$ at $303 \; K$ is $46.82 \; kbar$ and partial pressure of $O_2 = 0.920 \; bar$). (Assume solubility of $O_2$ in water is too small,nearly negligible).

What is the solubility of a gas in water at $25^{\circ} C$ if the partial pressure is $0.18 \ atm$? $(K_{H} = 0.16 \ mol \ dm^{-3} \ atm^{-1})$

Identify the reason for the solubility of polar solute in polar solvent from the following.

Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon $ . . . . . . $ .
$(i)$ Temperature $(ii)$ Nature of solute $(iii)$ Pressure $(iv)$ Nature of Solvent

Calculate the pressure of the gas if the solubility of the gas in water at $25^{\circ} C$ is $6.85 \times 10^{-4} \ mol \ dm^{-3}$. (Henry's law constant $K_H$ is $6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$) (in $bar$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo