For context, if you are unsure what this question is in reference to, feel free to take a look at this video by Bozeman science and go forward to about the 2:00 mark. In a typical cellular respiration lab, in particular the one outlined by the College Board, a respirometer and some peas are used to measure the rate of respiration. What confuses me is why KOH (potassium hydroxide) is necessary so that the interior volume of the respirometer won't change. Why wouldn't the solid formed by 2KOH combining with CO2 (K2CO3) cause a change in volume, since it occupies space in the respirometer just like the gasses? Sorry if this seems like a silly question, I have very poor background knowledge of chemistry.

For an animation, feel free to view this link from Pearson.



The solid precipitate does not contribute as much to the pressure inside the vial as the gas does. Intuitively, the gaseous molecule contributes to the pressure by collisions that exert force on the vessel (which would push the water out). By contrast, the solid precipitate only contributes to the pressure insofar as it decreases the volume of the vessel, which is negligible in comparison to the loss of kinetic energy per gas molecule.

Basically, just because they both occupy space, doesn't mean they occupy the same amount of space (or rather, they don't contribute equally to the pressure in the vial). Gases tend to occupy more space than liquid or solid phase molecules of identical size, due to their kinetic energy of motion and the force of their collisions on their surroundings. The KOH reacts with the CO2 to remove it from the pool of high energy gas molecules, converting them to low energy molecules in the solid. With the lower number of molecules contributing to the pressure (ideal gas law), the pressure is lowered and the force of the water becomes greater than the force of the gas, and the water enters the respirometer.

  • $\begingroup$ Cool. Is there anything I can add to improve the answer, like including a more quantitative explanation using the ideal gas law? $\endgroup$
    – Jory
    Apr 28 '16 at 5:54
  • $\begingroup$ No, I think I understand it. Thanks again! And sorry for forgetting to award you the bounty yesterday. $\endgroup$
    – AleksandrH
    Apr 28 '16 at 11:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.