# How to measure/know the moles of carbon dioxide (as dry ice) when in KOH solution

I am doing an experiment where I am dropping dry ice in a $\ce{KOH}$ solution. I want to know how I can figure out how many moles of $\ce{CO2}$ reacted with the $\ce{KOH}$. I am pouring $25 \,\mathrm{g}$ of $\ce{KOH}$ in $100 \,\mathrm{mL}$ of water and I am using $50 \,\mathrm{g}$ of dry ice. The experiment is conducted at room temperature.

I know one way to do this is by isolating the product of $\ce{K2CO3}$ but because $\ce{K2CO3}$ is soluble, it is going to be very hard to isolate this product and weight it and use simple stoichiometry to figure out the moles of $\ce{CO2}$ reacted.

I was thinking maybe the pH meter will help but again, I do not know how to use the pH to calculate how many moles of carbon dioxide reacted.

### Observations

During experimentation, I saw the pH decreased when I added the $\ce{CO2}$.Will this fact be useful to calculate moles of $\ce{K2CO3}$ formed? The pH decreased from 13 to 10.

• weigh-difference of the solution from before and after adding $CO_2$ and letting it react? – user37142 Mar 21 '17 at 13:21
• The pH meter will only help if you know there is some direct correlation between the addition of $CO_2$ and the pH of the solution during the reaction. Do you know definitively if this is the case? – J. Ari Mar 21 '17 at 14:17
• No I do not I was only guessing that CO2 in solution behaves as acid so maybe there might be an increase in pH – user510 Mar 21 '17 at 21:37
• @Fl.pf. wouldn't there be conversation of mass – user510 Mar 21 '17 at 21:38
• @user510 Yes? But that has nothing to do with anyhting here. You have 100 ml of water with 25 ml KOH = 100g + how much 25 g KOH weighs. After it has reacted with the CO2 you have 100 ml Water + rest KOH + K2CO3 - whatever leaves the reaction. Then you can easily calculate how much CO2 reacted.. – user37142 Mar 22 '17 at 8:45

In this particular case weighing would work with limited accuracy (I am going to explain the limitations later in the answer).

Since the only compound leaving the reaction is the unreacted $\ce{CO2}$, it is valid to use weighing. Everything else is either sticking around or is already included in the initial weighing. So by weighing everything prior to reaction and after "all unreacted $\ce{CO2}$ is vaporized", you can calculate the amount of "reacted $\ce{CO2}$".

(weight of reacted $\ce{CO2}$) = (Weight of initial $\ce{CO2}$) - (weight of unreacted $\ce{CO2}$)

(Weight of initial $\ce{CO2}$) = (50 g) - (condensed water from atmosphere [estimate])

(Weight of unreacted $\ce{CO2}$) = (deficit in pre vs post reaction weighing)

pre reaction weight: 50 g $\ce{CO2}$ + 25 g $\ce{KOH}$ +100 g water = 175 g

post reaction weight: weight of the whole solution after all unreacted $\ce{CO2}$ is escaped from the solution (make sure the whole solution is at room temperature and make sure there is no bubbling occurring).

I understand there are some practical limitations, mainly :

1- There might be unreacted/frozen $\ce{CO2}$ still hanging around, Try to shake well. Also when weighing post reaction, make sure your solution is slightly warm to assist unreacted $\ce{CO2}$ to evaporate fully from the reaction.

2-Since you have dry ice, some water from atmosphere will condense into your reaction, this will lead to underestimation of the "leaving compound". The way to overcome, is to keep your reaction vessel as small as possible and control its inlet by a cold trap to make sure no water from atmosphere gets in. Also account for amount of water usually present in 50 g dry ice, by examining the water content on a similar sample beforehand.

pH: Your initial pH is clearly not right. concentration of your $\ce{OH-}$ is way higher than just 0.1 M. Let's assume that it is due to high pH value being out of the calibration range of the pH-meter and let's assume that your pH=10 after addition of $\ce{CO2}$ is correct.

One will be able to calculate the amount of unreacted $\ce{CO2}$ from reaction balance. You are starting with 25 g (0.446 moles) of $\ce{KOH}$ and 50 g (1.14 moles) of $\ce{CO2}$. For every $\ce{CO2}$ reacting, 2 $\ce{KOH}$ are used up. So IF all(or most) of $\ce{KOH}$ is used up, you should end up with 0.446/2~ 0.22 moles of $\ce{K2CO3}$, which means the reacted CO2 is about 0.22 moles(~9.7 g), and unreacted is 0.92 moles(~40.3 g). Why do I make the assumption that all the $\ce{KOH}$ is reacted, notice your final pH is 10 , meaning that your [OH-] concentration is about 0.0001 M, your starting concentration was 4.4 M.

What stops the reaction from using all $\ce{KOH}$ to be used up and dropping the pH to neutral zone? You are using excess $\ce{CO2}$, so we expect all the $\ce{KOH}$ to be used up. This has to do with part of $\ce{CO2}$ escaping the reaction as gas, also since the temperature is freezing the water (by dry ice) you may not have complete agitation which hinders the reaction procession to some extent.

I still think that you should take the weighing method seriously. Using weighing methods will provide confirmation for your calculations.

• so you want me to evaporate the solution. I do not get how I will weight "evaporation of all unreacted CO2" – user510 Mar 24 '17 at 1:46
• No no, don't evaporate the whole solution. Just letting the sit and reach room temperature will make most of the CO2 to evaporate. – Kinformationist Mar 24 '17 at 3:38
• i edited my question after i saw some observation when i did the experiment. Can you let me know if the observations help to go the pH way? – user510 Mar 24 '17 at 17:49
• The pH should decrease after the CO2 addition, right? so please edit the question again. I will put up something for calculation. – Kinformationist Mar 24 '17 at 19:46
• i did: it went from 13 pH to 10 – user510 Mar 24 '17 at 20:15