2016年8月12日 · $$\begin{align} \ce{NaOH + CO2 &= NaHCO3}\\ \ce{NaHCO3 + NaOH &= Na2CO3 + H2O} \end{align}$$ Thus, only when the concentration of the alkali solution is quite low, the reaction proceeds via the formation of carbonic acid. But the acidic oxide is not completely neutralized in this case.

2021年8月10日 · You have trace concentration of a gas reacting with quiet (unstirred) solution of NaOH. So two steps are involved: (i) Absorption of carbon dioxide from a very dilute mixture (air) in NaOH. (ii) Irreversible reaction of carbon dioxide and NaOH to form sodium carbonate. All of this is mainly happening at the air-solution interface.

2019年10月10日 · The $\mathrm{pH}$ first increases and then starts decreasing but stop at the same level of initial $\mathrm{pH}$ value of the solution (The amount of $\ce{CO2}$: $\pu{50cc}=\pu{0.03g}$). Trials run for $\pu{60 min}$ after that, the samples are preserved at $\pu{4 ^\circ C}$ and $\mathrm{pH}$ measured again after $\pu{24 h}$ , but the value does ...

2019年1月23日 · $\ce{CO2}$ from the atmosphere reacts with the sample to increase the acidic nature of the solution and causes the disappearance of colour. This would explain speeding up of colour change by shaking. But, is the rate of absorption of CO2 into a still sample of $\ce{NaOH + Na2CO3}$ this fast? Other possibilities

Doing some rough back of the envelope calculations there would be about .016 moles O2 in the jar and with 300 mg CO2, that would be .006 moles used. If the CO2 respiration was a lot higher, that could definitely deplete all the O2. Even if all O2 was depleted in a sample, shouldn't that NaOH trap need less HCl to titrate than the blank? $\endgroup$

2023年5月13日 · $$\ce{2 NaHCO3(s) -> 2 NaOH(s) + 2 CO2(g)}$$ And how could you determine this without actually performing the experiment? I was thinking it may have to do with the fact that carbonate is a polyatomic ion and therefore stays that way, but I'm not sure.

2017年1月23日 · In sodium hydroxide any carbon dioxide is converted to the carbonate anion. In a strong sodium +1 solution sodium carbonate has limited solubility, and will ppt out. So until the NaOH is substantially reacted there really isn't any "dissolved" carbon dioxide. $\endgroup$ –

2016年12月17日 · $$\ce{2 NaOH + 2CO2 -> 2NaHCO3}$$ So in essence, for 1 mole of NaOH you react with one mole of $\ce{CO2}$ to produce one mole sodium bicarbonate. In totality, for every one mole $\ce{H2CO3}$, you react with one mole NaOH. For every direct reaction of NaOH with $\ce{CO2}$, the ratio is $1:1$. So I guess that's all reactions accounted for.

2014年2月14日 · When $\ce{CO2}$ escape from the solution (equilibrium $\mathrm{C_2}$ goes to the right) the concentration of the manly acid species goes down but the hydroxides from the equilibrium $\mathrm{B}$ are still present so at the end the solution in an open system with an atmosphere with a low partial pressure of $\ce{CO2}$ will have a higher pH, will ...

2019年1月31日 · You're right, both of these reactions take place. From Wikipedia, the Free Encyclopedia. In aqueous solution, carbonate, bicarbonate, carbon dioxide, and carbonic acid exist together in a dynamic equilibrium.