Why is it the $ceCO2$ is considerably much more soluble in water 보다 $ceO2$ is?

$ceCO2$ is nonpolar however dissolves in water which choose $ceCO2$ gift nonpolar, doesn"t make any sense.

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Is it the shortcut polarity and also that the bond polarity is an ext important in solubility?

If no than why is $ceCO2$ an ext soluble in water than $ceO2$.



Taken from my answer to your initial question

There are a pair of reasons why $ceCO2$ is an ext soluble in water 보다 $ceO2$. Since the two $ceC=O$ bonds in $ceCO2$ space polarized (whereas in $ceO2$ the bond is not polarized) it renders it less complicated for the polar water molecule come solvate it and to form hydrogen bonds. Both that these factors will stabilize a $ceCO2$ molecule much more than an $ceO2$ molecule in water; stabilization equates into higher solubility. One more factor improving the solubility of $ceCO2$ in water is the truth that $ceCO2$ reacts with water to set up an equilibrium with carbonic acid.$$ceCO2(aq) + H2O H2CO3(aq)$$This reaction will additionally enhance $ceCO2$"s solubility in water compared to oxygen i m sorry does not react v water.



That is since $ceCO2$ will react v water in equilibrium to kind $ceH2CO3$, which is an acid and thus will certainly dissociate to type ions that can conveniently fit in the polar solvent (i.e. Water). $ceO2$ on the other hand does not react with water since it is no a very great electrophile in comparison come $ceCO2$. The $pi$-bond is weak in $ceCO2$ (i.e. High in energy), partly since the $ceC=O$ link is polarized and also because of delocalisation in staying $pi$-bond that $ceCO2$. $ceO2$ does not have actually that luxury.


Though the entirety molecule has actually a network dipole moment of zero due to its geometry, the C=O bond has actually a dipole moment. Contrasted to the O=O shortcut which has no dipole moment.

Some folks throw about the formula H2CO3 as if it were a real compound. The presumption is that it is a weak acid, and thus does not ionize come a great extent. The truth of the matter is the there space no H2CO3 molecule in aqueous solution. Lock don"t exist. What we speak to "carbonic acid" is CO2 dissolved in water in equilibrium v slight amounts of H+ and also HCO3^-.

CO2(aq) + HOH(l) H+ + HCO3^- .............. Ka is small

A small Ka way that the equilibrium lies much to the left, and also that almost all of the CO2 continues to be CO2 and certainly doesn"t form molecular H2CO3.

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