# Why is the water solubility of CO worse than that of CO2?

I found many paper on the fitting of the solubility to some equations to be able to calculate the amount of $$\ce{CO}$$ in water, but I could hardly find any explanation on the low solubility. Only one paper [1] dealt with it.

Why is $$\ce{CO}$$ way less soluble in water than $$\ce{CO2}$$, despite, e.g., the first having a dipole moment and the latter doesn't? What am I missing?

[1] H. Sato, N. Matubayasi, M. Nakahara, F. Hirata, "Which carbon oxide is more soluble? Ab initio study on carbon monoxide and dioxide in aqueous solution," Chemical Physics Letters 2000, 323(3-4), 257–262 (https://doi.org/10.1016/S0009-2614(00)00508-X).

• Is this gonna be both Q&A? Jun 11, 2020 at 23:22
• Because if it's serious then your comparing it with wrong compound - comparison between CO and N2 would be interesting, but CO2 reacts with water, which changes everything. BTW @IvanNeretin don't even start that again... Jun 12, 2020 at 0:01
• @Mithoron I will if I have to. Jun 12, 2020 at 0:02
• I was going to comment on the polarity discussion here, but there's clearly some cross-talk I'm missing... :/
– Zhe
Jun 12, 2020 at 0:28
• Polar does not mean anything. It is among the most ambiguous words used in chemistry. Jun 12, 2020 at 5:32

The fact that $$\ce{CO2}$$ does not have a dipole moment does not really mean it's nonpolar. What $$\ce{CO2}$$ has is a quadrupole, two opposing dipoles that occupy different portions of space (opposite sides of the carbon atom).
Ordinarily we expect an electrostatic interaction with a quadrupole to be relatively weak, but in $$\ce{CO2}$$ the component dipoles are strong and at close range the water molecules can "home in" on one end of the molecule and interact selectively with the corresponding component dipole. Thereby the electrostatic interaction between water and the $$\ce{CO2}$$ quadrupole is stronger than the interaction between water and the weak $$\ce{CO}$$ dipole.