Why does CO2 lowers the pH of water below 7?

Ok so I understand that having a pH below 7 is considered to be acidic and I also understand that part of the definition of an acid is that it gives off H+ when dissolved in water(even though not all acids necessarily need to give off H+ when dissolved in water)


But I was the rather confused when I came upon this ( when they refer to the 2 gases they are talking about CO and CO2)
[![enter image description here](https://i.stack.imgur.com/1pbqR.png)](https://i.stack.imgur.com/1pbqR.png) 


What I dont understand is how the equation came to be ? ( the product side of the equation)


And how would It have looked if CO was dissolved in H2O, would it have still lowered the pH??




Carbon Dioxide (CO2) readily dissolve in water and form **Carbonic Acid (i.e H2CO3 (aq) )**
[![enter image description here](https://i.stack.imgur.com/axra3.png)](https://i.stack.imgur.com/axra3.png)


This is the formation of bonds.


[![enter image description here](https://i.stack.imgur.com/45oWk.gif)](https://i.stack.imgur.com/45oWk.gif)


Then Carbonic Acid (i.e H2CO3 (aq)  ) dissociate in water as follows.


[![enter image description here](https://i.stack.imgur.com/6hfV7.png)](https://i.stack.imgur.com/6hfV7.png)


So water gets H+ ions, so that cause water acidic.


The following shows dissociation of Carbonic Acid (i.e H2CO3 (aq) ) more clearly.


[![enter image description here](https://i.stack.imgur.com/b8pHj.jpg)](https://i.stack.imgur.com/b8pHj.jpg)


Carbon Monoxide (CO) do not readily react with water in room temperature in standard conditions. So it do not dissolve in water. 


 11

I once did a detailed calculation of this problem. Let me show it here. The partial pressure of $\ce{CO2}$ in the atomsphere is $3.8\times 10^{-4}\,\mathrm{atm}$. The solubility of an ideal gas in water is proportional to its partial pressure over the water surface. The proportionality constant is $0.034\,\mathrm{mol/(L\cdot atm)}$ at $25\,^\circ\mathrm{C}$. Among the dissolved $\ce{CO2}$ molecules in water, about $K\_h=1.7\times 10^{-3}$ (at $25\,^\circ\mathrm{C}$) of them combine with $\ce{H2O}$ molecules to form $\ce{H2CO3}$, which then dissociates and releases $\ce{H+}$ in two steps with $\,pK\_{a1}=3.58\,$ and $\,pK\_{a2}=10.32$, i.e.,


\begin{align}
&\ce{CO\_{2(aq)} + H2O <=> H2CO3},\quad K\_h=\frac{[\ce{H2CO3}]}{[\ce{CO\_{2(aq)}}]},\\
&\ce{H2CO3 <=> H+ + HCO3-},\quad K\_{a1}=\frac{[\ce{H+}][\ce{HCO3-}]}{[\ce{H2CO3}]},\\
&\ce{HCO3- <=> H+ + CO3^{2-}},\quad K\_{a2}=\frac{[\ce{H+}][\ce{CO3^{2-}}]}{[\ce{HCO3-}]}.
\end{align}


Therefore, we have


$$\ce{[CO\_{2(aq)}] + [H2CO3] + [HCO3-] + [CO3^{2-}]} = 0.034\times 3.8\times 10^{-4}\,\mathrm{mol/L},$$


and from charge neutrality


$$\ce{[H+] = [HCO3-] + 2[CO3^{2-}] + [OH-]}.$$


We then obtain the equation for $[\ce{H+}]$ given by


$$[\ce{H+}]-\frac{K\_w}{[\ce{H+}]}=\frac{\frac{K\_hK\_{a1}}{[\ce{H+}]}+\frac{2K\_hK\_{a1}K\_{a2}}{[\ce{H+}]^2}}{1+K\_h+\frac{K\_hK\_{a1}}{[\ce{H+}]}+\frac{K\_hK\_{a1}K\_{a2}}{[\ce{H+}]^2}}\times 1.29\times 10^{-5}\,\mathrm{mol/L},$$


where $K\_w=[\ce{H+}][\ce{OH-}]$ is the dissociation constant of water. The exact equation can be difficult to solve. Discarding insignificant terms, we obtain


$$[\ce{H+}]\approx\frac{K\_hK\_{a1}}{[\ce{H+}]}\times 1.29\times 10^{-5}\,\mathrm{mol/L},$$


which reduces to the familiar formula for the acidity of weak acids


$$[\ce{H+}]=\sqrt{K\_hK\_{a1}\times 1.29\times 10^{-5}\,\mathrm{mol/L}}=2.4\times 10^{-6}\,\mathrm{mol/L}.$$


Therefore the $\mathrm{pH}$ value of saturated carbonic acid is $\,\mathrm{pH}=5.6$, which is why we define rain water with $\,\mathrm{pH}<5.6\,$ as acid rain.


 4