Why does CO2 lowers the pH of water below 7?

Question

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)

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??

Answer 1

Carbon Dioxide (CO₂) readily dissolve in water and form Carbonic Acid (i.e H₂CO_{3 (aq)} )

This is the formation of bonds.

Then Carbonic Acid (i.e H₂CO_{3 (aq)} ) dissociate in water as follows.

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

The following shows dissociation of Carbonic Acid (i.e H₂CO_{3 (aq)} ) more clearly.

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

Answer 2

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.