Calculate the pH of rainwater (freshwater).

　Rainwater is assumed to be fresh water because it is condensed water vapor (*). Since it falls from the sky, carbon dioxide has reached dissolution equilibrium with the atmosphere and rainwater. The equation for dissolution equilibrium is,

K0 = 【H₂CO₃】／pCO₂ Equation (2).

The equation for the dissociation equilibrium of the carbonic acid component in water is,
K1 = 【HCO₃^-】・【H⁺】／【H₂CO₃】 Equation (3),

K2 = 【CO₃^2-】・【H⁺】／【HCO₃^-】 Equation (4).

The equation for the dissociation equilibrium of water is,

Kw = 【H⁺】・【OH^-】 Equation (5).

　In the previous problem, the relationship between the ratio of carbonic acid components and pH was illustrated. Rainwater that is just condensed water (before carbon dioxide is dissolved) is neutral (pH=7), so the ratio of CO₃^2- is not even 1%. If that is the case, let us assume that the ratio of CO₃^2- is small and the dissociation in equation (4) (【HCO₃^-】↔【CO₃^2-】+【H⁺】) need not be considered to make the calculation easier.

　We want to solve equations (2), (3), and (5) to find 【H⁺】, given the atmospheric CO₂ concentration (pCO₂), but we cannot solve this alone because there are 3 equations and 4 unknowns. I need to give some conditions.

Addition of conditional expressions: The ionic components in water are H⁺ for positive ions and HCO₃^- and OH^- for negative ions. If rainwater is electrically neutral (not charged), we can impose the condition that the positive and negative ionic charges are equal. (charge conservation)

電荷保存則：　【H⁺】= 【HCO3^-】 + 【OH^-】

Substitute 【HCO₃^-】 and 【HCO₃^-】 for this from Equations (3) and (5).

【H⁺】= K1 ・【H₂CO₃】／【H⁺】 + Kw／【H⁺】

Substitute 【H₂CO₃】 for this from Equation (2).

【H⁺】= K1 ・K0 ・pCO₂／【H⁺】 + Kw／【H⁺】

From this equation, if we transform the equation to find the hydrogen ion concentration,

【H⁺】= ( K1 K0×pCO₂ + Kw)^0.5Equation (9)

　Since the equilibrium constants for fresh water are K0 = 0.04 (mol/L/atm),  K1 = 4.0×10^-7 (mol/L), Kw = 10^-14 (mol/L), and pCO₂ = 3.8×10^-4 (atm) (**), substituting these into the above formula, 【H⁺】= 2.47×10^-6 (mol/L).

In terms of pH, the pH of rainwater = -Log(2.47 x 10^-6) = 5.61.

　Incidentally, the 【DIC】 of fresh water in contact with the atmosphere is,

【DIC】 =【HCO₃-】+【H₂CO₃】= 【H⁺】+K0・pCO₂ 

              = 2.465×10-6 + 0.04・3.8×10-4 ＝ 1.77×10-5 (mol /L) 

          = 0.0177 (mmol/L)

　Since the 【DIC】 of seawater is about 2 (mmol/L), seawater can dissolve nearly 100 times more DIC than fresh water. This is because, as explained earlier, seawater has alkalinity and contains enough DIC to counteract it. This is why seawater is a "huge reservoir of carbon". We will calculate the DIC concentration in seawater later.

　*If acidic substances (H₂SO₄, HCl, HNO₃, etc.) in the atmosphere are incorporated into cloud particles or raindrops, the pH of rainwater will drop dramatically. This calculation assumes the pH of clean rain where the effects of these pollutants are negligible. Rain with a pH lower than that of freshwater dissolved in carbonic acid (pH 5.61 in this calculation) is called "acid rain".

　**The concentration of CO₂ in the atmosphere, expressed as the ratio of the number of moles of CO₂ per unit mole (mole fraction) of the atmosphere, is currently about 380 ppm. Assuming CO₂ is an ideal gas, the mole fraction is equal to the partial pressure (atm).