With CO2 of 409.15 ppm and a temperature of 31.1 degrees Celsius, determine the pH of Water.

pH = 4.2

Explanation:

CO2 + H2O → H2CO3

∴ C CO2 = 409.15 mg/L * (g/1000mg) * (mol CO2/44.01g) = 9.296 E-3 mol/L

H2CO3 ↔ HCO3 - + H3O+

∴ Ka = 4.3 E-7 = [H3O+] * [HCO3-] / [H2CO3]

mass balanace:

⇒ C H2CO3 = C CO2 = 9.296 E-3 = [H2CO3] + [HCO3-]

⇒ [H2CO3] = 9.296 E-3 - [HCO3-]

charge balance:

⇒ [H3O+] = [HCO3-]

⇒ 4.3 E-7 = [H3O+]² / (9.296 E-3 - [H3O+])

⇒ [H3O+]² + 4.3 E-7 [H3O+] - 3.997 E-9 = 0

⇒ [H3O+] = 6.300 E-5 M

⇒ pH = - Log [H3O+]

⇒ pH = - Log (6.3 E-5)

⇒ pH = 4.2