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3 November, 23:27

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

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  1. 3 November, 23:33
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    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
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