Question

The CO2 concentration in the earth atmosphere is 390 ppm by volume. If you leave a beaker of pure water on the table until it reaches equilibrium with the air, what is the pH of the water? Known: Henry’s constant (KH’) for CO2 is 0.37 mol/L-atm, H2CO3 * ↔ H+ + HCO3- pKa = 6.35 * refers to both true H2CO3 and CO2 because they cannot be distinguished in water.

Answer 1

If you leave a beaker of pure water on the table until it reaches equilibrium with the air, the pH of the water is 10.19.

How to find pH?

When a beaker of pure water is left on the table, it will reach equilibrium with the air, and the CO₂ in the air will dissolve in the water. The CO₂ will then react with the water to form carbonic acid (H₂CO₃), which is a weak acid.

The carbonic acid will then dissociate to form hydrogen ions (H⁺) and bicarbonate ions (HCO₃⁻). The concentration of hydrogen ions will determine the pH of the water.

Calculate the concentration of CO₂ in water using Henry's law:

`CO_2_(concentration_in_water) = Henry_(constant) * CO_2_{concentration`

Where:

`CO_2_(concentration)_(in)_{water` = concentration of CO₂ in water (mol/L)

`Henry_{constant` = Henry's constant for CO₂ (0.37 mol/L-atm)

`CO_2_{concentration` = concentration of CO2 in air (390 ppm)

`CO_2_(concentration)_(in)_{water` = 0.37 mol/L-atm × 390e-6 mol/mol

= 0.0001463 mol/L

Calculate the concentration of H⁺ ions using the dissociation constant of carbonic acid (H₂CO₃):

`Ka = 10 ^ {-pKa`

`H_2CO_(3)_(concentration) = CO_2_(concentration)_(in)_(water) / (1 + Ka)`

`H_(ion)_(concentration) = Ka * H_2CO_(3)_{concentration`

Where:

Ka is the dissociation constant of carbonic acid (
`10^{-6.35`)

`Ka = 10 ^ {-6.35`

= 4.605e-7

`H_2CO_3{concentration` = 0.0001463 mol/L / (1 + 4.605e-7)

= 0.0001462 mol/L

`H_(ion)_{concentration` = 4.605e-7 × 0.0001462 mol/L

= 6.72e-10 mol/L

Calculate the pH of the water:

pH = -math.log10(
`H_(ion)_{concentration`)

pH = -math.log10(6.72e-10 mol/L)

= 10.19

Therefore, the pH of the water is 10.19.