Question

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibrium equation.The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. For a CO2 partial pressure of 3.2×10–4 bar in the atmosphere, what is the pH of water in equilibrium with the atmosphere? (For carbonic acid Ka1 = 4.46× 10–7 and Ka2 = 4.69× 10–11).

Answer 1

ANS:

The most common and important source of acidity in water is dissolved carbon dioxide.

The Carbon dioxide present in atmosphere enters the water maintaining the equilibrium with the atmosphere

CO₂ (aq) CO₂ (g)

The Dissolved CO₂ present in the form of H₂CO₃ may loose up to two protons through the acid equilibria

H₂CO₃ (aq) H⁺ (aq) + HCO^3- (aq)                               eqn 1 , Ka₁ = 4.46 * 10^-7

HCO^3- (aq) H⁺ (aq) + CO3^2- (aq)                                 eqn 2 , Ka₂ = 4.69 * 10^-11

To account for the fact that CO₂ (aq) is in equilibrium with H₂CO₃ (aq), the first acid equilibrium is normally given by

Ka1 = [H⁺] [HCO₃^-]/ [CO₂(aq)]

Now, to determine the pH of water ,first, we calculate the amount of CO2 dissolved in water under an atmosphere of pressure using the Henry’s Law which is:

[CO₂(aq)] =K_CO2. P_CO2

P_co2 is given as 3.2 *10^-4 atm

Now,Since CO2 makes up 0.0355% of the atmosphere (on the average) and

K_CO2 =2x10^-3 (Known)

Thus, [CO₂(aq)] = 3.38 *10^-2 mole/L atm * 3.2 *10^{-4 atm}

                          =1.081 * 10^-5 mole/L

Since CO₂ is in equilibrium with H₂CO₃ (aq), the first acid equilibrium is normally written in the following form is prominent:

Ka₁ = [H⁺] [HCO₃^-] / [CO₂(aq)] = 4.46 *10^-7

since the proton and bicarbonate concentrations are equal, we can also write Ka₁ as:

Ka₁ = [H⁺]² / [CO₂]

Now, [H⁺] = ( Ka₁ * [CO₂])^1/2 =0.000002195 =2.195 *10^-6

Thus pH is calculated using for,ula:

pH = - log [H⁺]