Question

What is the pH of a solution if 10^-3 mol of NaHCO₃ is added to DI water?

Chemical reactions:

Species:

Charge balance:

Proton condition:

What if the carbonate minerals as CO₂ is added?

CB:

PC:

What if the carbonate minerals as Na₂CO₃ is added?

CB:

PC:

Answer 1

1. Sodium bicarbonate is an amphoteric salt which on adding to water gives hydroxide ion and carbonic acid with the latter dissociating very weakly, leaving the solution as a whole alkaline. The salt being a product of neutralization of a weak acid and a strong base, the rate determining parameter the K_a1 of carbonic acid is considered here for the pH calculation of the salt as it is that species which abstracts a proton from water to give hydroxide anion and that value is 4.3 x 10^-7 giving a pK_a1 of 5.3665 (Refer to any Salt Hydrolysis material for doubts).

NaHCO₃ <-----> Na⁺ + HCO₃^-

NaHCO₃ + H₂O <-----> Na⁺ + H₂CO₃ + OH^-

CBE: [Na⁺] + [H₃O⁺] = [HCO₃^-] + [OH^-]

PBE: [H₃O⁺] + [H2CO3] = [HCO₃-] + 2[OH^-]

Now, the pH is caluclated by applying Bronsted - Lowry theory and the equilibrium established by the salt, we get the formula pH = 14 - 0.5 x (pK) + 0.5 x log [B] where K is the dissociation constant and B is the base's concentration.

pH = 14 - 0.5 x 5.3665 + 0.5 x log(0.001) = 10.6675

2. CO₂ when added to water gives carbonic acid, H₂CO₃ which is a weak acid as it releases protons.

H₂CO₃ <-----> HCO₃^- + H⁺ pKa1 = 4.3 x 10-7

HCO₃^- <-----> CO₃^2- + H⁺ pKa2 = 5.6 x 10-11

Overall: H₂CO₃ <-----> CO₃^2- + 2H⁺

CBE: 2[H⁺] = [CO₃^2-]

PBE:2[H₃O⁺] = [CO₃^2-] + [OH^-]

Here, as Ka1 > Ka2 and Ka1 >> Kw, first ionization of carbonic acid is the major contributor to the acidity of the solution. Applying the same formula as before, we get,

pH = 0.5 x (pK) - 0.5 x log[HA] = 0.5 x (5.3665) - 0.5 x (-3) = 4.1833

3. Na₂CO₃ when added to water releases CO₃^2-. This in turn forms HCO₃^- and leaves OH^- in the medium due to the K_a2 of carbonic acid being very low and close to K_w (= 10^-14). Hence, the pH is calculated as:

pH = 14 - 7 + 0.5x(pK_a1 + log[OH]) = 7 + 0.5 x (5.3665 + -3) = 8.3325

Na₂CO₃ <-----> Na⁺ + CO₃^2-

CO₃^2- + H₂O <-----> HCO₃^- + OH^-

HCO₃^- + H₂O <-----> H₂CO₃ + OH^-

CBE: [Na⁺] + 2[H₃O⁺] = [CO₃^2-] + [OH^-]

PBE: 2[H₃O⁺] + [H₂CO₃] = 2[OH^-]