Question

 

What is the pH of a solution if 10^-3 mol of NaHCO₃ is added to DI water? (ans=ph=8.79)

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. If however, the pKa of the carbonic acid system is changed, the pH changes accordingly with the same formula yielding the given pH. As the dissociation constant of carbonic acid is absent, the value from literature has been applied here.

pH = 7 - 0.5 x 5.3665 + 0.5 x log(0.001) = 8.9332

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 K_a1 > K_a2 and K_a1 >> K_w,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_a) - 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 - pOH = 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^-]