Answer:
pH = 8.34
Step-by-step explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ HCO₃⁻ + H⁺ Ka1 -Here, HCO₃⁻ is acting as a base-
HCO₃⁻⇄ CO₃²⁻ + H⁺ Ka2 -Here, is acting as an acid-
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
pH = 8.34