Question

Consider a solution that is 0.10 M in a weak triprotic acid which is represented by the gen- eral formula H3A with ionization constants K1 = 1.0×10−3, K2 = 1.0×10−8, and K3 = 1.0×10−12. What is the pH of the solution?

Answer 1

K₃ is very small,so K₂ and K₁ is resulting infinding pH
H₂A^- = H⁺ + HA^-
K₂ = [H⁺]²/H₂A^-

H₂A^- = [H⁺]²/K₂

H₃A = H⁺ + [H⁺]²/K₂

K₁ = ([H⁺]²/K₂)²/[H₃A]

[H⁺]⁴ = K₁*K₂²[H₃A]

[H⁺] = (K₁*K₂²[H₃A])^1/4 = (1.0×10^−3*(1.0×10^−8)² * 0.10)^1/4 = 1x10¹⁰

pH = -log [H⁺] = 10