Question

Given a diprotic acid, H2A, with two ionization constants of Ka1 = 2.4x10^-4 and Ka2 = 4.9x10^-11, calculate the pH for 0.182 M solution of NaHA.

Answer 1

Answer:

pH = 6.964 which rounds to 7.0 if 2 sig figs

Explanation:

Ka₁ * Ka₂ = x

2.4x10⁻⁴ * 4.9x10⁻¹¹ = 1.176x10⁻¹⁴

√1.176x10⁻¹⁴ = 1.084435337x10⁻⁷

pH= -log[H⁺]

pH= -log[1.084435337x10⁻⁷]

pH= 6.964796339