Calculate the pH of a 0.250M H3PO4 solution. Ka1 = 7.5E-3, Ka2 = 6.3E-8, and Ka3 = 3.6E-3?

Ka3 should be "3.6E-13" instead of "3.6E-3" instead.

As Ka1 ≫ Ka2 ≫ Ka3, the pH of the solution depends almost entirely on the first dissociation only.

Consider the first dissociation of H₃PO₄:

____________ H₃PO₄(aq) __ ⇌ __ H₂PO₄⁻(aq) __ + __ H⁺(aq) …… Ka1 = 7.5 × 10⁻³

Initial _______ 0.250 M ___________ 0 M __________ 0 M

Change _______ -y M ____________ +y M __________ +y M

Equilibrium _ (0.250 - y) M ________ y M ___________ y M

At equilibrium:

Ka1 = [H₂PO₄⁻] [H⁺] / [H₃PO₄]

7.5 × 10⁻³ = y² / (0.250 - y)

y² = (1.875 × 10⁻³) - (7.5 × 10⁻³)

y² + (7.5 × 10⁻³) - (1.875 × 10⁻³) = 0

y = {-(7.5 × 10⁻³) ± √[(7.5 × 10⁻³)² - 4(-1.875 × 10⁻³)]} / 2

y = 3.97 × 10⁻² or y = -4.72 × 10⁻² (rejected)

(As y ≈ 16% of 0.250, we CANNOT use the approximation that 0.250 ≫ y !)

[H⁺] = 3.97 × 10⁻² M

[H] = -log[H⁺] = -log(3.97 × 10⁻² M) = 1.40