Acid-Base Equilibria

80) m = 50.0 g Na₃PO₄           V = 1.00 L          K_a = 4.2 x 10^-13

      Na₃PO₄(aq) 3Na⁺(aq) + PO₄^3-(aq)

      [PO₄^3-] = n/V = 50.0 g Na₃PO₄ x 1 mol Na₃PO₄/163.94 g Na₃PO₄ x 1 mol PO₄^3-/ 1 mol Na₃PO₄/1.00 L

      [PO₄^3-] = 0.305 M

                   PO₄^3-(aq) + H₂O(l) HPO₄^2-(aq) + OH^-(aq)

      [ ]_i           0.305                                   0                  0

      [ ]_c                 -x                                    +x                +x

      [ ]_e              0.305-x                               x                  x

      K_b= [HPO₄^2-] x [OH^-]/[PO₄^3-]

      K_w = K_a x K_b = 1.00 x 10^-14

      K_b = K_w /K_a = 1.00 x 10^-14/4.2 x 10^-13 = 0.024

      0.024 = x · x/(0.305-x) ≈ x²/0.305

      [OH^-] = 8.6 x 10^-2 M

      %ion = [HPO₄^2-]/[[PO₄^3-] x 100%

      %ion = 8.6 x 10^-2 M/0.305 M x 100% = 28%

      %ion > ≈ 5%, therefore the assumption 0.305-x ≈ 0.305 is not valid and the quadratic formula must be used.

      0.024 = x²/(0.305-x)

      x² + 0.024x - 0.0073 = 0

      x = (-b ± (b² - 4ac)^1/2)/2a

      [OH^-] = (-0.024 ± ((0.024)² - 4 x (-0.0073))^1/2)/2 = 0.073 M

      pOH = -log[OH^-] = -log(0.073) = 1.14

      pH + pOH = 14.00

      pH = 14.00 - 1.14 = 12.86