## Acid-Base Equilibria

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

Na3PO4(aq) 3Na+(aq) + PO43-(aq)

[PO43-] = n/V = 50.0 g Na3PO4 x 1 mol Na3PO4/163.94 g Na3PO4 x 1 mol PO43-/ 1 mol Na3PO4/1.00 L

[PO43-] = 0.305 M

PO43-(aq) + H2O(l) HPO42-(aq) + OH-(aq)

[ ]i           0.305                                   0                  0

[ ]c                 -x                                    +x                +x

[ ]e              0.305-x                               x                  x

Kb= [HPO42-] x [OH-]/[PO43-]

Kw = Ka x Kb = 1.00 x 10-14

Kb = Kw /Ka = 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 = [HPO42-]/[[PO43-] 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