Question

3. Show how phosphoric, H₃PO₄, dissociates in water and calculate the pH of a 0.030 M solution of the acid. K_a1 = 7.5 x 10^-3; K _a2 = 6.2 x 10^-8; K_a3 = 4.8 x 10^-13

Answer 1

3)

H3PO4 + H2O (l) --------------> H3O+   +   H2PO4-

0.030                            0             0

0.030 - x                       x              x

Ka1 = [H3O+][H2PO4-] / [H3PO4]

7.5 x 10^-3 = x^2 / 0.030 - x

x^2 + 7.5 x 10^-3 x - 2.25 x 10^-4 = 0

x = 0.0117

[H3O+] = 0.0117 M

pH = -log [H3O+] = -log (0.0117)

pH = 1.93