Question

Calculate the pH and the equilibrium concentrations of H2PO4-, HPO42- and PO43- in a 0.0353 M aqueous phosphoric acid solution. For H3PO4, Ka1 = 7.5×10-3, Ka2 = 6.2×10-8, and Ka3 = 3.6×10-13

pH = ?

[H2PO4-] = ?M

[HPO42-] = ?M

[PO43-] = ?M

Answer 1

H3PO4 -----------> H+ +   H2PO4-

0.0353                      0               0

0.0353 - x                x               x

Ka1 = x^2 / 0.0353 - x

7.5×10^-3 = x^2 / 0.0353 - x

x = 0.01295

[H+] = 0.01295 M

pH = -log (0.0129)

pH = 1.89

[H2PO4-] = 0.0129 M

[HPO42-] = 6.20 ×10^-8 M

[PO43-] = 1.72 x 10^-18 M