Question

Ka=10^-pKa

Phosphoric acid, H_3PO_4(ag), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the pH, and the concentrations of all species in a 0.400 M phosphoric acid solution.

Answer 1

H3PO4     ---------------->   H2PO4- + H+

0.40                                       0          0

0.40 - x                                    x           x

Ka1 = [x][x]/[0.40-x]

6.92 x 10^-3 = x^2/(0.40-x)

x = 0.0493

[H3PO4] = 0.40 - 0.0493 = 0.351M

[H2PO4-] = x = 0.0493 M

[H+] = x = 0.0493M

almost maximum H+ ions comes from first ionisation so

pH = -log [H+] = -log(0.0493)

pH= 1.31

[OH-] = kw/[H+] = 1 x10^-14/0.029

[OH-] = 2.03 x10^-13 M

H2PO4- -----------------> HPO4^2- + H+

0.0493                               0               0.0493

0.0493-y                           y                0.0493+y

Ka2 =[HPO4^2][ H+]/[H2PO4-]

6.2 x10^-8 = (y)(0.0493+y)/(0.0493-y)

by solving this y = 6.2 x10^-8

second ionisation constant value always =[HPO4^-2]

[H2PO4-]= 0.029-y = 0.029-6.2 x10^-8

[HPO4^-2] =6.17 x10^-8 M

   [H2PO4-]     = 0.0493 M

HPO4^2 ---------------------> PO4^-3 + H+

6.2 x10^-8                            0              0.0493

6.2 x10^-8- z                        z                0.0493 +z

Ka3 = [ PO4^-3][H+]/[HPO4^2]

4.7 x 10^-13 = (z) x(0.0493 +z)/ (6.2 x10^-8- z )

z= 5.91 x 10^-19 M

[PO4^-3] = z= 5.91 x 10^-19 M