Question

Phosphoric acid, H3PO4(aq), 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 0.500 M phosphoric acid solution.

pKa1= 2.16 pKa2 = 7.21 pKa3 = 12.32

What's: H3PO4, H₂PO₄^-, HPO₄^-2, PO4^-3, H⁺ ,OH^- , pH

Answer 1

Solution :-

Using the give pka values lets calculate the ka values

Ka= antilog [-pka]

Ka1 = antilog[-2.16] = 6.918*10^-3

Ka2 = antilog[-7.21] =6.166*10^-8

Ka3=antilog [-12.32]=4.786*10^-13

Now using the ka values lets calculate the concentration of the H+

H3PO4 ---- > H+   + H2PO4^-

Ka1 = [H+] [H2PO4^-] /[H3PO4]

Ka1 = [x][x]/[0.500-x]

6.918*10^-3 = x^2 / 0.500-x

6.918*10^-3 * 0.500-x =x^2

x= 0.0555 M = [H+] [H2PO4-]

now lets use ka2

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

Ka2 = [x][x]/[0.0555-x]

6.166*10^-8 = x^2 / 0.0550-x

6.166*10^-8 * 0.0550-x =x^2

x= 0.0555 M = [H+] [HPO4^2-]

x= 5.82*10^-5 = [H+] =[HPO4^2-]

now lets use the ka3

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

Ka3 = [x][x]/[5.82*10^-5-x]

4.786*10^-13 = x^2 / 5.82*10^-5-x

4.786*10^-13 * 5.82*10^-5-x =x^2

x= 5.28*10^-9 M = [H+] [PO4^3-]

Therefore the concentration of the each species are as follows

[H3PO4] =0.500 -0.0555 = 0.445 M

[H2PO4^2-] = 0.05550 – 5.82*10^-5 = 5.49*10^-2 M

[HPO4^2-] = 5.82*10^-5 M

[PO4^3-] = 5.28*10^-9 M

[H+] =0.0555 +5.82*10^-5 = 0.0555 M

[OH-] = kw / [H+]

           = 1*10^-14 /0.0555 M

          = 1.80*10^-13 M

pH= -log [H+]

   = -log [0.0555]

     = 1.26