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(M) of all species in a 0.300 M phosphoric acid solution.

pKa1= 2.16 pKa2=7.21 pKa3=12.32

[H3PO4]=

[H2PO4-]=

[HPO4^2-]=

[PO4^3-]

[H+]=

[OH-]=

pH=

Answer 1

Answer – We are given, [H₃PO₄] = 0.300 M

pK_a1 = 2.16, pK_a2 = 7.21 and pK_a3 = 12.32

First dissociation-

First we need to put ICE table for H₃PO₄

     H₃PO₄+ H₂O ------> H₂PO₄^- + H₃O⁺

I     0.30                      0            0

C      -x                      +x          +x

E   0.30-x                  +x          +x

K_a1 = [H₂PO₄^-] [H₃O⁺] / [H₃PO₄]

Now we need to calculate K_a1 from the pK_a1

We know,

pK_a1 = -log K_a1

so, K_a1 = 10^-pKa1

K_a1 = 10^-2.16

       = 6.92*10^-3

6.92*10^-3 = x *x / (0.30-x)

We need to arrange this equation in the quadratic form –

6.92*10^-3 (0.30-x) = x²

0.00207 - 6.92*10^-3 x =x²

x² + 6.92*10^-3 x – 0.00207 = 0

using the quadratic equation –

x = -b +/- √b²-4a*c / 2a

solving this for x by plugging the values

x = 0.0422 M

x = [H₂PO₄^-] = [H₃O⁺] = 0.0422

[H₃PO₄] = 0.30-x

               = 0.30-0.0422

               = 0.258 M

Second dissociation

   H₂PO₄^- + H₂O -----> HPO₄^2- + H₃O⁺

I     0.0422                  0               0

C      -x                     +x             +x

E   0.0422-x              +x            +x

K_a2 = [HPO₄^2-] [H₃O⁺] / [H₂PO₄^-]

K_a2 = 10^-pKa2

K_a2 = 10^-7.21

       = 6.16*10^-8

6.16*10^-8 = x *x / (0.0422-x)

We can neglect x in the 0.0422-x, since Ka2 value is too small

6.16*10^-8 *0.0422 = x²

x² = 2.60*10^-9

x = 5.10*10^-5 M

x = [HPO₄^2-] = [H₃O⁺] = 5.10*10^-5 M

[H₂PO₄^-] = 0.0422-x

               = 0.0422 - 5.10*10^-5 M

               = 0.0421 M

Third dissociation –

       HPO₄^2- + H₂O -----> PO₄^3- + H₃O⁺

I     5.10*10^-5                  0               0

C      -x                        +x             +x

E   5.10*10^-5 -x            +x            +x

K_a3 = [PO₄^3-] [H₃O⁺] / [HPO₄^2-]

K_a3 = 10^-pKa3

K_a3 = 10^-12.32

       = 4.78*10^-13

4.78*10^-13 = x *x / (5.10*10^-5 -x)

We can neglect x in the 5.10*10^-5-x, since Ka3 value is too small

4.78*10^-13 *5.10*10^-5 = x²

x² = 2.44*10^-17

x = 4.94*10^-9 M

x = [PO₄^3-] = 4.94*10^-9 M

[HPO₄^2-] = 5.10*10^-5 -x

                = 5.10*10^-5 -4.94*10^-9 M

               = 5.10*10^-5 M

Total [H₃O⁺] = 0.0422 + 5.10*10^-5 M + 4.94*10^-9 M

                      = 0.0422 M

We know,

[H₃O⁺] [OH^-] = 1*10^-14

So, [OH^-] = 1*10^-14 / [H₃O⁺]

                 = 1*10^-14 / 0.0422 M

               = 2.37*10^-13 M

Now pH –

pH = -log [H₃O⁺]

      = -log 0.0422

      = 1.37