Question

A solution was prepared by dissolving 0.0258 mol disodium phosphate in 500.0mL of deionized water. Phosphoric acid has the following pKa values: pKa1 = 2.148 , pKa2 = 7.198 , pKa3 = 12.375.

a) What was the pH of the resulting solution?

b) What was the concentration of H2PO4- in the solution?

Answer 1

Solution:

a) The following equilibrium will exists:

H₃PO4 -----------> H⁺ + H₂PO^4-

H₂PO4^- ------------> H⁺ + HPO₄^2-

HPO₄^2- ----------> H⁺ + PO₄^3-

Disodium phosphate will dissociate in the deionized water as follows:

Na₂HPO -----------> 2Na⁺ + HPO₄^2-

0.0258             0.0258             0.0258

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

0.0258             0          0 (Initial)

0.0258-x          x          x (Equilibrium)

Given: pKa₃ = 12.375

Or, pKa₃ = -log(Ka₃)

Or, 12.375 = -log(Ka₃)

Or, log(Ka₃) = - 12.375

Or, Ka₃ = 10^-12.375 = 4.2x10^-13

Now, Ka₃ = [H⁺][PO₄^3-]/[HPO₄^2-]

Or, 4.2x10^-13 = x²/(0.0258-x)

Or, 4.2x10^-13 = x²/(0.0258);      {neglecting ‘x’ in the denominator since very weak acid}

Or, x² = 1.087x10^-14

Or, x = 1.04x10^-7

Or, [H⁺] = 1.04x10^-7 M

Therefore, pH = -log[H⁺]

Or, pH = -log(1.04x10^-7) = 6.98

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

y                                1.04x10^-7 (0.0258-x)        [At Equilibrium]

y                         1.04x10^-7   0.0258             (neglecting ‘x’ since it is very small)

Therefore, Ka₂ = [H⁺][ HPO₄^2-]/[ H₂PO4^-];       pKa₂ = 7.198 = -log(Ka₂)

                                                                        Or, Ka₂ = 10^-7.198 = 6.33x10^-8

Or,          6.33x10^-8 = (0.0258)x1.04x10^-7/y

Or,       y = 0.04238

Or,       [H₂PO4^-] = 0.04238 M