Question

What pH results when 10^-4 mole/L of Na₂HPO₄ is added to distilled water? Compare the results to that found if, rather than distilled water, a salt solution is used where the final solution ionic strength is 10^-2.

Answer 1

pH is dependent upon the concentration of H+ in solution

So when 10^-4 mol/L Na2HPO4 is added to distilled water

HPO4^2- + H2O <==> H2PO4- + OH-

Kb = 1 x 10^-14/6.31 x 10^-8 = x^2/1 x 10^-4

x = [OH-] = 3.98 x 10^-6 M

pOH = -log[OH-] = 5.40

pH = 14 - pOH = 8.60

When the same is added to a salt solution with ionic strength = 10^-2 0.051

activity ocefficient of H2PO4- = inv.log(-0.51 x sq.rt.(0.01)/(1 + 3.3 x 0.45 x sq.rt.(0.01)) = 0.903

activity ocefficient of HPO4^2- = inv.log(-0.51 x sq.rt.(0.01)/(1 + 3.3 x 0.40 x sq.rt.(0.01)) = 0.901

activity ocefficient of OH- = inv.log(-0.51 x sq.rt.(0.01)/(1 + 3.3 x 0.35 x sq.rt.(0.01)) = 0.900

Kb = 1 x 10^-14/6.31 x 10^-8 = x^2(0.903 x 0.900)/1 x 10^-4(0.901)

x = [OH-] = 4.19 x 10^-6 M

pOH = -log[OH-] = 5.38

pH = 14 - pOH = 8.62